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2. Chapter 2 Technical Mathematics Physics, 6th Edition 1
Chapter 2. Technical Mathematics Signed Numbers 2-1. +7 2-8. -17
2-15. +2 2-22. +12 2-2. +4 2-9. +6 2-16. -2 2-23. +8 2-3. +2 2-10.
-32 2-17. -4 2-24. -4 2-4. -2 2-11. -36 2-18. -3 2-25. 0 2-5. -10
2-12. +24 2-19. +2 2-26. +220 2-6. -33 2-13. -48 2-20. -4 2-27. +32
2-7. -5 2-14. +144 2-21. -3 2-28. -32 2-29. (a) -60 C; (b) -170 C;
(c) 360 C 2-30. L = 2 mm[(-300 C) (-50 C)] = 2 mm(-25) = -50 mm;
Decrease in length. Algebra Review 2-31. x = (2) + (-3) + (-2) =
-3; x = -3 2-32. x = (2) (-3) (-2) = +7; x = +7 2-33. x = (-3) +
(-2) - (+2) = -7; x = -7 2-34. x = -3[(2) (-2)] = -3(2 + 2) = -12;
x = -12 2-35. x b c a x 3 2 2 3 2 2 1 2 ( ) ; 2-36. x x 2 3 2 2 3 2
1 2 ( ) ; 2-37. x = (-3)2 (-2)2 = 9 4 = 5; x = 5 2-38. x b ac x ( )
( )( ) 3 2 2 3 4 3 4 ; 2-39. x x 2 3 2 2 2 1 3 4 4 3( )( ) ( ) ( );
2-40. x = (2)2 + (-3)2 + (-2)2 ; x = 17 2-41. x a b c 2 2 2 17
2-42. x = (2)(-3)[(-2) (+2)]2 ; x = -6(-4)2 = -96 2-43. Solve for
x: 2ax b = c; 2ax = b + c; x b c a x 2 3 2 2 2 5 4( ) ;
3. Chapter 2 Technical Mathematics Physics, 6th Edition 2 2-44.
ax bx c a b x c x c a b x 4 4 4 4 2 2 3 8; ( ) ; ( ) ( ) ; 2-45. 3
2 3 2 2 3 2 3 3 2 1ax ab c cx b x b c x ; ; ( ) ( ) ; 2-46. 4 2 16
4 2 16 4 16 2 4 2 2 16 3 2 32 ac b x b ac x b x ac b x ; ; ( )( ) (
) ; 2-47. 5m 16 = 3m 4 2-48. 3p = 7p - 16 5m 3m = -4 + 16 3p 7p =
-16 2m = 12; m = 6 -4p = - 16; p = +4 2-49. 4m = 2(m 4) 2-50. 3(m
6) = 6 4m = 2m 8 3m 18 = 6 2m = -8; m = -4 3m = 24; m = +8 2-51. x
x 3 4 3 12 36 ( )( ) ; 2-52. p p 3 2 6 1 3 1 ; 2-53. 96 48 96 48 2
x x ; 2-54. 14 = 2(b 7); 14 = 2b 14; b = 14 2-55. R2 = (4)2 + (3)2
= 16 + 9 2-56. 1 2 1 1 6 6 2 6 6 6 p p p p p ; R R2 25 5 3p = 6 +
p; p = 3 2-57. V = IR; R V I 2-58. PV nRT T PV nR ; 2-59. F ma a F
m ; 2-60. s = vt + d; d = s vt 2-61. F mv R FR mv R mv F 2 2 2 ; ;
2-62. s = at2 ; 2s = at2 ; a s t 2 2
4. Chapter 2 Technical Mathematics Physics, 6th Edition 3 2-63.
2 2 2 0 2 2 0 2 as v v a v v s f f ; 2-64. C Q V V Q C 2 2 2 2 ;
2-65. 1 1 1 1 2 2 1 1 2 R R R R R R R R R ; 2-66. mv Ft mv F t ; (
) ;R R R R R R R R R R 1 2 1 2 1 2 1 2 t mv F 2-67. mv mv Ft mv Ft
mv2 1 2 1 ; 2-68. PV T PV T PV T PV T1 1 1 2 2 2 1 1 2 2 2 1 ; v Ft
mv m 2 1 T PV T PV 2 2 2 1 1 1 2-69. v = vo + at; v v0 = at 2-70.
c2 = a2 + b2 ; b2 = c2 - a2 a v v t 0 b c a 2 2 Exponents and
Radicals 2-71. 212 2-72. 35 23 2-73. x10 2-74. x5 2-75. 1/a 2-76.
a/b2 2-77. 1/22 2-78. a2 /b2 2-79. 2x5 2-80. 1/a2 b2 2-81. m6 2-82.
c4 /n6 2-83. 64 x 106 2-84. (1/36) x 104 2-85. 4 2-86. 3 2-87. x3
2-88. a2 b3 2-89. 2 x 102 2-90. 2 x 10-9 2-91. 2a2 2-92. x + 2
Scientific Notation 2-93. 4.00 x 104 2-94. 6.70 x 101 2-95. 4.80 x
102 2-96. 4.97 x 105 2-97. 2.10 x 10-3 2-98. 7.89 x 10-1 2-99. 8.70
x 10-2 100. 9.67 x 10-4 2-101. 4,000,000
5. Chapter 2 Technical Mathematics Physics, 6th Edition 4
2-102. 4670 2-103. 37.0 2-104. 140,000 2-105. 0.0367 2-106. 0.400
2-107. 0.006 2-108. 0.0000417 2-109. 8.00 x 106 2-110. 7.40 x 104
2-111. 8.00 x 102 2-112. 1.80 x 10-8 2-113. 2.68 x 109 2-114. 7.40
x 10-3 2-115. 1.60x 10-5 2-116. 2.70 x 1019 2-117. 1.80 x 10-3
2-118. 2.40 x 101 2-119. 2.00 x 106 2-120. 2.00 x 10-3 2-121. 2.00
x 10-9 2-122. 5.71 x 10-1 2-123. 2.30 x 105 2-124. 6.40x 102 2-125.
2.40 x 103 2-126. 5.60 x 10-5 2-127. 6.90 x 10-2 2-128. 3.30 x 10-3
2-129. 6.00 x 10-4 2-130. 6.40 x 106 2-131. 8.00x 106 2-132. -4.00
x 10-2 Graphs 2-133. Graph of speed vs. time: When t = 4.5 s, v =
144 ft/s; When v = 100 m/s, t = 3.1 s. 2-134. Graph of advance of
screw vs. turns: When screw advances 2.75 in., N = 88 turns. 2-135.
Graph of wavelength vs. frequency: 350 kHz 857 m; 800 kHz 375 m.
2-136. Electric Power vs. Electric Current: 3.20 A 10.4 W; 8.0 A
64.8 W. Geometry 2-137. 900 . 1800 , 2700 , and 450 2-138. 2-139a.
A = 170 , B = 350 , C = 380 2-139b. A = 500 Rule 2; B = 400 Rule 2.
2-140a. A = 500 Rule 3; B = 1300 2-140b. B = 700 , C = 420 Rule 2 A
C B D
6. Chapter 2 Technical Mathematics Physics, 6th Edition 5 Right
Triangle Trigonometry 2-141. 0.921 2-147. 19.3 2-153. 684 2-159.
54.20 2-165. 36.90 2-142. 0.669 2-148. 143 2-154. 346 2-160. 6.730
2-166. 76.00 2-143. 1.66 2-149. 267 2-155. 803 2-161. 50.20 2-167.
31.20 2-144. 0.559 2-150. 32.4 2-156. 266 2-162. 27.10 2-145. 0.875
2-151. 235 2-157. 2191 2-163. 76.80 2-146. 0.268 2-152. 2425 2-158.
1620 2-164. 6.370 Solve triangles for unknown sides and angles
(Exercises 168 175): 2-168. tan = 18/35, = 35.80 ; R 18 252 2 R =
30.8 ft 2-169. tan = 600/400, = 56.30 ; R 40 802 2 R = 721 m.
2-170. y = 650 sin 210 = 233 m; x = 650 cos 210 = 607 m. 2-171. sin
= 200/500, = 23.60 ; 500 2002 2 2 x , x = 458 km. 2-172. sin =
210/400, = 31.70 ; 500 2002 2 2 m , m = 340 m. 2-173. x = 260 cos
510 = 164 in.; y = 260 sin 510 = 202 in. 2-174. tan = 40/80, =
26.60 ; R 40 802 2 R = 89.4 lb 2-175. = 1800 - 1200 = 600 ; y = 300
sin 600 = 260 m; x = 300 cos 600 = 150 m, left
7. Chapter 2 Technical Mathematics Physics, 6th Edition 6
Challenge Problems 2-176. 30.21 0.59 in. = 29.62 in. 2-178. T = Tf
T0 = -150 C (290 C); T = -44 C0 . 2-179. Tf T0 = -340 C; Tf - 200 C
= -340 C; Tf = -140 C 2-180. Six pieces @ 4.75 in. = 6(4.75 in.) =
28.5 in.; Five cuts @ 1/16 = 5/16 = 0.3125 in. Original length =
28.5 in. + 0.3125 in. = 28.8 in. 2-181. V = r2 h; Solve for h: h V
r 2 2-182. 2 2 ; mv mv F R R F 2-183. Solve for x and evaluate: a =
2, b = -2, c = 3, and d = -1 xb + cd = a(x + 2) xb + cd = ax + 2a
xb ax = 2a cd (b - a)x = 2a cd x a cd b a 2 ; x a cd b a x 2 2 2 3
1 2 2 7 4 7 4 ( ) ( )( ) ( ) ( ) ; 2-184. c b a b c a b2 2 2 2 2 2
2 50 20 539 ; . b = 53.9 2-185. F Gm m R F 1 2 2 6 67 500 ( . ; . x
10 )(4 x 10 )(3 x 10 ) (4 x 10 ) x 10 -11 -8 -7 -2 2 -22 2-186. L =
L0 + L0(t t0); L = 21.41 cm + (2 x 10-3 /C0 )( 21.41 cm )(1000 C -
200 C); L = 24.84 cm. 2-187. Construct graph of y = 2x and verify
that x = 3.5 when y = 7 (from the graph). 2-188. (a) A + 600 = 900
; A = 300 . A + C = 900 ; C = 600 . B = 600 by rule 2. (b) D + 300
= 900 ; D = 600 . A = 600 (alt. int. angles); B = 300 ; C = 1200
.
8. Chapter 2 Technical Mathematics Physics, 6th Edition 7
Critical Thinking Problems 2-189. A = (-8) (-4) = -4; B = (-6) +
(14) = 8; C = A B = (-4) (8) = -12; C = - 12 cm. B A = (8) (-4) =
+12. There is a difference of 24 cm between B A and A B. 2-190. T L
g T L g L gT 2 4 4 2 2 2 2 Let L = 4Lo; Since 4 2 , the period will
be doubled when the length is quadrupled. Let gm = ge /6, Then, T
would be changed by a factor of 1 1 6 6 2 45 / . Thus, the period T
on the moon would be 2(2.45) or 4.90 s. 2-191. (a) Area = LW =
(3.45 x 10-4 m)(9.77 x 10-5 m); Area = 3.37 x 10-8 m2 . Perimeter
(P) = 2L + 2W = 2(L + W); P = 2(3.45 x 10-4 + 9.77 x 10-5 ) = 8.85
x 10-4 m. (b) L = L0/2 and W = 2W0: A = (L0/2)(2W0) = L0W0; No
change in area. P P0 = [2(L0/2) + 2(2W0)] - [2L0 + 2W0] = 2W0 L0 P
= 2(9.77 x 10-5 ) 3.45 x 10-4 P = -1.50 x 10-4 m. The area doesnt
change, but the perimeter decreases by 0.150 mm. 2-192. Graph shows
when T = 420 K, P = 560 lb/in.2 ; when T = 600 K, P = 798 lb/in.2
2-193. Graph shows when V = 26 V, I = 377 mA; when V = 48 V, I =
696 mA.
9. Chapter 3 Technical Measurement and Vectors Physics, 6th
Edition 8 Chapter 3. Technical Measurement and Vectors Unit
Conversions 3-1. A soccer field is 100 m long and 60 m across. What
are the length and width of the field in feet? 100 cm 1 in. 1 ft
(100 m) 328 ft 1 m 2.54 cm 12 in. L = 328 ft 100 cm 1 in. 1 ft (60
m) 197 ft 1 m 2.54 cm 12 in. W = 197 ft 3-2. A wrench has a handle
8 in. long. What is the length of the handle in centimeters? 2.54
cm (8 in.) 20.3 cm 1 in. L = 20.3 cm 3-3. A 19-in. computer monitor
has a viewable area that measures 18 in. diagonally. Express this
distance in meters. . 2.54 cm 1 m (18 in.) 0.457 m 1 in. 100 cm L =
0.457 m 3-4. The length of a notebook is 234.5 mm and the width is
158.4 mm. Express the surface area in square meters. 1 m 1 m Area =
(234.5 mm)(158.4 mm) 0.037 1000 mm 1000 mm A = 0.0371 m2 3-5. A
cube has 5 in. on a side. What is the volume of the cube in SI
units and in fundamental USCS units? . 3 3 3 3 32.54 cm 1 m (5 in.)
125 in. 0.00205 m 1 in. 100 cm V V = 0.00205 m3 3 3 31 ft (125 in.
) 0.0723 ft 12 in. V V = 0.0723 ft3
10. Chapter 3 Technical Measurement and Vectors Physics, 6th
Edition 9 3-6. The speed limit on an interstate highway is posted
at 75 mi/h. (a) What is this speed in kilometers per hour? (b) In
feet per second? (a) mi 1.609 km 75 h 1 mi 121 km/h (b) mi 1 h 5280
ft 75 h 3600 s 1 mi = 110 ft/s 3-7. A Nissan engine has a piston
displacement (volume) of 1600 cm3 and a bore diameter of 84 mm.
Express these measurements in cubic inches and inches. Ans. 97.6
in.3 , 3.31 in. (a) 3 3 1 in. 1600 cm 2.54 cm = 97.6 in.3 (b) 1 in.
84 mm = 25.4 mm = 3.31 in. 3-8. An electrician must install an
underground cable from the highway to a home located 1.20 mi into
the woods. How many feet of cable will be needed? 5280 ft 1.2 mi
6340 ft 1 mi L = 6340 ft 3-9. One U.S. gallon is a volume
equivalent to 231 in.3 . How many gallons are needed to fill a tank
that is 18 in. long, 16 in. wide, and 12 in. high? Ans. 15.0 gal. V
= (18 in.)(16 in.)(12 in.) = 3456 in.3 3 3 1 gal (3456 in. ) 15.0
gal 231 in. V V = 15.0 gal 3-10. The density of brass is 8.89 g/cm3
. What is the density in kg/m3 ? 3 3 3 g 1 kg 100 cm kg 8.89 8890
cm 1000 g 1 m m = 8890 kg/m3 Addition of Vectors by Graphical
Methods 3-11. A woman walks 4 km east and then 8 km north. (a) Use
the polygon method to find her resultant displacement. (b) Verify
the result by the parallelogram method. Let 1 cm = 1 km; Then: R =
8.94 km, = 63.40 4 km 8 km R
11. Chapter 3 Technical Measurement and Vectors Physics, 6th
Edition 10 3-12. A land-rover, on the surface of Mars, moves a
distance of 38 m at an angle of 1800 . It then turns and moves a
distance of 66 m at an angle of 2700 . What is the displacement
from the starting position? Choose a scale, e.g., 1 cm = 10 m Draw
each vector to scale as shown. Measure R = 7.62 cm or R = 76.2 m
Measure angle = 60.10 S of W = 1800 + 60.10 = 240.10 R = 76.2 m,
240.10 3-13. A surveyor starts at the southeast corner of a lot and
charts the following displacements: A = 600 m, N; B = 400 m, W; C =
200 m, S; and D = 100 m, E. What is the net displacement from the
starting point? . Choose a scale, 1 cm = 100 m Draw each vector
tail to tip until all are drawn. Construct resultant from origin to
finish. R = 500 m, = 53.10 N of E or = 126.90 . 3-14. A downward
force of 200 N acts simultaneously with a 500-N force directed to
the left. Use the polygon method to find the resultant force. Chose
scale, measure: R = 539 N, = 21.80 S. of E. 3-15. The following
three forces act simultaneously on the same object. A = 300 N, 300
N of E; B = 600 N, 2700 ; and C = 100 N due east. Find the
resultant force using the polygon method. Choose a scale, draw and
measure: R = 576 N, = 51.40 S of E 67 m 38 m, 1800 B D R C A R 200
N 500 N B C R A
12. Chapter 3 Technical Measurement and Vectors Physics, 6th
Edition 11 3-16. A boat travels west a distance of 200 m, then
north for 400 m, and finally 100 m at 300 S of E. What is the net
displacement? (Set 1 cm = 100 N) Draw and measure: R = 368 N, =
108.00 3-17. Two ropes A and B are attached to a mooring hook so
that an angle o 60 exists between the two ropes. The tension in
rope A is 80 lb and the tension in rope B is 120 lb. Use the
parallelogram method to find the resultant force on the hook. Draw
and measure: R = 174 lb 3-18. Two forces A and B act on the same
object producing a resultant force of 50 lb at 36.90 N of W. The
force A = 40 lb due west. Find the magnitude and direction of force
B? Draw R = 50 lb, 36.90 N of W first, then draw 40 lb, W. F = 30
lb, 900 Trigonometry and Vectors 3-19. Find the x and y-components
of: (a) a displacement of 200 km, at 340 . (b) a velocity of 40
km/h, at 1200 ; and (c) A force of 50 N at 330o . (a) Dx = 200 cos
340 = 166 km Dy = 200 sin 340 = 112 km (b) vx = -40 cos 600 = -20.0
km/h vy = 40 sin 600 = +34.6 km/h (c) Fx = 50 cos 300 = 43.3 N; Fy
= - 50 sin 300 = -25.0 N R C B A B A R 36.90 40 lb FR 300 340 600
(a) (b) (c)
13. Chapter 3 Technical Measurement and Vectors Physics, 6th
Edition 12 3-20. A sled is pulled with a force of 540 N at an angle
of 400 with the horizontal. What are the horizontal and vertical
components of this force? Fx = 540 cos 400 = 414 N Fy = 540 sin 400
= 347 N 3-21. The hammer in Fig. 3-26 applies a force of 260 N at
an angle of 150 with the vertical. What is the upward component of
the force on the nail? F = 260 lb, = 750 ; Fxy = 260 sin Fy = 251
N. 3-22. A jogger runs 2.0 mi west and then 6.0 mi north. Find the
magnitude and direction of the resultant displacement. R ( ) ( )2
62 2 6.32 mi tan = 6 2 ; = 71.60 N of W * 3-23. A river flows south
with a velocity of 20 km/h. A boat has a maximum speed of 50 km/h
in still water. In the river, at maximum throttle, the boat heads
due west. What is the resultant speed and direction of the boat? R
(50) ( ) . ; 2 2 20 539 km / h; tan = 20 50 = 21.8 S of W0 * 3-24.
A rope, making an angle of 300 with the horizontal, drags a crate
along the floor. What must be the tension in the rope, if a
horizontal force of 40 lb is required to drag the crate? Fx = F cos
300 ; F Fx cos30 40 0 lb cos30 ;0 F = 46.2 N 540 N 400 (a) F R 50
km/h 20 km/h R 300 Fx F R = 53.9 km/h, 21.80 S of E
14. Chapter 3 Technical Measurement and Vectors Physics, 6th
Edition 13 * 3-25. An vertical lift of 80 N is needed to lift a
window. A long pole is used to lift the window. What force must be
exerted along the pole if it makes an angle of 340 with the wall?
Fy = F sin 300 ; F Fy sin sin ; 34 40 340 lb 0 F = 96.5 N * 3-26.
The resultant of two forces A and B is 400 N at 2100 . If force A
is 200 N at 2700 , what are the magnitude and direction of force B?
( = 210 - 1800 = 300 ) B = -400 N cos 300 = -346 N: B = 346 N, 1800
The Component Method of Vector Addition 3-27. Find the resultant of
the following perpendicular forces: (a) 400 N, 00 ; (b) 820 N, 2700
; and (b) 500 N, 900 . Draw each vector, then find R: Ax = +400 N;
Bx = 0; Cx = 0: Rx = +400 N Ay = 0; By = -820 N; Cy = +500 N; Ry =
0 820 N + 500 N = -320 N 2 2 400 320R ; 320 tan ; 400 R = 512 N,
38.70 S of E 3-28. Four ropes, all at right angles to each other,
pull on a ring. The forces are A = 40 lb, E; B = 80 lb, N; C = 70
lb, W; and D = 20 lb, S. Find the resultant force on the ring. Ax =
+40 lb; Bx = 0; Cx = -70 lb Dx = 0: Rx = +40 lb 70 lb = -30 lb Ay =
0; By = +80 lb; Cy = 0; Dy = -20 lb ; Ry = 0 + 80 lb - 20 lb = +60
lb 2 2 30 60R ; tan ; 60 30 R = 67.1 N, 116.60 F80 N 340 B 400 N
200 N 300 R C B A A = 40 lb, E
15. Chapter 3 Technical Measurement and Vectors Physics, 6th
Edition 14 *3-29. Two forces act on the car in Fig. 3-27. Force A
is 120 N, west and force B is 200 N at 600 N of W. What are the
magnitude and direction of the resultant force on the car? Ax =
-120; Bx = - (200 N) cos 600 = -100 N Rx = 120 N - 100 N; Rx = -220
N Ay = 0, By = 200 sin 600 = +173 N; Ry = 0 + 173 N = 173 N; Thus,
Rx = -100 N, Ry = +173 N and 2 2 (220) (173) 280 NR Resultant
direction: 0173 tan ; 38.2 N of W 210 ; R = 280 N, 141.80 *3-30.
Suppose the direction of force B in Problem 3-29 is reversed (+1800
) and other parameters are unchanged. What is the new resultant?
(This result is the vector difference A B). The vector B will now
be 600 S of E instead of N of E. Ax = -120 N; Bx = +(200 N) cos 600
= +100 N Rx = 120 N + 100 N; Rx = -20 N Ay = 0, By = -200 sin 600 =
-173 N; Ry = 0 - 173 N = -173 N; Thus, Rx = -20 N, Ry = -173 N and
2 2 ( 20) (173) 174 NR Resultant direction: tan ; . 173 20 8340 S
of W ; R = 174 N, 253.40 *3-31. Determine the resultant force on
the bolt in Fig. 3-28. ( Ax = 0 ) Bx = -40 cos 200 = -37.6 lb; Bx =
-50 cos 600 = -25.0 lb Rx = 0 37.6 lb 25.0 lb; Rx = -62.6 lb Ay =
+60 lb; By = 40 sin 200 = 13.7 lb; Cy = 50 sin 60 = -43.3 lb Ry =
60 lb 13.7 lb 43.3 lb; Ry = +30.4 lb 2 2 ( 62.6) (30.4)R R = 69.6
lb 30.4 tan ; 62.6 = 25.90 N of W 600 B A A 600 B 600 200 A = 60 lb
B = 40 lb C = 50 lb
16. Chapter 3 Technical Measurement and Vectors Physics, 6th
Edition 15 *3-32. Determine the resultant of the following forces
by the component method of vector addition: A = (200 N, 300 ); B =
(300 N, 3300 ; and C = (400 N, 2500 ). Ax = 200 cos 300 = 173 N; Bx
= 300 cos 300 = 260 N Cx = -400 cos 700 = -137 N; Rx = Fx = 296 N
Ay = 200 sin 300 = 100 N; By = 300 sin 300 = -150 N Cy = -400 sin
700 = -376 N; Ry = Fy = -430 N 2 2 (296) ( 430)R ; 426 tan ; 296 R
= 519 N, 55.20 S of E *3-33. Three boats exert forces on a mooring
hook as shown in Fig. 3-29. Find the resultant of these three
forces. Ax = 420 cos 600 = +210 N; Cx = -500 cos 400 = -383 N Bx =
0; Rx = 210 N + 0 383 N; Rx = -173 N Ay = 420 sin 600 = 364 N; By =
150; Cy = 500 sin 400 = 321 N Ry = Fy = 835 N; R ( ) (835)173 2 2 ;
tan ; 835 173 R = 853 N, 78.30 N of W Challenge Problems 3-34. Find
the horizontal and vertical components of the following vectors: A
= (400 N, 370 ); B = 90 m, 3200 ); and C = (70 km/h, 1500 ). Ax =
400 cos 370 = 319 N; Ay = 400 sin 370 = 241 N Bx = 90 cos 400 =
68.9 N; By = 90 sin 400 = 57.9 Cx = -70 cos 300 = -60.6 N; Cy = 70
sin 300 = 25.0 N 700 300 300 C B A 400 600 C B A500 N 420 N 150 N
370 90 N 300 400 C B A 70 N 400 N
17. Chapter 3 Technical Measurement and Vectors Physics, 6th
Edition 16 3-35. A cable is attached to the end of a beam. What
pull at an angle of 400 with the horizontal is needed to produce an
effective horizontal force of 200 N? P cos 400 = 200 N; P = 261 N
3-36. A fishing dock runs north and south. What must be the speed
of a boat heading at an angle of 400 E of N if its velocity
component along the dock is to be 30 km/h? v cos 400 = 30 km/h; v =
39.2 km/h 3-37. Find the resultant R = A + B for the following
pairs of forces: A = (520 N, south); B = 269 N, west; (b) A = 18
m/s, north; B = 15 m/s, west. (a) R ( ) (520)269 5852 2 N tan ; 520
295 R = 585 N, = 62.60 S of W (b) R ( ) ( )15 182 2 ; tan ; 18 15 R
= 23.4 N, 50 .20 N of W *3-38. Determine the vector difference (A
B) for the pairs of forces in Problem 3-37. (a) R ( ) (520)269 5852
2 N tan ; 520 295 R = 585 N, 62.60 S of E (b) R ( ) ( )15 182 2 ;
tan ; 18 15 R = 23.4 N, 50.20 N of E *3-39. A traffic light is
attached to the midpoint of a rope so that each segment makes an
angle of 100 with the horizontal. The tension in each rope segment
is 200 N. If the resultant force at the midpoint is zero, what must
be the weight of the traffic light? Rx = Fx = 0; T sin 100 + T sin
100 W = 0; 2(200) sin 100 = W: W = 69.5 N P 400 (a) (b) R B A 15
m/s 18 m/s R B A 269 N 520 N (a) (b) R A -B =15 m/s 18 m/s R A -B =
269 N 520 N Change B into the vector B, then ADD: W TT
18. Chapter 3 Technical Measurement and Vectors Physics, 6th
Edition 17 *3-40. Determine the resultant of the forces shown in
Fig. 3-30 Rx = 420 N 200 cos 700 410 cos 530 = 105 lb Ry = 0 + 200
sin 700 410 sin 700 = -139.5 lb R R Rx y 2 2 R = 175 lb; = 306.90
*3-41. Determine the resultant of the forces shown in Fig. 3-31. Rx
= 200 cos 300 300 cos 450 155 cos 550 = 128 N Ry = 0 + 200 sin 700
410 sin 700 = -185 N; R R Rx y 2 2 R = 225 N; = 124.60 *3-42. A
200-N block rests on a 300 inclined plane. If the weight of the
block acts vertically downward, what are the components of the
weight down the plane and perpendicular to the plane? Choose x-axis
along plane and y-axis perpendicular. Wx = 200 sin 300 ; Wx = 173
N, down the plane. Wy = 200 sin 600 ; Wx = 100 N, normal to the
plane. *3-43. Find the resultant of the following three
displacements: A = 220 m, 600 ; B = 125 m, 2100 ; and C = 175 m,
3400 . Ax = 220 cos 600 = 110 m; Ay = 220 sin 600 = 190.5 m Bx =
125 cos 2100 = -108 m; By = 125 sin 2100 = -62.5 m Cx = 175 cos
3400 = 164.4 m; Cy = 175 sin 3400 = -59.9 m Rx = 110 m 108 m +
164.4 m; Ry = 190.5 m 62.5 m 59.9 m ; Rx = 166.4 m; Ry = 68.1 m R (
. ) ( . )166 4 681 1802 2 m tan . . ; . 681 166 4 22 30 ; R = 180
m, = 22.30 300 200 600 A B C 550 300450 C = 155 N B = 300 N A = 200
N 530 700 C = 410 lb B = 200 lb A = 420 lb 300 W 300
19. Chapter 3 Technical Measurement and Vectors Physics, 6th
Edition 18 C = 60 N B = 40 N A = 80 N Critical Thinking Questions
3-44. Consider three vectors: A = 100 m, 00 ; B = 400 m, 2700 ; and
C = 200 m, 300 . Choose an appropriate scale and show graphically
that the order in which these vectors is added does not matter,
i.e., A + B + C = C + B + A. Is this also true for subtracting
vectors? Show graphically how A C differs from C A. 3-45. Two
forces A = 30 N and B = 90 N can act on an object in any direction
desired. What is the maximum resultant force? What is the minimum
resultant force? Can the resultant force be zero? Maximum resultant
force occurs when A and B are in same direction. A + B = 30 N + 90
N = 120 N Minimum resultant force occurs when A and B are in
opposite directions. B - A = 90 N 30 N = 60 N No combination gives
R = 0. *3-46. Consider two forces A = 40 N and B = 80 N. What must
be the angle between these two forces in order to produce a
resultant force of 60 N? Since R = C = 60 N is smaller than 80 N,
the angle between A and B must be > 900 . Applying the law of
cosines to the triangle, we can find and then . A + B + C C + B + A
A C C - A
20. Chapter 3 Technical Measurement and Vectors Physics, 6th
Edition 19 C2 = A2 + B2 2AB Cos ; (60)2 = (80)2 + (40)2 2(80)(40)
Cos ; = 46.60 . The angle between the direction of A and the
direction of B is = 1800 - ; = 133.40 . *3-47. What third force F
must be added to the following two forces so that the resultant
force is zero: A = 120 N, 1100 and B = 60 N, 2000 ? Components of
A: Ax = 120 Cos 1100 = -40.0 N; Ay = 120 Sin 1100 = 113 N
Components of B: Bx = 60 Cos 2000 = -56.4 N; By = 60 Sin 2000 =
-20.5 N Rx = 0; Rx = Ax + Bx + Fx = 0; Rx = -40.0 N 56.4 N + Fx =
0; Or Fx = +97.4 N Ry = 0; Ry = Ay + By + Fy = 0; Ry = 113 N 20.5 N
+ Fy = 0; Or Fy = -92.2 N F ( . ) ( . )97 4 92 2 1312 2 N tan . . ;
. 92 2 97 4 4330 And = 3600 43.40 Thus, the force F has a magnitude
and direction of: F = 134 N, = 316.60 *3-48. An airplane needs a
resultant heading of due west. The speed of the plane is 600 km/h
in still air. If the wind has a speed of 40 km/h and blows in a
direction of 300 S of W, what direction should the aircraft be
pointed and what will be its speed relative to the ground? From the
diagram, Rx = R, Ry = 0, So that Ay + By = 0. Ay = 600 sin y = -40
sin 300 = -20 km/h 600 sin - 20 = 0; 600 sin = 20 sin ; . 20 600
1910 N of W (direction aircraft should be pointed) Noting that R =
Rx and that Ax +Bx = Rx, we need only find sum of x-components. Ax
= -600 cos 599.7 km/hx = 40 Cos 300 = -34.6 km/h R = -599.7 km/h
34.6 km/h; R = -634 km/h. Thus, the speed of the plane relative to
the ground is 634 km/h, 00 ; and the plane must be pointed in a
direction of 1.910 N of W. R B = 40 km/h A = 600 km/h
21. Chapter 3 Technical Measurement and Vectors Physics, 6th
Edition 20 A = 200 lb 200 E F *3-49. What are the magnitude F and
direction of the force needed to pull the car of Fig. 3-32 directly
east with a resultant force of 400 lb? Rx = 400 lb and Ry = 0; Rx =
Ax + Fx = 400 lb 200 Cos 200 + Fx = 400 lb Fx = 400 lb 200 Cos 200
= 212 lb Ry = 0 = Ay + Fy; Ay = -200 sin 200 = -68.4 lb Fy = -Ay =
+68.4 lb; So, Fx = 212 lb and Fy = +68.4 lb F ( ) ( . ) ;212 68 42
2 ; tan = 68.4 212 R = 223 lb, 17.90 N of E
22. Chapter 4. Translational Equilibrium and Friction Physics,
6th Ed. 21 BA W B 400 Bx By W B A 300 600 W B A 600 W B A 300 600
Chapter 4. Translational Equilibrium and Friction. Note: For all of
the problems at the end of this chapter, the rigid booms or struts
are considered to be of negligible weight. All forces are
considered to be concurrent forces. Free-body Diagrams 4-1. Draw a
free-body diagram for the arrangements shown in Fig. 3-18. Isolate
a point where the important forces are acting, and represent each
force as a vector. Determine the reference angle and label
components. (a) Free-body Diagram (b) Free-body with rotation of
axes to simplify work. 4-2. Study each force acting at the end of
the light strut in Fig. 3-19. Draw the appropriate free- body
diagram. There is no particular advantage to rotating axes.
Components should also be labeled on diagram. Solution of
Equilibrium Problems: 4-3. Three identical bricks are strung
together with cords and hung from a scale that reads a total of 24
N. What is the tension in the cord that supports the lowest brick?
What is the tension in the cord between the middle brick and the
top brick? Each brick must weight 8 N. The lowest cord supports
only one brick, whereas the middle cord supports two bricks. Ans. 8
N, 16 N.
23. Chapter 4. Translational Equilibrium and Friction Physics,
6th Ed. 22 W B A 600 4-4. A single chain supports a pulley whose
weight is 40 N. Two identical 80-N weights are then connected with
a cord that passes over the pulley. What is the tension in the
supporting chain? What is the tension in each cord? Each cord
supports 80 N, but chain supports everything. T = 2(80 N) + 40 N =
200 N. T = 200 N *4-5. If the weight of the block in Fig. 4-18a is
80 N, what are the tensions in ropes A and B? By - W = 0; B sin 400
80 N = 0; B = 124.4 N Bx A = 0; B cos 400 = A; A = (124.4 N) cos
400 A = 95.3 N; B = 124 N. *4-6. If rope B in Fig. 4-18a will break
for tensions greater than 200 lb, what is the maximum weight W that
can be supported? Fy = 0; By W = 0; W = B sin 400 ; B = 200 N W =
(200 N) sin 400 ; W = 129 lb *4-7. If W = 600 N in Fig. 18b, what
is the force exerted by the rope on the end of the boom A in Fig.
18b? What is the tension in rope B? Fx = 0; A Wx = 0; A = Wx = W
cos 600 A = (600 N) cos 600 = 300 N Fy = 0; B Wy = 0; B = Wy = W
sin 600 B = (600 N) sin 600 = 520 N A = 300 N; B = 520 N Wy Wx 80
N80 N 40 N By Bx B 400A W Bx B 400A W
24. Chapter 4. Translational Equilibrium and Friction Physics,
6th Ed. 23 W B = 800 N A 600 300 W F N *4-8. If the rope B in Fig.
18a will break if its tension exceeds 400 N, what is the maximum
weight W? Fy = By - W = 0; By = W B sin 400 = 400 N ; B = 622 N Fx
= 0 Bx A = 0; B cos 400 = A; A = (622 N) cos 400 A = 477 N. *4-9.
What is the maximum weight W for Fig. 18b if the rope can sustain a
maximum tension of only 800 N? (Set B = 800 N). Draw diagram, then
rotate x-y axes as shown to right. Fy = 0; 800 N W Sin 600 = 0; W =
924 N. The compression in the boom is A = 924 Cos 600 A = 462 N.
*4-10. A 70-N block rests on a 300 inclined plane. Determine the
normal force and find the friction force that keeps the block from
sliding. (Rotate axes as shown.) Fx = N Wx = 0; N = Wx = (70 N) cos
300 ; N = 60.6 N Fx = F Wy = 0; F = Wy = (70 N) sin 300 ; F = 35.0
N *4-11. A wire is stretched between two poles 10 m apart. A sign
is attached to the midpoint of the line causing it to sag
vertically a distance of 50 cm. If the tension in each line segment
is 2000 N, what is the weight of the sign? (h = 0.50 m) tan =
(0.5/5) or = 5.710 ; 2(2000 N) sin = W W = 4000 sin 5.71; W = 398
N. *4-12. An 80-N traffic light is supported at the midpoint of a
30-m length of cable between to poles. Find the tension in each
cable segment if the cable sags a vertical distance of 1 m. h = 1
m; Tan = (1/15); = 3.810 T sin + T sin = 80 N; 2T sin = 80 N 15 m 5
m W = ? h 2000 N 2000 N 5 m 15 m W = 80 N h T T Bx B 400A W By
25. Chapter 4. Translational Equilibrium and Friction Physics,
6th Ed. 24 Solution to 4-12 (Cont.): T 80 381 6010 N 2 N sin . ; T
= 601 N *4-13. The ends of three 8-ft studs are nailed together
forming a tripod with an apex that is 6ft above the ground. What is
the compression in each of these studs if a 100-lb weight is hung
from the apex? Three upward components Fy hold up the 100 lb
weight: 3 Fy = 100 lb; Fy = 33.3 lb sin = (6/8); = 48.90 F sin
48.90 = 33.3 lb; F 333 44 4 . . lb sin 48.9 lb0 F = 44.4 lb,
compression *4-14. A 20-N picture is hung from a nail as in Fig.
4-20, so that the supporting cords make an angle of 600 . What is
the tension of each cord segment? According to Newtons third law,
the force of frame on nail (20 N) is the same as the force of the
nail on the rope (20 N , up). Fy = 0; 20 N = Ty + Ty; 2Ty = 20 N;
Ty = 10 N Ty = T sin 600 ; So T sin 600 = 10 N, and T = 11.5 N.
Friction 4-15. A horizontal force of 40 N will just start an empty
600-N sled moving across packed snow. After motion is begun, only
10 N is needed to keep motion at constant speed. Find the
coefficients of static and kinetic friction. s k 40 10N 600 N
0.0667 N 600 N 0.0167 s = 0.0667; k = 0.016 4-16. Suppose 200-N of
supplies are added the sled in Problem 4-13. What new force is
needed to drag the sled at constant speed? N= 200 N + 600 N = 800
N; Fk = kN = (0.0167)(800 N); Fk = 13.3 N F Fy h 600 600 T T 20
N
26. Chapter 4. Translational Equilibrium and Friction Physics,
6th Ed. 25 4-17. Assume surfaces where s = 0.7 and k = 0.4. What
horizontal force is needed to just start a 50-N block moving along
a wooden floor. What force will move it at constant speed? Fs = sN
= (0.7)(50 N) = 35 N ; Fk = sN = (0.4)(50 N) = 20 N 4-18. A
dockworker finds that a horizontal force of 60 lb is needed to drag
a 150-lb crate across the deck at constant speed. What is the
coefficient of kinetic friction? k F N ; k 60 lb 150 lb 0.400 k =
0.400 4-19. The dockworker in Problem 4-16 finds that a smaller
crate of similar material can be dragged at constant speed with a
horizontal force of only 40 lb. What is the weight of this crate?
Fk = sN = (0.4)W = 40 lb; W = (40 lb/0.4) = 100 lb; W = 100 lb.
4-20. A steel block weighing 240 N rests on level steel beam. What
horizontal force will move the block at constant speed if the
coefficient of kinetic friction is 0.12? Fk = sN = (0.12)(240 N) ;
Fk = 28.8 N. 4-21. A 60-N toolbox is dragged horizontally at
constant speed by a rope making an angle of 350 with the floor. The
tension in the rope is 40 N. Determine the magnitude of the
friction force and the normal force. Fx = T cos 350 Fk = 0; Fk =
(40 N) cos 350 = 32.8 N Fy = N + Ty W = 0; N = W Ty = 60 N T sin
350 N = 60 N (40 N) sin 350 ; N = 37.1 N Fk = 32.8 N 4-22. What is
the coefficient of kinetic friction for the example in Problem
4-19? k F N 32 8. ; N 37.1 N k = 0.884 F N T 350 W
27. Chapter 4. Translational Equilibrium and Friction Physics,
6th Ed. 26 4-23. The coefficient of static friction for wood on
wood is 0.7. What is the maximum angle for an inclined wooden plane
if a wooden block is to remain at rest on the plane? Maximum angle
occurs when tan = s; s = tan = 0.7; = 35.00 4-24. A roof is sloped
at an angle of 400 . What is the maximum coefficient of static
friction between the sole of the shoe and the roof to prevent
slipping? Tan = k; k = Tan 400 =0.839; k = 0.839 *4-25. A 200 N
sled is pushed along a horizontal surface at constant speed with a
50-N force that makes an angle of 280 below the horizontal. What is
the coefficient of kinetic friction? Fx = T cos 280 Fk = 0; Fk =
(50 N) cos 280 = 44.1 N Fy = N - Ty W = 0; N = W + Ty = 200 N + T
sin 280 N = 200 N + (50 N) sin 350 ; N = 223 N k F N 441. N 223 N k
= 0.198 *4-26. What is the normal force on the block in Fig. 4-21?
What is the component of the weight acting down the plane? Fy = N -
W cos 430 = 0; N = (60N) cod 430 = 43.9 N Wx = (60 N) sin 350 ; Wx
= 40.9 N *4-27. What push P directed up the plane will cause the
block in Fig. 4-21 to move up the plane with constant speed? [From
Problem 4-23: N = 43.9 N and Wx = 40.9 N] Fk = kN = (0.3)(43.9 N);
Fk = 13.2 N down plane. Fx = P - Fk Wx = 0; P = Fk + Wx; P = 13.2 N
+ 40.9 N; P = 54.1 N P Fk N 280 W W F N P 430
28. Chapter 4. Translational Equilibrium and Friction Physics,
6th Ed. 27 *4-28. If the block in Fig. 4-21 is released, it will
overcome static friction and slide rapidly down the plane. What
push P directed up the incline will retard the downward motion
until the block moves at constant speed? (Note that F is up the
plane now.) Magnitudes of F , Wx, and N are same as Prob. 4-25. Fx
= P +Fk Wx = 0; P = Wx - Fk; P = 40.9 N - 13.2 N P = 27.7 N
directed UP the inclined plane Challlenge Problems *4-29. Determine
the tension in rope A and the compression B in the strut for Fig.
4-22. Fy = 0; By 400 N = 0; B 400 462 N N sin600 Fx = 0; Bx A = 0;
A = B cos 600 A = (462 N) cos 600 ; A = 231 N and B = 462 N *4-30.
If the breaking strength of cable A in Fig. 4-23 is 200 N, what is
the maximum weight that can be supported by this apparatus? Fy = 0;
Ay W = 0; W = (200 N) sin 400 = 129 N The maximum weight that can
be supported is 129 N. *4-31. What is the minimum push P parallel
to a 370 inclined plane if a 90-N wagon is to be rolled up the
plane at constant speed. Ignore friction. Fx = 0; P - Wx = 0; P =
(90 N) sin 370 P = 54.2 N W F N P 430 B 400 N A 600 By A W 200 N B
400 Ay N P 370 W = 90 N
29. Chapter 4. Translational Equilibrium and Friction Physics,
6th Ed. 28 340 N A B W 4-32. A horizontal force of only 8 lb moves
a cake of ice slides with constant speed across a floor (k = 0.1).
What is the weight of the ice? Fk = kN = (0.3) W; Fk = 8 lb; (0.1)W
= 8 lb; W = 80 lb. *4-33. Find the tension in ropes A and B for the
arrangement shown in Fig. 4-24a. Fx = B Wx = 0; B = Wx = (340 N)
cos 300 ; B = 294 N Fy = A Wx = 0; A = Wy = (340 N) sin 300 ; A =
170 N A = 170 N; B = 294 N *4-34. Find the tension in ropes A and B
in Fig. 4-24b. Fy = By 160 N = 0; By = 160 N ; B sin 500 = 294 N B
160 500 N sin ; B = 209 N Fx = A Bx = 0; A = Bx = (209 N) cos 500 ;
A = 134 N *4-35. A cable is stretched horizontally across the top
of two vertical poles 20 m apart. A 250-N sign suspended from the
midpoint causes the rope to sag a vertical distance of 1.2 m. What
is the tension in each cable segment?. h = 1.2 m; tan . ; . 12 10
6840 2Tsin 6.840 = 250 N; T = 1050 N *4-36. Assume the cable in
Problem 4-31 has a breaking strength of 1200 N. What is the maximum
weight that can be supported at the midpoint? 2Tsin 6.840 = 250 N;
2(1200 N) sin 6.840 = W W = 289 N Wy Wy Wx 300 W = 160 N B A500 W =
250 N h T T 10 m 10 m
30. Chapter 4. Translational Equilibrium and Friction Physics,
6th Ed. 29 *4-37. Find the tension in the cable and the compression
in the light boom for Fig. 4-25a. Fy = Ay 26 lb = 0; Ay = 26 lb ; A
sin 370 = 26 lb A 26 lb sin ; 370 A = 43.2 lb Fx = B Ax = 0; B = Ax
= (43.2 lb) cos 370 ; B = 34.5 lb *4-38. Find the tension in the
cable and the compression in the light boom for Fig. 4-25b. First
recognize that = 900 - 420 = 480 , Then W = 68 lb Fy = By 68 lb =
0; By = 68 lb ; B sin 480 = 68 lb B 68 lb sin ; 480 A = 915 lb Fx =
Bx A = 0; A = Bx = (91.5 lb) cos 480 ; B = 61.2 lb *4-39. Determine
the tension in the ropes A and B for Fig. 4-26a. Fx = Bx Ax = 0; B
cos 300 = A cos 450 ; B = 0.816 A Fy = A sin 450 B sin 300 420 N =
0; 0.707 A 0.5 B = 420 N Substituting B = 0.816A: 0.707 A
(0.5)(0.816 A) = 420 N Solving for A, we obtain: A = 1406 N; and B
= 0.816A = 0.816(1406) or B = 1148 N Thus the tensions are : A =
1410 N; B = 1150 N *4-40. Find the forces in the light boards of
Fig. 4-26b and state whether the boards are under tension or
compression. ( Note: A = 900 - 300 = 600 ) Fx = Ax Bx = 0; A cos
600 = B cos 450 ; A = 1.414 B Fy = B sin 450 + A sin 600 46 lb = 0;
0.707 B + 0.866 A = 46 lb Substituting A = 1.414B: 0.707 B +
(0.866)(1.414 B) = 46 lb Solving for B: B = 23.8 lb; and A = 1.414B
= 01.414 (23.8 lb) or A = 33.7 lb A = 33.7 lb, tension; B = 23.8
lb, compression W = 26 lb A B370 B W 68 lb A 480 By 420 N A B 300
450 W 46 lb A B 600450 W
31. Chapter 4. Translational Equilibrium and Friction Physics,
6th Ed. 30 Critical Thinking Questions 4-41. Study the structure
drawn in Fig. 4-27 and analyze the forces acting at the point where
the rope is attached to the light poles. What is the direction of
the forces acting ON the ends of the poles? What is the direction
of the forces exerted BY the poles at that point? Draw the
appropriate free-body diagram. Imagine that the poles are bolted
together at their upper ends, then visualize the forces ON that
bolt and BY that bolt. *4-42. Determine the forces acting ON the
ends of the poles in Fig 3-27 if W = 500 N. Fx = Bx Ax = 0; B cos
300 = A cos 600 ; B = 0.577 A Fy = A sin 600 B sin 300 500 N = 0;
0.866 A 0.5 B = 500 N Substituting B = 0.577 A: 0.866 A (0.5)(
0.577 A) = 500 N Solving for A, we obtain: A = 866 N; and B = 0.577
A = 0.577(866) or B = 500 N Thus the forces are : A = 866 N; B =
500 N Can you explain why B = W? Would this be true for any weight
W? Try another value, for example W = 800 N and solve again for B.
W A B 300 600 Forces ON Bolt at Ends (Action Forces): The force W
is exerted ON the bolt BY the weight. The force B is exerted ON
bolt BY right pole. The force A is exerted ON bolt BY the middle
pole. To understand these directions, imagine that the poles snap,
then what would be the resulting motion. Wr Ar Br 300 600 Forces BY
Bolt at Ends (Reaction Forces): The force Wr is exerted BY the bolt
ON the weight. The force Br is exerted ON bolt BY right pole. The
force Ar is exerted BY bolt ON the middle pole. Do not confuse
action forces with the reaction forces. W A B 300 600
32. Chapter 4. Translational Equilibrium and Friction Physics,
6th Ed. 31 *4-43. A 2-N eraser is pressed against a vertical
chalkboard with a horizontal push of 12 N. If s = 0.25, find the
horizontal force required to start motion parallel to the floor?
What if you want to start its motion upward or downward? Find the
vertical forces required to just start motion up the board and then
down the board? Ans. 3.00 N, up = 5 N, down = 1 N. For horizontal
motion, P = Fs = sN P = 0.25 (12 N); P = 3.00 N For vertical
motion, P 2 N Fk = 0 P = 2 N + 3 N; P = 5.00 N For down motion: P +
2 N Fs = 0; P = - 2 N + 3 N; P = 1.00 N *4-44. It is determined
experimentally that a 20-lb horizontal force will move a 60-lb lawn
mower at constant speed. The handle of the mower makes an angle of
400 with the ground. What push along the handle will move the mower
at constant speed? Is the normal force equal to the weight of the
mower? What is the normal force? k 20 0 333 lb 60 lb . Fy = N Py -
W= 0; W = 60 lb N = P sin 400 + 60 lb; Fk = kN = 0.333 N Fy = Px -
Fk = 0; P cos 400 0.333N = 0 P cos 400 0.333 (P sin 400 + 60 lb) =
0; 0.766 P = 0.214 P + 20 lb; 0.552 P = 20 lb; P 20 0552 36 2 lb lb
. . ; P = 36.2 lb The normal force is: N = (36.2 lb) sin 400 + 60
lb N = 83.3 lb 12 N P 2 N F N 2 N F F P P Fk N 400 W 12 N F 2 N P
N
33. Chapter 4. Translational Equilibrium and Friction Physics,
6th Ed. 32 W = 70N N F 400 *4-45. Suppose the lawn mower of Problem
4-40 is to be moved backward. What pull along the handle is
required to move with constant speed? What is the normal force in
this case? Discuss the differences between this example and the one
in the previous problem. k 20 0 333 lb 60 lb . Fy = N + Py - W= 0;
W = 60 lb N = 60 lb - P sin 400 ; Fk = kN = 0.333 N Fy = Px - Fk =
0; P cos 400 0.333N = 0 P cos 400 0.333 (60 lb - P sin 400 ) = 0;
0.766 P - 20 lb + 0.214 P = 0; 0.980 P = 20 lb; P 20 0 980 20 4 lb
lb . . ; P = 20.4 lb The normal force is: N = 60 lb (20.4 lb) sin
400 N = 46.9 lb *4-46. A truck is removed from the mud by attaching
a line between the truck and the tree. When the angles are as shown
in Fig. 4-28, a force of 40 lb is exerted at the midpoint of the
line. What force is exerted on the truck? = 200 T sin 200 + T sin
200 = 40 lb 2 T sin 200 = 40 lb T = 58.5 lb *4-47. Suppose a force
of 900 N is required to remove the move the truck in Fig. 4-28.
What force is required at the midpoint of the line for the angles
shown?. 2 T sin 200 = F; 2(900 N) sin 200 = F; F = 616 N *4-48. A
70-N block of steel is at rest on a 400 incline. What is the static
friction force directed up the plane? Is this necessarily the
maximum force of static friction? What is the normal force? F = (70
N) sin 400 = 45.0 N N = (70 N) cos 400 = 53.6 N P Fk N 400 W F h T
T
34. Chapter 4. Translational Equilibrium and Friction Physics,
6th Ed. 33 *4-49. Determine the compression in the center strut B
and the tension in the rope A for the situation described by Fig.
4-29. Distinguish clearly the difference between the compression
force in the strut and the force indicated on your free-body
diagram. Fx = Bx Ax = 0; B cos 500 = A cos 200 ; B = 1.46 A Fy = B
sin 500 A sin 200 500 N = 0; 0.766 B 0.342 A = 500 N Substituting B
= 1.46 A: 0.766 (1.46 A) (0.342 A) = 500 N Solving for A, we
obtain: A = 644 N; and B = 1.46 A = 1.46 (644) or B = 940 N Thus
the tensions are : A = 644 N; B = 940 N *4-50. What horizontal push
P is required to just prevent a 200 N block from slipping down a
600 inclined plane where s = 0.4? Why does it take a lesser force
if P acts parallel to the plane? Is the friction force greater,
less, or the same for these two cases? (a) Fy = N Wy Py = 0; Wy =
(200 N) cos 600 = 100 N Py = P sin 600 = 0.866 P; N = 100 N + 0.866
P F = N = 0.4(100 N + 0.866 P); F = 40 N + 0.346 P Fx = Px Wx + F =
0; P cos 600 - (200 N) sin 600 + (40 N + 0.346 P) = 0 0.5 P 173.2 N
+ 40 N + 0.346 P = 0 Solving for P gives: P = 157 N (b) If P were
parallel to the plane, the normal force would be LESS, and
therefore the friction force would be reduced. Since the friction
force is directed UP the plane, it is actually helping to prevent
slipping. You might think at first that the push P (to stop
downward slipping) would then need to be GREATER than before, due
to the lesser friction force. However, only half of the push is
effective when exerted horizontally. If the force P were directed
up the incline, a force of only 133 N is required. You should
verify this value by reworking the problem. WA B 200 500 x W600 600
F N P600
35. Chapter 4. Translational Equilibrium and Friction Physics,
6th Ed. 34 *4-51. Find the tension in each cord of Fig. 4-30 if the
suspended weight is 476 N. Consider the knot at the bottom first
since more information is given at that point. Cy + Cy = 476 N; 2C
sin 600 = 476 N C 476 275 N 2sin60 N0 Fy = A sin 300 - (275 N) sin
600 = 0 A = 476 N; Fx = A cos 300 C cos 600 B = 0; 476 cos 300 275
cos 600 B = 0 B = 412 N 137 N = 275 N; Thus: A = 476 N, B = 275 N,
C = 275 N *4-52. Find the force required to pull a 40-N sled
horizontally at constant speed by exerting a pull along a pole that
makes a 300 angle with the ground (k = 0.4). Now find the force
required if you push along the pole at the same angle. What is the
major factor that changes in these cases? (a) Fy = N + Py - W= 0; W
= 40 N N = 40 N - P sin 300 ; Fk = kN Fx = P cos 300 - kN = 0; P
cos 400 - 0.4(40 N - P sin 300 ) =0; 0.866 P 16 N + 0.200 P = 0; P
= 15.0 N (b) Fy = N - Py - W= 0; N = 40 N + P sin 300 ; Fk = kN Fx
= P cos 300 - kN = 0; P cos 400 - 0.4(40 N + P sin 300 ) =0; 0.866
P 16 N - 0.200 P = 0; P = 24.0 N Normal force is greater! 476 N A C
C 600 600 B C 275 N 600 300 300 P Fk N 300 W P Fk N W
36. Chapter 4. Translational Equilibrium and Friction Physics,
6th Ed. 35 **4-53. Two weights are hung over two frictionless
pulleys as shown in Fig. 4-31. What weight W will cause the 300-lb
block to just start moving to the right? Assume s = 0.3. Note: The
pulleys merely change the direction of the applied forces. Fy = N +
(40 lb) sin 450 + W sin 300 300 lb = 0 N = 300 lb 28.3 lb 0.5 W; F
= sN Fx = W cos 300 - sN (40 lb) cos 450 = 0 0.866 W 0.3(272 lb 0.5
W) 28.3 lb = 0; W = 108 lb **4-54. Find the maximum weight than can
be hung at point O in Fig. 4-32 without upsetting the equilibrium.
Assume that s = 0.3 between the block and table. We first find F
max for the block F = sN = 0.3 (200 lb) = 60 lb Now set A = F = 60
lb and solve for W: Fx = B cos 200 A = 0; B cos 200 = 60 lb; B =
63.9 lb Fy = B sin 200 W = 0; W = B sin 200 = (63.9 lb) sin 200 ; W
= 21.8 lb F 40 lb N W 450 300 300 lb W 200 F B A A
37. Chapter 5 Torque and Rotational Equilibrium Physics, 6th
Edition 36 Chapter 5. Torque and Rotational Equilibrium Unit
Conversions 5-1. Draw and label the moment arm of the force F about
an axis at point A in Fig. 5-11a. What is the magnitude of the
moment arm? Moment arms are drawn perpendicular to action line: rA
= (2 ft) sin 250 rA = 0.845 ft 5-2. Find the moment arm about axis
B in Fig. 11a. (See figure above.) rB = (3 ft) sin 250 rB = 1.27 ft
5-3. Determine the moment arm if the axis of rotation is at point A
in Fig. 5-11b. What is the magnitude of the moment arm? rB = (2 m)
sin 600 rB = 1.73 m 5-4. Find the moment arm about axis B in Fig.
5-11b. rB = (5 m) sin 300 rB = 2.50 m Torque 5-5. If the force F in
Fig. 5-11a is equal to 80 lb, what is the resultant torque about
axis A neglecting the weight of the rod. What is the resultant
torque about axis B? Counterclockwise torques are positive, so that
A is - and B is +. (a) A = (80 lb)(0.845 ft) = -67.6 lb ft (b) B =
(80 lb)(1.27 ft) = +101 lb ft 5-6. The force F in Fig. 5-11b is 400
N and the angle iron is of negligible weight. What is the resultant
torque about axis A and about axis B? Counterclockwise torques are
positive, so that A is + and B is -. (a) A = (400 N)(1.732 m) =
+693 N m; (b) B = (400 N)(2.50 m) = -1000 N m 3 ft 2 ft rB B A 250
F rA 250 2 m 5 m rB rA 600 B 300 A F
38. Chapter 5 Torque and Rotational Equilibrium Physics, 6th
Edition 37 5-7. A leather belt is wrapped around a pulley 20 cm in
diameter. A force of 60 N is applied to the belt. What is the
torque at the center of the shaft? r = D = 10 cm; = (60 N)(0.10 m)
= +6.00 N m 5-8. The light rod in Fig. 5-12 is 60 cm long and
pivoted about point A. Find the magnitude and sign of the torque
due to the 200 N force if is (a) 900 , (b) 600 , (c) 300 , and (d)
00 . = (200 N) (0.60 m) sin for all angles: (a) = 120 N m (b) = 104
N m (b) = 60 N m (d) = 0 5-9. A person who weighs 650 N rides a
bicycle. The pedals move in a circle of radius 40 cm. If the entire
weight acts on each downward moving pedal, what is the maximum
torque? = (250 N)(0.40 m) = 260 N m 5-10. A single belt is wrapped
around two pulleys. The drive pulley has a diameter of 10 cm, and
the output pulley has a diameter of 20 cm. If the top belt tension
is essentially 50 N at the edge of each pulley, what are the input
and output torques? Input torque = (50 N)(0.10 m) = 5 N m Output
torque = (50 N)(0.20 m) = 10 N m Resultant Torque 5-11. What is the
resultant torque about point A in Fig. 5-13. Neglect weight of bar.
= +(30 N)(6 m) - (15 N)(2 m) - (20 N)(3 m) = 90.0 N m,
Counterclockwise. F A 200 N r 60 cm 30 N 2 m 15 N 20 N A 4 m 3
m
39. Chapter 5 Torque and Rotational Equilibrium Physics, 6th
Edition 38 5-12. Find the resultant torque in Fig. 5-13, if the
axis is moved to the left end of the bar. = +(30 N)(0) + (15 N)(4
m) - (20 N)(9 m) = -120 N m, counterclockwise. 5-13. What
horizontal force must be exerted at point A in Fig 5-11b to make
the resultant torque about point B equal to zero when the force F =
80 N? = P (2 m) (80 N)(5 m) (sin 300 ) = 0 2 P = 200 N; P = 100 N
5-14. Two wheels of diameters 60 cm and 20 cm are fastened together
and turn on the same axis as in Fig. 5-14. What is the resultant
torque about a central axis for the shown weights? r1 = (60 cm) =
0.30 m ; r2 = (30 cm) = 0.15 m = (200 N)(0.30 m) (150 N)(0.15 m) =
37.5 N m; = 37.5 N m, ccw 5-15. Suppose you remove the 150-N weight
from the small wheel in Fig. 5-14. What new weight can you hang to
produce zero resultant torque? = (200 N)(0.30 m) W (0.15 m) = 0; W
= 400 N 5-16. Determine the resultant torque about the corner A for
Fig. 5-15. = +(160 N)(0.60 m) sin 400 - (80 N)(0.20 m) = 61.7 N m
16.0 N m = 45.7 N m R = 45.7 N m 5-17. Find the resultant torque
about point C in Fig. 5-15. = - (80 N)(0.20 m) = -16 N m 30 N 2 m
15 N 20 N A 4 m 3 m 2 m 5 m rB B 300 P F = 80 N C B A 80 N 400 20
cm 60 cm r 400 160 N C 80 N 400 20 cm 60 cm r 160 N
40. Chapter 5 Torque and Rotational Equilibrium Physics, 6th
Edition 39 *5-18. Find the resultant torque about axis B in Fig.
5-15. Fx = 160 cos 400 ; Fy = 160 sin 400 = (123 N)(0.2 m) + (103
N)(0.6 m) = 37.2 N m Equilibrium 5-19. A uniform meter stick is
balanced at its midpoint with a single support. A 60-N weight is
suspended at the 30 cm mark. At what point must a 40-N weight be
hung to balance the system? (The 60-N weight is 20 cm from the
axis) = 0; (60 N)(20 cm) (40 N)x = 0 40 x = 1200 N cm or x = 30 cm:
The weight must be hung at the 80-cm mark. 5-20. Weights of 10 N,
20 N, and 30 N are placed on a meterstick at the 20 cm, 40 cm, and
60 cm marks, respectively. The meterstick is balanced by a single
support at its midpoint. At what point may a 5-N weight be attached
to produce equilibrium. = (10 N)(30 cm) + (20 N)(10 cm) (30 N)(10
cm) (5 N) x = 0 5 x = (300 + 200 300) or x = 40 cm The 5-N weight
must be placed at the 90-cm mark 5-21. An 8-m board of negligible
weight is supported at a point 2 m from the right end where a 50-N
weight is attached. What downward force at the must be exerted at
the left end to produce equilibrium? F (6 m) (50 N)(2 m) = 0 6 F =
100 N m or F = 16.7 N Fx Fy B 80 N 400 20 cm 60 cm 160 N 20 cm x 40
N60 N 10 cm 30 cm 10 N 20 N x 5 N30 N 50 N F 6 m 2 m
41. Chapter 5 Torque and Rotational Equilibrium Physics, 6th
Edition 40 5-22. A 4-m pole is supported at each end by hunters
carrying an 800-N deer which is hung at a point 1.5 m from the left
end. What are the upward forces required by each hunter? = A (0)
(800 N)(1.5 m) + B (4.0 m) = 0 4B = 1200 N or B = 300 N Fy = A + B
800 lb = 0; A = 500 N 5-23. Assume that the bar in Fig. 5-16 is of
negligible weight. Find the forces F and A provided the system is
in equilibrium. = (80 N)(1.20 m) F (0.90 m) = 0; F = 107 N Fy = F A
80 N = 0; A = 107 N 80 N = 26.7 N F = 107 N, A = 26.7 N 5-24. For
equilibrium, what are the forces F1 and F2 in Fig. 5-17. (Neglect
weight of bar.) = (90 lb)(5 ft) F2 (4 ft) (20 lb)(5 ft) = 0; F2 =
87.5 lb Fy = F1 F2 20 lb 90 lb = 0 F1 = F2 +110 lb = 87.5 lb + 110
lb, F1 = 198 lb 5-25. Consider the light bar supported as shown in
Fig. 5-18. What are the forces exerted by the supports A and B? = B
(11 m) (60 N)(3 m) (40 N)( 9 m) = 0; B = 49.1 N Fy = A + B 40 N 60
N = 0 A = 100 N B = 100 N 49.1 N; B = 50.9 N 5-26. A V-belt is
wrapped around a pulley 16 in. in diameter. If a resultant torque
of 4 lb ft is required, what force must be applied along the belt?
R = (16 in.) = 8 in. R = (8/12 ft) = 0.667 ft F (0.667 ft) = 4 lb
ft; F = 6.00 lb F 800 N BA 2.5 m1.5 m Axis 80 N F A 90 cm30 cm Axis
20 lb F2 5 ft Axis 1 ft 90 lb F1 4 ft B 3 m Axis 40 N 2 m 60 N A 6
m
42. Chapter 5 Torque and Rotational Equilibrium Physics, 6th
Edition 41 5-27. A bridge whose total weight is 4500 N is 20 m long
and supported at each end. Find the forces exerted at each end when
a 1600-N tractor is located 8 m from the left end. = B (20 m) (1600
N)(8 m) (4500 N)( 10 m) = 0; B = 2890 N Fy = A + B 1600 N 4500 N =
0 A = 6100 N B = 6100 N 2890 N; B = 3210 N 5-28. A 10-ft platform
weighing 40 lb is supported at each end by stepladders. A 180-lb
painter is located 4 ft from the right end. Find the forces exerted
by the supports. = B(10 ft) (40 lb)(5 ft) (180 lb)( 6 ft) = 0; B =
128 lb Fy = A + B 40 lb 180 lb = 0 A = 220 lb B = 220 lb 128 lb; A
= 92.0 lb *5-29. A horizontal, 6-m boom weighing 400 N is hinged at
the wall as shown in Fig. 5-19. A cable is attached at a point 4.5
m away from the wall, and a 1200-N weight is attached to the right
end. What is the tension in the cable? = 900 370 = 530 ; Ty = T sin
530 = (T sin 530 )(4.5 m) (400 N)(3 m) (1200 N)(6 m) = 0; 3.59 T =
1200 N + 7200 N; T = 2340 N *5-30. What are the horizontal and
vertical components of the force exerted by the wall on the boom?
What is the magnitude and direction of this force? Fx = H Tx = 0; H
T cos 530 = 0; H = (2340 N) cos 530 ; H = 1408 N Fy = V + T sin 530
400 N 1200 N = 0; V = 1600 N (2340 N) sin 530 = -269 N Thus, the
components are: H = 1408 N and V = -269 N. The resultant of these
is: R H V 2 2 1434 N; tan = -269 1408 = 10.8 S of E0 R = 1434 N,
349.20 B 10 m Axis 4500 N 2 m 1600 N A 8 m B 4 ft Axis 180 lb 1 ft
40 lb A 5 ft 1.5 mH Ty Ty B 1.5 m Axis 1200 N400 N V 3 m
43. Chapter 5 Torque and Rotational Equilibrium Physics, 6th
Edition 42 Center of Gravity 5-31. A uniform 6-m bar has a length
of 6 m and weighs 30 N. A 50-N weight is hung from the left end and
a 20-N force is hung at the right end. How far from the left end
will a single upward force produce equilibrium? Fy = F 50 N 30 N 20
N = 0; F = 100 N = F x (30 N)(3 m) (20 N)(6 m) = 0 (100 N) x = 210
N m; x = 2.10 m 5-32. A 40-N sphere and a 12-N sphere are connected
by a light rod 200 mm in length. How far from the middle of the
40-N sphere is the center of gravity? Fy = F 40 N 12 N = 0; F = 52
N = F x (40 N)(0) (12 N)(0.20 m) = 0 (52 N) x = 2.40 N m; x =
0.0462 m or x = 46.2 mm 5-33. Weights of 2, 5, 8, and 10 N are hung
from a 10-m light rod at distances of 2, 4, 6, and 8 m from the
left end. How far from the left in is the center of gravity? Fy = F
10 N 8 N 5 N 2 N = 0; F = 25 N Fx (2 N)(2 m) (5 N)(4 m) (8 N)(6 m)
(10 N)(8 m) = 0 (25 N) x = 152 N m; x = 6.08 m 5-34. Compute the
center of gravity of sledgehammer if the metal head weighs 12 lb
and the 32- in. supporting handle weighs 2 lb. Assume that the
handle is of uniform construction and weight. Fy = F 2 lb 12 lb =
0; F = 14 lb Fx (12 lb)(0) (2 lb)(16 in.) = 0; Fx = 32 lb in. (14
lb) x = 32 lb in.; x = 2.29 in. from head. Axis F 20 N30 N50 N x 3
m3 m F 12 N40 N 200 mm x 10 N5 N 8 N2 N 2 m 2 m 2 m 2 m 2 m x F F
16 in. 16 in. x 2 lb12 lb
44. Chapter 5 Torque and Rotational Equilibrium Physics, 6th
Edition 43 Challenge Problems 5-35. What is the resultant torque
about the hinge in Fig. 4-20? Neglect weight of the curved bar. =
(80 N)(0.6 m) (200 N)(0.4 m) sin 400 = 48.0 N m 51.4 N m; = 3.42 N
m 5-36. What horizontal force applied to the left end of the bar in
Fig. 4-20 will produce rotational equilibrium? From Prob. 5-33: = -
3.42 N m. Thus, if = 0, then torque of +3.42 N m must be added. F
(0.6 m) cos 400 = +3.45 N m; F = 7.45 N 5-37. Weights of 100, 200,
and 500 lb are placed on a light board resting on two supports as
shown in Fig. 4-21. What are the forces exerted by the supports? =
(100 lb)(4 ft) + B(16 ft) (200 lb)(6 ft) (500 lb)(12 ft) = 0; B =
425 lb Fy = A + B 100 lb 200 lb 500 lb = 0 A = 800 lb B = 800 lb
425 lb; A = 375 lb The forces exerted by the supports are : A = 375
N and B = 425 N 5-38. An 8-m steel metal beam weighs 2400 N and is
supported 3 m from the right end. If a 9000-N weight is placed on
the right end, what force must be exerted at the left end to
balance the system? = A (5 m) + (2400 N)(1 m) (9000 N)( 3 m) = 0; A
= 4920 N Fy = A + B 2400 N 9000 N = 0 B = 11,400 N A = 11,400 N
4920 N; A = 6480 N 400500 F 200 N 60 cm 40 cm r 400 400 80 N 200
N60 cm 40 cm r 400 80 N Axis 100 lb 200 lb 500 lb A B 6 ft6 ft 4
ft4 ft A 9000 N F 4 m 3 m1 m 2400 N
45. Chapter 5 Torque and Rotational Equilibrium Physics, 6th
Edition 44 *5-39. Find the resultant torque about point A in Fig.
5-22. = (70 N)(0.05 m) sin 500 (50 N)(0.16 m) sin 550 = 2.68 N m
6.55 N m = 3.87 N m = 3.87 N m *5-40. Find the resultant torque
about point B in Fig. 5-22. = (70 N)(0) (50 N)(a + b) ; First find
a and b. a = (0.05 m) cos 500 = 0.0231 m; b = (0.16 m) sin 550 =
0.131 m = (50 N)(0.0231 m + 0.131 m) = 8.16 N m = 8.16 N m Critical
Thinking Questions *5-41. A 30-lb box and a 50-lb box are on
opposite ends of a 16-ft board supported only at its midpoint. How
far from the left end should a 40-lb box be placed to produce
equilibrium? Would the result be different if the board weighed 90
lb? Why, or why not? = (30 lb)(8 ft) + (40 lb)(x) (50 lb)(8 ft) =
0; x = 4.00 ft Note that the weight acting at the center of the
board does NOT contribute to torque about the center, and
therefore, the balance point is not affected, regardless of the
weight. 5-42. On a lab bench you have a small rock, a 4-N
meterstick and a single knife-edge support. Explain how you can use
these three items to find the weight of the small rock. Measure
distances a and b; determine F and then calculate the weight W from
equilibrium methods. 0.5 m F 4 N W ba b a70 N 50 N B 5 cm 16 cm 500
550 r r 70 N 50 N B 5 cm A 16 cm 500 550 x F W 40 lb 8 ft8 ft 50
lb30 lb
46. Chapter 5 Torque and Rotational Equilibrium Physics, 6th
Edition 45 *5-43. Find the forces F1, F2, and F3 such that the
system drawn in Fig. 5-23 is in equilibrium. Note action-reaction
forces R and R. First, lets work with top board: (about R) = 0;
Force R is upward. R = (300 lb)(6 ft) (50 lb)(2 ft) F1(8 ft) = 0 F1
= 213 lb Now, Fy = 0 gives: 213 lb + R 300 lb 50 lb = 0; R = 138 lb
= R Next we sum torques about F2 with R = 138 lb is directed in a
downward direction: F = (138 lb)(3 ft) + F3(7 ft) (200 lb)(5 ft) =
0; From which: F3 = 83.9 lb Fy = 0 = F2 + 83.9 lb 138 lb 200 lb; F2
= 254 lb The three unknown forces are: F1 = 213 lb, F2 = 254 lb, F3
= 83.9 lb *5-44. (a) What weight W will produce a tension of 400 N
in the rope attached to the boom in Fig. 5-24?. (b) What would be
the tension in the rope if W = 400 N? Neglect the weight of the
boom in each case. (a) m) sin 300) W (6 m) cos 300 = 0 W = 154 N
(b) = T(4 m) sin 300 (400 N)(6 m) cos 300 = 0 T = 600 N *5-45.
Suppose the boom in Fig. 5-24 has a weight of 100 N and the
suspended weight W is equal to 400 N. What is the tension in the
cord? m) sin 300) (400 N)(6 m) cos 300 (100 N)(3 m) cos 300 = 0 T =
1169 N 50 lb 2 ft 5 ft 2 ft6 ft 3 ft 2 ft 300 lb 200 lb F3F2 F1 R R
Axis 300 4 m 2 m 400 N W 300 100 NAxis 300 4 m 2 m T W 300
47. Chapter 5 Torque and Rotational Equilibrium Physics, 6th
Edition 46 *5-46. For the conditions set in Problem 5-5, what are
the horizontal and vertical components of the force exerted by the
floor hinge on the base of the boom? Fx = H 1169 N = 0; or H = 1169
N Fy = V 100 N 400 N = 0; or V = 500 N H = 1169 N and V = 500 N
**5-47. What is the tension in the cable for Fig. 5-25. The weight
of the boom is 300 N but its length is unknown. (Select axis at
wall, L cancels.) TL N L Lsin ( ) sin sin75 300 2 30 546 30 00 0 0
T sin 750 = 75.0 N + 273 N; T = 360 N **5-48. What are the
magnitude and direction of the force exerted by the wall on the
boom in Fig. 5-25? Again assume that the weight of the board is 300
N. Refer to the figure and data given in Problem 5-7 and recall
that T = 360 N. Fx = H - (360 N) cos 450 = 0; H = 255 N Fy = V +
(360 N) sin 450 300 N 546 N = 0; V = 591 N H = 255 N and V = 591 N
*5-49. An car has a distance of 3.4 m between front and rear axles.
If 60 percent of the weight rests on the front wheels, how far is
the center of gravity located from the front axle? = 0.6W(0) +
0.4W(3.4 m) F x = 0 But F = W: 1.36 W W x = 0 x = 1.36 m from front
axle V H 100 NAxis 300 4 m 2 m 1169 N 400 N 300 600 300 450 T = 360
N T H 546 N L r 750 300 N 450 300 V 0.4W xF 0.6W3.4 m Axis
48. Chapter 6 Uniform Acceleration Physics, 6th Edition 47
Chapter 6. Uniform Acceleration Problems: Speed and Velocity 6-1. A
car travels a distance of 86 km at an average speed of 8 m/s. How
many hours were required for the trip? s vt 86,000 m 1 h 10,750 s 8
m/s 3600 s t t = 2.99 h 6-2. Sound travels at an average speed of
340 m/s. Lightning from a distant thundercloud is seen almost
immediately. If the sound of thunder reaches the ear 3 s later, how
far away is the storm? t s t 20 m 340 m / s 0.0588 s t = 58.8 ms
6-3. A small rocket leaves its pad and travels a distance of 40 m
vertically upward before returning to the earth five seconds after
it was launched. What was the average velocity for the trip? v s t
40 80m + 40 m 5 s m 5 s v = 16.0 m/s 6-4. A car travels along a
U-shaped curve for a distance of 400 m in 30 s. Its final location,
however is only 40 m from the starting position. What is the
average speed and what is the magnitude of the average velocity?
Average speed: v s t 400 m 30 s v = 13.3 m/s Average velocity: v D
t m 30 s 40 v = 1.33 m/s, E s = 400 m D = 40 m
49. Chapter 6 Uniform Acceleration Physics, 6th Edition 48 6-5.
A woman walks for 4 min directly north with a average velocity of 6
km/h; then she walks eastward at 4 km/h for 10 min. What is her
average speed for the trip? t1 = 4 min = 0.0667 h; t2 = 10 min =
0.167 h s1 = v1t1 = (6 km/h)(0.0667 h) = 0.400 km s1 = v2t2 = (4
km/h)(0.167 h) = 0.667 km v s s t t 1 2 1 2 0.4 km + 0.667 km
0.0667 h + 0.167 h v = 4.57 km/h 6-6. What is the average velocity
for the entire trip described in Problem 6-5? D ( . ; tan . . 0 667
0 4 0 667 km) + (0.400 km) km km 2 2 D = 0.778 km, 31.00 v 0 778
333 . . km 0.0667 h + 0.167 h km / h v = 3.33 km/h, 31.00 6-7. A
car travels at an average speed of 60 mi/h for 3 h and 20 min. What
was the distance? t = 3 h + 0.333 h = 3.33 h; s = vt = (60
mi/h)(3.33 h); s = 200 mi 6.8 How long will it take to travel 400
km if the average speed is 90 km/h? t s t 400 km 90 km / h t = 4.44
h *6-9. A marble rolls up an inclined ramp a distance of 5 m, then
stops and returns to a point 5 m below its starting point. The
entire trip took only 2 s. What was the average speed and what was
the average velocity? (s1 = 5 m, s2 = -10 m) speed = 5 m + 10 m 2 s
v = 7.50 m/s velocity = D t 5 m - 10 m 2 s v = 2.5 m/s, down plane.
D s2 s1 6 km/h, 4 min 4 km/h, 10 min D s1 s2 CB A E
50. Chapter 6 Uniform Acceleration Physics, 6th Edition 49
Uniform Acceleration 6-10. The tip of a robot arm is moving to the
right at 8 m/s. Four seconds later, it is moving to the left at 2
m/s. What is the change in velocity and what is the acceleration. v
= vf - vo = (2 m/s) (8 m/s) v = 10 m/s a v t 10 m / s 4 s a = 2.50
m/s2 6-11. An arrow accelerates from zero to 40 m/s in the 0.5 s it
is in contact with the bow string. What is the average
acceleration? a v v t f o 40 m / s - 0 0.5 s a = 80.0 m/s2 6-12. A
car traveling initially at 50 km/h accelerates at a rate of 4 m/s2
for 3 s. What is the final speed? vo = 50 km/h = 13.9 m/s; vf = vo
+ at vf = (13.9 m/s) + (4 m/s2 )(3 s) = 25.9 m/s; vf = 25.9 m/s
6-13. A truck traveling at 60 mi/h brakes to a stop in 180 ft. What
was the average acceleration and stopping time? vo = 60 mi/h = 88.0
ft/s 2as = vf 2 vo 2 2 2 2 0 (88.0 ft/s) 2 2(180 ft) f ov v a s a =
21.5 ft/s2 0 0 2 2(180 ft) ; 2 88.0 ft/s + 0 f f v v x x t t v v t
= 4.09 s
51. Chapter 6 Uniform Acceleration Physics, 6th Edition 50
6-14. An arresting device on a carrier deck stops an airplane in
1.5 s. The average acceleration was 49 m/s2 . What was the stopping
distance? What was the initial speed? vf = vo + at; 0 = vo + ( 49
m/s2 )(1.5 s); vo = 73.5 m/s s = vf t - at2 ; s = (0)(1.5 s) (-49
m/s2 )(1.5 s)2 ; s = 55.1 m 6-15. In a braking test, a car
traveling at 60 km/h is stopped in a time of 3 s. What was the
acceleration and stopping distance? ( vo = 60 km/h = 16.7 m/s) vf =
vo + at; (0) = (16.7 m/s) + a (3 s); a = 5.56 m/s2 0 16.6 m/s + 0 3
s 2 2 fv v s t ; s = 25.0 m 6-16. A bullet leaves a 28-in. rifle
barrel at 2700 ft/s. What was its acceleration and time in the
barrel? (s = 28 in. = 2.33 ft) 2as = vo 2 - vf 2 ; a v v s f 2 0 2
2 (2700 ft / s) 0 2(2.33 ft) 2 ; a = 1.56 x 106 m/s2 s v v t s v v
f f 0 02 2 2 2 33 ; t = ft) 0 + 2700 ft / s ( . ; t = 1.73 ms 6-17.
The ball in Fig. 6-13 is given an initial velocity of 16 m/s at the
bottom of an inclined plane. Two seconds later it is still moving
up the plane, but with a velocity of only 4 m/s. What is the
acceleration? vf = vo + at; a v v t f 0 4 m / s - (16 m / s) 2 s ;
a = -6.00 m/s2 6-18. For Problem 6-17, what is the maximum
displacement from the bottom and what is the velocity 4 s after
leaving the bottom? (Maximum displacement occurs when vf = 0) 2as =
vo 2 - vf 2 ; s v v a f 2 0 2 2 0 (16 m / s) 2(-6 m / s ) 2 2 ; s =
+21.3 m vf = vo + at = 16 m/s = (-6 m/s2 )(4 s); vf = - 8.00 m/s,
down plane
52. Chapter 6 Uniform Acceleration Physics, 6th Edition 51
6-19. A monorail train traveling at 80 km/h must be stopped in a
distance of 40 m. What average acceleration is required and what is
the stopping time? ( vo = 80 km/h = 22.2 m/s) 2as = vo 2 - vf 2 ; a
v v s f 2 0 2 2 0 (22.2 m / s) 2(40 m) 2 ; a = -6.17 m/s2 s v v t s
v v f f 0 02 2 2 40 ; t = m) 22.2 m / s + 0 ( ; t = 3.60 m/s
Gravity and Free-Falling Bodies 6-20. A ball is dropped from rest
and falls for 5 s. What are its position and velocity? s = vot +
at2 ; s = (0)(5 s) + (-9.8 m/s2 )(5 s)2 ; s = -122.5 m vf = vo + at
= 0 + (-9.8 m/s2 )(5 s); v = -49.0 m/s 6-21. A rock is dropped from
rest. When will its displacement be 18 m below the point of
release? What is its velocity at that time? s = vot + at2 ; (-18 m)
= (0)(t) + (-9.8 m/s2 )t2 ; t = 1.92 s vf = vo + at = 0 + (-9.8
m/s2 )(1.92 s); vf = -18.8 m/s 6-22. A woman drops a weight from
the top of a bridge while a friend below measures the time to
strike the water below. What is the height of the bridge if the
time is 3 s? s = vot + at2 = (0) + (-9.8 m/s2 )(3 s)2 ; s = -44.1 m
6-23. A brick is given an initial downward velocity of 6 m/s. What
is its final velocity after falling a distance of 40 m? 2as = vo 2
- vf 2 ; v v asf 0 2 2 40(-6 m / s) 2(-9.8 m / s m)2 2 )( ; v =
28.6 m/s; Since velocity is downward, v = - 28.6 m/s
53. Chapter 6 Uniform Acceleration Physics, 6th Edition 52
6-24. A projectile is thrown vertically upward and returns to its
starting position in 5 s. What was its initial velocity and how
high did it rise? s = vot + at2 ; 0 = vo(5 s) + (-9.8 m/s2 )(5 s)2
; vo = 24.5 m/s It rises until vf = 0; 2as = vo 2 - vf 2 ; s 0 ( )
24.5 m / s) 2(-9.8 m / s 2 2 ; s = 30.6 m 6-25. An arrow is shot
vertically upward with an initial velocity of 80 ft/s. What is its
maximum height? (At maximum height, vf = 0; a = g = -32 ft/s2 ) 2as
= vo 2 - vf 2 ; s v v a f 2 0 2 2 0 - (80 ft / s) 2(-32 ft / s 2 2
) ; s = 100 ft 6-26. In Problem 6-25, what are the position and
velocity of the arrow after 2 s and after 6 s? s = vot + at2 = (80
ft/s)(2 s) + (-32 ft/s2 )(2 s)2 ; s = 96 ft vf = vo + at = (80
ft/s) + (-32 ft/s2 )(2 s); vf = 16 ft/s s = vot + at2 = (80 ft/s)(6
s) + (-32 ft/s2 )(6 s)2 ; s = -96 ft vf = vo + at = (80 ft/s) +
(-32 ft/s2 )(6 s); vf = -112 ft/s 6-27. A hammer is thrown
vertically upward to the top of a roof 16 m high. What minimum
initial velocity was required? 2as = vo 2 - vf 2 ; v v asf0 2 2 16
(0) 2(-9.8 m / s m)2 2 )( ; vo = 17.7 m/s Horizontal Projection
6-28. A baseball leaves a bat with a horizontal velocity of 20 m/s.
In a time of 0.25 s, how far will it have traveled horizontally and
how far has it fallen vertically? x = vox t = (20 m/s)(2.5 s) ; x =
50.0 m y = voy + gt2 = (0)(2.5 s) + (-9.8 m/s2 )(0.25 s)2 y =
-0.306 m
54. Chapter 6 Uniform Acceleration Physics, 6th Edition 53 0
6-29. An airplane traveling at 70 m/s drops a box of supplies. What
horizontal distance will the box travel before striking the ground
340 m below? First we find the time to fall: y = voy t + gt2 t y g
2 2 9 8 ( . 340 m) m / s2 t = 8.33 s ; x = vox t = (70 m/s)(8.33 s)
; x = 583 m 6-30. At a lumber mill, logs are discharged
horizontally at 15 m/s from a greased chute that is 20 m above a
mill pond. How far do the logs travel horizontally? y = gt2 ; t y g
2 2 9 8 ( . 20 m) m / s2 ; t = 2.02 s x = vox t = (15 m/s)(8.33 s)
; x = 30.3 m 6-31. A steel ball rolls off the edge of a table top 4
ft above the floor. If it strikes the floor 5 ft from the base of
the table, what was its initial horizontal speed? First find time
to drop 4 ft: t y g 2 2 32 ( 4 ft) ft / s2 ; t = 0.500 s x = vox t
; v x t x0 5 05 ft s. ; vox = 10.0 ft/s 6-32. A bullet leaves the
barrel of a weapon with an initial horizontal velocity of 400 m/s.
Find the horizontal and vertical displacements after 3 s. x = vox t
= (400 m/s)(3 s) ; x = 1200 m y = voy + gt2 = (0)(3 s) + (-9.8 m/s2
)(3 s)2 y = -44.1 m 6-33. A projectile has an initial horizontal
velocity of 40 m/s at the edge of a roof top. Find the horizontal
and vertical components of its velocity after 3 s. vx = vox = 40
m/s vy = voy t + gt = 0 + (-9.8 m/s2 )(3s); vy = -29.4 m/s
55. Chapter 6 Uniform Acceleration Physics, 6th Edition 54 The
More General Problem of Trajectories 6-34. A stone is given an
initial velocity of 20 m/s at an angle of 580 . What are its
horizontal and vertical displacements after 3 s? vox = (20 m/s) cos
580 = 10.6 m/s; voy = (20 m/s) sin 580 = 17.0 m/s x = voxt = (10.6
m/s)(3 s); x = 31.8 m y = voyt + gt2 = (17.0 m/s)(3 s) +(-9.8 m/s2
)(3 s)2 ; y = 6.78 m 6-35. A baseball leaves the bat with a
velocity of 30 m/s at an angle of 300 . What are the horizontal and
vertical components of its velocity after 3 s? vox = (30 m/s) cos
300 = 26.0 m/s; voy = (30 m/s) sin 300 = 15.0 m/s vx = vox = 26.0
m/s ; vx = 26.0 m/s vy = voy + gt = (15 m/s) + (-9.8 m/s2 )(3 s) ;
vy = -14.4 m/s 6-36. For the baseball in Problem 6-33, what is the
maximum height and what is the range? ymax occurs when vy = 0, or
when: vy = voy + gt = 0 and t = - voy/g t v g t oy 30 30 9 8 153 0
sin . ; . m / s s; Now we find ymax using this time. ymax = voyt +
gt2 = (15 m/s)(1.53 s) + (-9.8 m/s2 )(1.53 s)2 ; ymax = 11.5 m The
range will be reached when the time is t = 2(1.53 s) or t = 3.06 s,
thus R = voxt= (30 m/s) cos 300 (3.06 s); R = 79.5 m 6-37. An arrow
leaves the bow with an initial velocity of 120 ft/s at an angle of
370 with the horizontal. What are the horizontal and vertical
components of is displacement two seconds later? vox = (120 ft/s)
cos 370 = 104 ft/s; voy = (120 ft/s) sin 300 = 60.0 ft/s
56. Chapter 6 Uniform Acceleration Physics, 6th Edition 55
6-37. (Cont.) The components of the initial velocity are: vox = 104
ft/s; voy = 60.0 ft/s x = voxt = (104 ft/s)(2 s); x = 208 ft y =
voyt + gt2 = (60.0 m/s)(2 s) +(-32 ft/s2 )(2 s)2 ; y = 56.0 ft
*6-38. In Problem 6-37, what are the magnitude and direction of
arrows velocity after 2 s? vx = vox = 104 ft/s ; vx = 104 ft/s vy =
voy + gt = (60 m/s) + (-32 ft/s2 )(2 s) ; vy = -4.00 ft/s *6-39. A
golf ball in Fig. 6-14 leaves the tee with a velocity of 40 m/s at
650 . If it lands on a green located 10 m higher than the tee, what
was the time of flight, and what was the horizontal distance to the
tee? vox = (40 m/s) cos 650 = 16.9 m/s; voy = (40 m/s) sin 650 =
36.25 m/s y = voyt + gt2 : 10 ft = (36.25 m/s) t + (-9.8 m/s2 )t2
Solving quadratic (4.9t2 36.25t + 10 = 0) yields: t1 = 0.287 s and
t2 = 7.11 s The first time is for y = +10 m on the way up, the
second is y = +10 m on the way down. Thus, the time from tee to
green was: t = 7.11 s Horizontal distance to tee: x = voxt = (16.9
m/s)(7.11 s); x = 120 m *6-40. A projectile leaves the ground with
a velocity of 35 m/s at an angle of 320 . What is the maximum
height attained. vox = (35 m/s) cos 320 = 29.7 m/s; voy = (35 m/s)
sin 320 = 18.55 m/s ymax occurs when vy = 0, or when: vy = voy + gt
= 0 and t = - voy/g t v g t oy 1855 9 8 189 0 . . ; . m / s s2 ;
Now we find ymax using this time. ymax = voyt + gt2 = (18.55
m/s)(1.89 s) + (-9.8 m/s2 )(1.89 s)2 ; ymax = 17.5 m
57. Chapter 6 Uniform Acceleration Physics, 6th Edition 56
*6-41. The projectile in Problem 6-40 rises and falls, striking a
billboard at a point 8 m above the ground. What was the time of
flight and how far did it travel horizontally. vox = (35 m/s) cos
320 = 29.7 m/s; voy = (35 m/s) sin 320 = 18.55 m/s y = voyt + gt2 :
8 m = (18.55 m/s) t + (-9.8 m/s2 )t2 Solving quadratic (4.9t2
18.55t + 8 = 0) yields: t1 = 0.497 s and t2 = 3.36 s The first time
is for y = +8 m on the way up, the second is y = +8 m on the way
down. Thus, the time from tee to green was: t = 3.29 s Horizontal
distance to tee: x = voxt = (29.7 m/s)(3.29 s); x = 97.7 m
Challenge Problems 6-42. A rocket travels in space at 60 m/s before
it is given a sudden acceleration. Its velocity increases to 140
m/s in 8 s, what was its average acceleration and how far did it
travel in this time? a v v t f 0 (140 m / s) - (60 m / s) 8 s ; a =
10 m/s2 s v v t f F HG I KJ0 2 140 8 m / s+ 60 m / s 2 sb g; t =
800 s 6-43. A railroad car starts from rest and coasts freely down
an incline. With an average acceleration of 4 ft/s2 , what will be
the velocity after 5 s? What distance does it travel? vf = vo + at
= 0 + (4 ft/s2 )(5 s); vf = 20 ft/s s = vot + at2 = 0 + (4 ft/s2
)(5 s)2 ; s = 50 ft
58. Chapter 6 Uniform Acceleration Physics, 6th Edition 57
*6-44. An object is projected horizontally at 20 m/s. At the same
time, another object located 12 m down range is dropped from rest.
When will they collide and how far are they located below the
release point? A: vox = 20 m/s, voy = 0; B: vox = voy = 0 Ball B
will have fallen the distance y at the same time t as ball A. Thus,
x = voxt and (20 m/s)t = 12 m; t = 0.600 s y = at2 = (-9.8 m/s2
)(0.6 s)2 ; y = -1.76 m 6-45. A truck moving at an initial velocity
of 30 m/s is brought to a stop in 10 s. What was the acceleration
of the car and what was the stopping distance? a v v t f 0 0 - 30 m
/ s 10 s ; a = -3.00 m/s2 s v v t f F HG I KJ0 2 30 10 m / s + 0 2
sb g; s = 150 m 6-46. A ball is thrown vertically upward with an
initial velocity of 23 m/s. What are its position and velocity
after 2s, after 4 s, and after 8 s? Apply s = vot + at2 and vf = vo
+ at for time of 2, 4, and 8 s: (a) s = (23 m/s)(2 s) + (-9.8 m/s2
)(2 s)2 ; s = 26.4 m vf = (23 m/s) + (-9.8 m/s2 )(2 s) ; vf = 3.40
m/s (b) s = (23 m/s)(4 s) + (-9.8 m/s2 )(4 s)2 ; s = 13.6 m vf =
(23 m/s) + (-9.8 m/s2 )(4 s) ; vf = -16.2 m/s (c) s = (23 m/s)(8 s)
+ (-9.8 m/s2 )(8 s)2 ; s = -130 m vf = (23 m/s) + (-9.8 m/s2 )(8 s)
; vf = -55.4 m/s y BA 12 m
59. Chapter 6 Uniform Acceleration Physics, 6th Edition 58
6-47. A stone is thrown vertically downward from the top of a
bridge. Four seconds later it strikes the water below. If the final
velocity was 60 m/s. What was the initial velocity of the stone and
how high was the bridge? vf = vo + at; v0 = vf at = (-60 m/s) -
(-9.8 m/s)(4 s); vo = -20.8 m/s s = vot + at2 = (-20.8 m/s)(4 s) +
(-9.8 m/s)(4 s)2 ; s = 162 m 6-48. A ball is thrown vertically
upward with an initial velocity of 80 ft/s. What are its position
and velocity after (a) 1 s; (b) 3 s; and (c) 6 s Apply s = vot +
at2 and vf = vo + at for time of 2, 4, and 8 s: (a) s = (80 ft/s)(1
s) + (-32 ft/s2 )(1 s)2 ; s = 64.0 ft vf = (80 ft/s) + (-32 ft/s2
)(2 s) ; vf = 16.0 ft/s (b) s = (80 ft/s)(3 s) + (-32 ft/s2 )(3 s)2
; s = 96.0 ft vf = (80 ft/s) + (-32 ft/s2 )(3 s) ; vf = -16.0 ft/s
(c) s = (80 ft/s)(6 s) + (-32 ft/s2 )(6 s)2 ; s = 64.0 ft vf = (80
ft/s) + (-32 ft/s2 )(6 s) ; vf = -96.0 ft/s 6-49. An aircraft
flying horizontally at 500 mi/h releases a package. Four seconds
later, the package strikes the ground below. What was the altitude
of the plane? y = gt2 = (-32 ft/s2 )(4 s)2 ; y = -256 ft *6-50. In
Problem 6-49, what was the horizontal range of the package and what
are the components of its final velocity? vo = 500 mi/h = 733 ft/s;
vx = vox = 733 ft/s; voy = 0; t = 4 s x = vxt = (733 ft/s)(4 s); x
= 2930 ft vy = voy + at = 0 + (-32 ft/s)(4 s); vy = -128 ft/s; vx =
733 m/s
60. Chapter 6 Uniform Acceleration Physics, 6th Edition 59
*6-51. A putting green is located 240 ft horizontally and 64 ft
vertically from the tee. What must be the magnitude and direction
of the initial velocity if a ball is to strike the green at this
location after a time of 4 s? x = voxt; 240 ft = vox (4 s); vox =
60 m/s s = vot + at2 ; 64 ft = voy(4 s) + (-32 ft/s2 )(4 s)2 ; voy
= 80 ft/s v v vx y 2 2 60( (80ft / s) ft / s)2 2 ; tan 80 ft / s 60
ft / s v = 100 ft/s, = 53.10 Critical Thinking Questions 6-52. A
long strip of pavement is marked off in 100-m intervals. Students
use stopwatches to record the times a car passes each mark. The
following data is listed: Distance, m 0 10 m 20 m 30 m 40 m 50 m
Time, s 0 2.1 s 4.3 s 6.4 s 8.4 s 10.5 s Plot a graph with distance
along the y-axis and time along the x-axis. What is the
significance of the slope of this curve? What is the average speed
of the car? At what instant in time is the distance equal to 34 m?
What is the acceleration of the car? Data taken directly from the
graph (not drawn): Ans. Slope is v, 4.76 m/s, 7.14 s, 0. 6-53. An
astronaut tests gravity on the moon by dropping a tool from a
height of 5 m. The following data are recorded electronically.
Height, m 5.00 m 4.00 m 3.00 m 2.00 m 1.00 m 0 m Time, s 0 1.11 s
1.56 s 1.92 s 2.21 s 2.47 s
61. Chapter 6 Uniform Acceleration Physics, 6th Edition 60
6-53. (Cont.) Plot the graph of this data. Is it a straight line?
What is the average speed for the entire fall? What is the
acceleration? How would you compare this with gravity on earth?
Data taken directly from the graph (not drawn): Ans. Slope is v,
4.76 m/s, 7.14 s, 0. *6-54. A car is traveling initially North at
20 m/s. After traveling a distance of 6 m, the car passes point A
where it's velocity is still northward but is reduced to 5 m/s. (a)
What are the magnitude and direction of the acceleration of the
car? (b) What time was required? (c) If the acceleration is held
constant, what will be the velocity of the car when it returns to
point A? (a) vo = 20 m/s, vf = 5 m/s, x = 6 m 2as = vo 2 - vf 2 ; 2
2 2 2 0 (5 m/s) (20 m/s) 2 2(6 m) fv v a s ; a = -31.2 m/s2 (b) 0 0
2 2(6 m) ; 2 20 m/s + 5 m/s f f v v s s t t v v ; t = 0.480 s (c)
Starts at A with vo = + 5 m/s then returns to A with zero net
displacement (s = 0): 2as = vo 2 - vf 2 ; 0 = (5 m/s)2 vf 2 ; vf (5
m / s) m / s2 5 ; vf = - 5 m/s *6-55. A ball moving up an incline
is initially located 6 m from the bottom of an incline and has a
velocity of 4 m/s. Five seconds later, it is located 3 m from the
bottom. Assuming constant acceleration, what was the average
velocity? What is the meaning of a negative average velocity? What
is the average acceleration and final velocity? vo = + 4 m/s; s =
-3 m; t = 5 s Find vavg s = vavg t; v 3 m 5 s ; vavg = -0.600 m/s
Negative average velocity means that the velocity was down the
plane most of the time. x = 6 mx = 0 A v = 5 m/sv = 20 m/s 4 m/s 6
m 3 m s = 0
62. Chapter 6 Uniform Acceleration Physics, 6th Edition 61
*6-55. (Cont.) s = vot + at2 ; -3 m = (4 m/s)(5 s) + a (5 s)2 ; a =
-1.84 m/s2 vf = vo + at = 4 m/s + (-1.84 m/s2 )(5 s); vf = -5.20
m/s *6-56. The acceleration due to gravity on an distant planet is
determined to be one-fourth its value on the earth. Does this mean
that a ball dropped from a height of 4 m above this planet will
strike the ground in one-fourth the time? What are the times
required on the planet and on earth? The distance as a function of
time is given by: s = at2 so that one-fourth the acceleration
should result in twice the drop time. t s g e e 2 2(4 m) 9.8 m / s2
te = 0.904 s t s g p p 2 2(4 m) 2.45 m / s2 tp = 1.81 s *6-57.
Consider the two balls A and B shown in Fig. 6-15. Ball A has a
constant acceleration of 4 m/s2 directed to the right, and ball B
has a constant acceleration of 2 m/s2 directed to the left. Ball A
is initially traveling to the left at 2 m/s, while ball B is
traveling to the left initially at 5 m/s. Find the time t at which
the balls collide. Also, assuming x = 0 at the initial position of
ball A, what is their common displacement when they collide?
Equations of displacement for A and B: s = so + vot + at2 (watch
signs) For A: sA = 0 + (-2 m/s)t + (+4 m/s2 ) t2 For B: sB = 18 m +
(-5 m/s)t + (-2 m/s2 ) t2 ; Next simplify and set sA = sB - 2t +
2t2 = 18 5t - t2 3t2 + 3t 18 = 0 t1 = - 3 s, t2 = +2 s Accept t =
+3 s as meaningful answer, then substitute to find either sA or sB:
sA = -2(2 s) + 2(2 s)2 ; x = + 4 m v = - 5 m/s2 v = - 2 m/s + aa =
+4 m/s2 x = 18 mx = 0 A B ab = -2 m/s2
63. Chapter 6 Uniform Acceleration Physics, 6th Edition 62
*6-58. Initially, a truck with a velocity of 40 ft/s is located
distance of 500 ft to the right of a car. If the car begins at rest
and accelerates at 10 ft/s2 , when will it overtake the truck? How
far is the point from the initial position of the car? Equations of
displacement for car and truck: s = so + vot + at2 (watch signs)
For car: sC = 0 + (+10 ft/s2 ) t2 ; Truck: sT = 500 ft + (40 ft/s)t
+ 0; Set sC = sT 5t2 = 500 + 40t or t2 8t 100 = 0; t1 = -6.77 s; t2
= +14.8 s Solve for either distance: sC = (10 ft/s2 )(14.8 s)2 ; s
= 1092 ft *6-59. A ball is dropped from rest at the top of a 100-m
tall building. At the same instant a second ball is thrown upward
from the base of the building with an initial velocity of 50 m/s.
When will the two balls collide and at what distance above the
street? For A: sA = 100 m + v0At + gt2 = 100 m + 0 + (-9.8 m/s2 )
t2 For B: sB = 0 + (50 m/s)t + (-9.8 m/s2 ) t2 Set sA = sB 100 4.9
t2 = 50 t 4.9 t2 ; 50 t = 100; t = 2.00 s Solve for s: sA = 100 m
(4.9 m/s2 )(2 s)2 ; s = 80.4 m *6-60. A balloonist rising
vertically with a velocity of 4 m/s releases a sandbag at the
instant when the balloon is 16 m above the ground. Compute the
position and velocity of the sandbag relative to the ground after
0.3 s and 2 s. How many seconds after its release will it strike
the ground? The initial velocity of the bag is that of the balloon:
voB = + 4 m/s From ground: s = soB + voBt + gt2 ; s = 18 m + (4
m/s)t + (-9.8 m/s2 )t2 s = 18 m + (4 m/s)(0.3 s) (4.9 m/s2 )(0.3
s)2 ; s = 16.8 m v = 0 s = 0 s = 500 ft v = 40 ft/s + A B s = 0 s =
100 m
64. Chapter 6 Uniform Acceleration Physics, 6th Edition 63
*6-61. An arrow is shot upward with a velocity of 40 m/s. Three
seconds later, another arrow is shot upward with a velocity of 60
m/s. At what time and position will they meet? Let t1 = t be time
for first arrow, then t2 = t - 3 for second arrow. s1 = (40 m/s)t1
+ (-9.8 m/s2 )t1 2 ; s1 = 40t 4.9t2 s2 = (60 m/s)t2 + (-9.8 m/s2
)t2 2 ; s2 = 60(t 3) - 4.9(t 3)2 s1 = s2; 40t 4.9t2 = 60t 180
4.9(t2 6t + 9) The solution for t gives: t = 4.54 s Now find
position: s1 = s2 = (40 m/s)(4.54 s) (4.9 m/s2 )(4.54 s)2 ; s =
80.6 m *6-62. Someone wishes to strike a target, whose horizontal
range is 12 km. What must be the velocity of an object projected at
an angle of 350 if it is to strike the target. What is the time of
flight? y = voyt + gt2 = 0; ( vo sin 350 )t = (4.9 m/s2 )t2 or t v
t v 0574 4 9 01170 0 . . ; . R = voxt = 12 km; (vo cos 350 )t =
12,000 m; t v 14 649 0 , Set t = t 0574 4 9 14 6490 0 . . ,v v ;
From which vo = 354 m/s and t = 41.4 s *6-63. A wild boar charges
directly toward a hunter with a constant speed of 60 ft/s. At the
instant the boar is 100 yd away, the hunter fires an arrow at 300
with the ground. What must be the velocity of the arrow if it is to
strike its target? y = 0 = (v0 sin 300 )t + (-32 ft/s2 )t2 ; Solve
for t t v v 05 2 32 0 031250 0 . ( ) . ; t = 0.03125 vo s1 =( v0
cos 300 ) t = (0.866 vo)(0.03125 vo); s1 = 0.0271 vo 2 s1 = s2 60
m/s 40 m/s v = -60 ft/s s1 = s2 s = 300 fts = 0 vo 300
65. Chapter 6 Uniform Acceleration Physics, 6th Edition 64
*6-63. (Cont.) s1 = 0.0271 vo 2 ; t = 0.03125 vo vB = - 60 ft/s;
soB = 300 ft s2 = soB + vBt = 300 ft + (-60 ft/s)t s2 = 300 60
(0.03125 vo) = 300 1.875 vo Now, set s1 = s2 and solve for vo
0.0271 vo 2 = 300 1.875 vo or vo 2 + 69.2 vo 11,070 = 0 The
quadratic solution gives: vo = 76.2 ft/s s1 = s2 s = 300 fts = 0 vo
300
66. Chapter 7 Newtons Second Law Physics, 6th Edition 65
Chapter 7. Newtons Second Law Newtons Second Law 7-1. A 4-kg mass
is acted on by a resultant force of (a) 4 N, (b) 8 N, and (c) 12 N.
What are the resulting accelerations? (a) a 4N 4 kg 1 m/s2 (b) a 8N
4 kg 2 m/s2 (c) a 12N 4 kg 3 m/s2 7-2. A constant force of 20 N
acts on a mass of (a) 2 kg, (b) 4 kg, and (c) 6 kg. What are the
resulting accelerations? (a) a 20N 2 kg 10 m/s2 (b) a 20N 4 kg 5
m/s2 (c) a 20N 6 kg 3.33 m/s2 7-3. A constant force of 60 lb acts
on each of three objects, producing accelerations of 4, 8, and 12
N. What are the masses? m 60 lb 4 ft / s2 15 slugs m 60 lb 8 ft /
s2 7.5 slugs m 60 lb 12 ft / s2 5 slugs 7-4. What resultant force
is necessary to give a 4-kg hammer an acceleration of 6 m/s2 ? F =
ma = (4 kg)(6 m/s2 ); F = 24 N 7-5. It is determined that a
resultant force of 60 N will give a wagon an acceleration of 10
m/s2 . What force is required to give the wagon an acceleration of
only 2 m/s2 ? m 60 6 N 10 m / s slugs2 ; F = ma = (6 slugs)(2 m/s2
); F = 12 N 7-6. A 1000-kg car moving north at 100 km/h brakes to a
stop in 50 m. What are the magnitude and direction of the force?
Convert to SI units: 100 km/h = 27.8 m/s 2 2 0 27 8 2 2 2 2 2 2 as
v v a v v s af o f o ; ( ) ( . (50 m) ; m / s) 7.72 m / s 2 2 F =
ma = (1000 kg)(7.72 m/s2 ); F = 772 N, South.
67. Chapter 7 Newtons Second Law Physics, 6th Edition 66 The
Relationship Between Weight and Mass 7-7. What is the weight of a
4.8 kg mailbox? What is the mass of a 40-N tank? W = (4.8 kg)(9.8
m/s2 ) = 47.0 N ; m 40 N 9.8 m / s2 = 4.08 kg 7-8. What is the mass
of a 60-lb child? What is the weight of a 7-slug man? m 60 lb 32 ft
/ s2 = 1.88 slugs ; W = (7 slugs)(32 ft/s2 ) = 224 lb 7-9. A woman
weighs 180 lb on earth. When she walks on the moon, she weighs only
30 lb. What is the acceleration due to gravity on the moon and what
is her mass on the moon? On the Earth? Her mass is the same on the
moon as it is on the earth, so we first find the constant mass: me
180 5625 lb 32 ft / s slugs;2 . mm = me = 5.62 slugs ; Wm = mmgm gm
30 lb 5.625 slugs ; gm = 5.33 ft/s2 7-10. What is the weight of a
70-kg astronaut on the surface of the earth. Compare the resultant
force required to give him or her an acceleration of 4 m/s2 on the
earth with the resultant force required to give the same
acceleration in space where gravity is negligible? On earth: W =
(70 kg)(9.8 m/s2 ) = 686 N ; FR = (70 kg)(4 m/s2 ) = 280 N
Anywhere: FR = 280 N The mass doesnt change. 7-11. Find the mass
and the weight of a body if a resultant force of 16 N will give it
an acceleration of 5 m/s2 . m 16 50 N m / s2 . = 3.20 kg ; W =
(3.20 kg)(9.8 m/s2 ) = 31.4 N
68. Chapter 7 Newtons Second Law Physics, 6th Edition 67 7-12.
Find the mass and weight of a body if a resultant force of 200 lb
causes its speed to increase from 20 ft/s to 60 ft/s in a time of 5
s. a m 60 ft / s - 20 ft / s 5 s ft / s lb 8 ft / s 2 2 8 200 ; =
25.0 slugs W = mg = (25.0 slugs)(32 ft/s2 ); W = 800 lb 7-13. Find
the mass and weight of a body if a resultant force of 400 N causes
it to decrease its velocity by 4 m/s in 3 s. a v t a 4 3 133 m / s
s m / s2 ; . ; m 400 133 N m / s2 . ; m = 300 kg W = mg = (300
kg)(9.8 m/s2 ); W = 2940 N Applications for Single-Body Problems:
7-14. What horizontal pull is required to drag a 6-kg sled with an
acceleration of 4 m/s2 if a friction force of 20 N opposes the
motion? P 20 N = (6 kg)(4 m/s2 ); P = 44.0 N 7-15. A 2500-lb
automobile is speeding at 55 mi/h. What resultant force is required
to stop the car in 200 ft on a level road. What must be the
coefficient of kinetic friction? We first find the mass and then
the acceleration. (55 mi/h = 80.7 m/s) m as v vf 2500 2 2 0 2lb 32
ft / s 78.1 slugs; Now recall that:2 ft / s) 2(200 ft) and - 16.3 m
/ s 2 2 a v v s a f 2 0 2 2 0 7( ) (80. ; F = ma = (78.1
slugs)(-16.3 ft/s2 ); F = -1270 lb k 1270 lb ; ; 2500 lb k kF N k =
0.508 6 kg 20 N P
69. Chapter 7 Newtons Second Law Physics, 6th Edition 68 7-16.
A 10-kg mass is lifted upward by a light cable. What is the tension
in the cable if the acceleration is (a) zero, (b) 6 m/s2 upward,
and (c) 6 m/s2 downward? Note that up is positive and that W = (10
kg)(9.8 m/s2 ) = 98 N. (a) T 98 N = (10 kg)(0 m/s and T = 98 N (b)
T 98 N = (10 kg)(6 m/s and T = 60 N + 98 N or T = 158 N (c) T 98 N
= (10 kg)(-6 m/s and T = - 60 N + 98 N or T = 38.0 N 7-17. A 64-lb
load hangs at the end of a rope. Find the acceleration of the load
if the tension in the cable is (a) 64 lb, (b) 40 lb, and (c) 96 lb.
(a) 2 64 lb ; 64 lb 64 lb = 32 ft/s W T W a a g ; a = 0 (b) 2 64 lb
; 40 lb 64 lb = 32 ft/s W T W a a g ; a = -12.0 ft/s2 (b) 2 64 lb ;
96 lb 64 lb = 32 ft/s W T W a a g ; a = 16.0 ft/s2 7-18. An 800-kg
elevator is lifted vertically by a strong rope. Find the
acceleration of the elevator if the rope tension is (a) 9000 N, (b)
7840 N, and (c) 2000 N. Newtons law for the problem is: T mg = ma
(up is positive) (a) 9000 N (800 kg)(9.8 m/s2 ) = (800 kg)a ; a =
1.45 m/s2 (a) 7840 N (800 kg)(9.8 m/s2 ) = (800 kg)a ; a = 0 (a)
2000