Chapter 14 Static Equilibrium

8/12/2019 Chapter 14 Static Equilibrium

1/24

CHAPTER 14 STATIC EQUILIBRIUM

ActivPhysicscan help with these Problems:

Activities 7.17.6

Section 14-1: Conditions for Equilibrium

Problem

1. Five forces act on a rod, as shown in Fig. 14-25.Write the torque equations that must be satisfiedfor the rod to be in static equilibrium taking thetorques (a) about the top of the rod and (b) aboutthe center of the rod.

figure14-25 Problem 1.

Solution

All of the forces lie in the plane of Fig. 14-25, so all of

the torques about any point on the rod are into or outof the page. Suppose the latter direction, out of thepage or counterclockwise, is positive. Moreover, all ofthe forces are perpendicular to the rod, so their leverarms about any point on the rod (recall that themagnitude of the torque is force times lever arm) caneasily be read-off from Fig. 14-25. (a) About the topof the rod, F4 and F5 contribute zero torque, andEquation 14-2 becomes 0 = 1

3F3 23 F2+ F1.

(b) About the center of the rod, the perpendiculardistances to F2 and F3 are

16

, and to F1, F4 and F5are 12 , so 0 =

12

(F1+ F4 F5) 16 (F2+ F3).

Problem

2. A body is subject to three forces: F1 = 2 + 2N,applied at the point x = 2 m, y= 0 m; F2 = 23N, applied atx = 1 m, y= 0; and F3 = 1N,applied atx = 7 m, y= 1 m. (a) Show explicitlythat the net force on the body is zero. (b) Showexplicitly that the net torque about the origin iszero. (c) To confirm the assertion followingEquation 14-4 that the net torque must be zeroabout any other point, evaluate the net torqueabout the point (3 m, 2 m), the point (7 m, 1 m),and about any other point of your choosing.

Solution

(a)

Fi= (2 + 2 2 3 + )N = 0. (b) (

i)0 =[2 (2 + 2) + () (2 3) + (7 + )]Nm = (4 + 3 7)kNm = 0. (c) For any pointr0 = (x0 + y0 )m,

(ri r0) F= {[(2 x0)

y0] (2 + 2) + [(1 x0 ) y0] (2 3) +[(7 x0 ) + (1 y0)] }Nm = [(4 + 3 7) +x0(2 + 3 1) + y0(2 2 + 0)]k Nm = 0.

Problem

3. Suppose the force F3 in the preceding problem isdoubled so the forces no longer balance and thebody is therefore accelerating. Show that (a) thetorque about the point (7 m, 1 m) is still zero, butthat (b) the torque about the origin is no longerzero. What is the torque about the origin?

Solution

(a) Sincer3 = (7+ ) m is the point of applicationofF3, the total torque about r3 is just due to F1 andF2: ( i)3 = (r1 r3) F1+ (r2 r3) F2=[(2 + 7 ) (2 + 2) + ( + 7 ) (23)]Nm = [(92)k (12)(k) + 6 (3) k +(12)(k)] Nm = 0. (b) (i)0 = (ri Fi) =[2 (2 + 2) + () (2 3) + (7 + ) (2)] Nm = [4k + 3k 14k] Nm = 7kNm.

Problem

4. A rod of mass mand length is falling freely in ahorizontal orientation, with no torque about itscenter of mass. Find the magnitude of the torqueabout either end. Why does your answer notviolate the point made in Problem 8?


2/24

208 CHAPTER 14

Solution

The magnitude of the torque on a body of mass Mina constant gravity field is|grav| = |rcm Mg| =rcmM g sin . With origin at either end, rcm=

1

2 , andsin = 1 for horizontal orientation, hence grav=12

Mg.With origin at the CM (rcm= 0) the torque iszero, but this does not contradict the result ofProblem 8 since

F= Mg = 0 in free fall.

Problem

5. In Fig. 14-26 the forces shown all have the samemagnitudeF. For each of the cases shown, is itpossible to place a third force so the three forcesmeet both conditions for static equilibrium? If so,specify the force and a suitable application point; ifnot, why not?


Solution

The conditions for static equilibrium, under the actionof three forces, can be written as: F3 = (F1+ F2)andr3 F3 = (r1 F1+ r2 F2). (a) In this case,F1 = F, r1 = (2 m), F2 = F, and r2 = (1 m).Thus,F3 = F( + ), which is a force of magnitude

2F , 45 down into the third quadrant (x= 225 or135 CCW from the xaxis). The point ofapplication,r3, can be found from the secondcondition,r3 F3 = (x3 + y3 ) (F F) =(x3+ y3)Fk= r1 F1 r2 F2= 0 (1 m)

F= (1 m)Fk. Thus,x3+ y3 = 1 m, or the line ofaction ofF3 passes through the point of application ofF2 (the point (0, 1 m)). Any point on this line is asuitable point of application forF3 (e.g. the point(0, 1 m)). (b) In this case, F1 = F2 so F3 = 0, butr1 F1+ r2 F2 = (r2 r1) F2= 0 so r3 F3= 0. Thus there is no single force that can beadded to produce static equilibrium.

Problem

6. Are there any other application points for the forceF3 in Problem 2 that will ensure that both staticequilibrium conditions are met?

Solution

Equation 14-1 does not involve the points of

application of the forces. Equation 14-2 can besatisfied for any point of application r3, for F3,providedr3 F3 = r3 F3, or (x3 + y3 ) =(7 + ) , which implies x3 = 7. Thus, any pointalong the line x = 7 is a possible point of applicationforF3, satisfying the equilibrium conditions.

Problem

7. Four forces act on a body, as shown in Fig. 14-27.Write the set of scalar equations that must hold forthe body to be in equilibrium, evaluating thetorques (a) about point Oand (b) about point P.

F3

F2F1

F4

P

O

1

2


Solution

All of the forces lie in the same plane, which includesthe points Oand P, so there are two independentcomponents of the force condition (Equation 14-1)and one component of the torque condition(Equation 14-2). Taking the xaxis to the right, the yaxis up and the zaxis out of the page in Figure 14-27,we have:

Fx= 0 = F1+ F2sin + F3,

Fy = 0 =

F2cos + F4, (

z)0 = 0 = 1F2 2F3sin +2F4cos , and (

z)P = 2F1sin + (2 1)

F2 = 0. (The lever arms of all the forces about eitherOor Pshould be evident from Fig. 14-27).


3/24

CHAPTER 14 209

Problem

8. In this problem you prove the statement in Sec-tion 14-1 that the choice of pivot point does not

matter when applying the conditions for staticequilibrium. Figure 14-28 shows an object on whichthe net force is assumed to be zero. Also, the nettorque about the point Oshown is zero. Youre toshow that the net torque about any other point Pisalso zero. To do so, write the net torque about PasP =

rPi Fi, where the vectorsrPare from P

to the force application points, and the index ilabels the different forces. Note in Fig. 14-28 thatrPi= rOi+ R, whereR is a vector from Pto O.Use this result in your expression for P and applythe distributive law to get two separate sums. Usethe assumptions that Fnet = 0 and O = 0 to argue

that both terms are zero. This completes the proof.

figure

14-28 Problem 8.

Solution

With reference to Fig. 14-28, we can write p=rPi Fi=

(rOi+ R)Fi =

rOi Fi+

RFi= O+ RFnet= O. IfFnet= 0, thetotal torque about any two points is the same.

Section 14-2: Center of Gravity

Problem

9. Figure 14-29ashows a thin, uniform square plate ofmass mand side . The plate is in a vertical plane.Find the magnitude of the gravitational torque onthe plate about each of the three points shown.

Solution

The center of gravity is at the center of a uniformplate. In calculating the gravitational torque, one mayconsider the entire weight as acting at the center ofgravity. (a) rA=

2/2 at 135 from the weight of the

plate, so A= (

2/2)mg sin 135 = 12

mg .(b) rB is

figure14-29(a) Problem 9 Solution.

colinear with the weight, so B = 0. (c) C=12

mg sin90 = 12

mg (but note that C= A).(We also assumed that Band Care at the centers oftheir respective sides. Alternatively, the torques canbe found from the lever arms shown.)

Problem

10. Figure 14-29bshows a thin, uniform plate ofmassmin the shape of an equilateral triangle of

side. The plate is in a vertical plane. Find themagnitude of the gravitational torque on the plateabout each of the three points shown.

Solution

The center of gravity is at the center of the triangle, aperpendicular distance of/2

3 from any side.

(a) The lever arm of the weight about point A is /2so A=

12

mg.(b) The lever arm about point Biszero, and B = 0. (c) The lever arm about point Cis/4 (Cis halfway up from the base) so C=

14

mg.(Of course, A and Care in opposite directions. Thetorque could also be found from Equation 12-12.)

Problem

11. Three identical books of lengthL are stacked overthe edge of a table as shown in Fig. 14-30. Thetop book overhangs the middle one by 1

2L, so it

just barely avoids falling. The middle bookoverhangs the bottom one by 1

4L. How much of

the bottom book can be allowed to overhang theedge of the table without the books falling?


4/24

210 CHAPTER 14

figure14-29(b) Problem 10 Solution.

Solution

In equilibrium, the farthest right the center of mass ofthe combination of three books can lie is directly abovethe edge of the table. (This is unstable equilibrium,since the slightest disturbance to the right would causethe books to fall.) The center of mass of each book isat its center, so if we take the origin at the edge withpositive to the right, this condition becomes

0 =xcm

= 1

3m

mx1+ m(x1+

1

4L) + m(x1+

1

4L +

1

2L)

,

wherex1 is the horizontal position of the center of thebottom book, and the centers of the other books aredisplaced as given. Therefore, 3x1+ L= 0, orx1 = 13 L. If the center of the bottom book is 13 L tothe left of the edge, then only 1

2L 1

3L= 1

6L can

overhang on the right. (An argument based on torquesis equivalent, since at the farthest right position, thenormal contact force on the books acts essentially justat the tables edge.)

figure14-30 Problem 11 Solution.

Problem

12. A 60-kg uniform tabletop 2.4 m long is supportedby a pivot 80 cm from the left end, and by a scale

at the right end (Fig. 14-31). How far from theleft end should a 40-kg child sit if the scale is toread zero?

Solution

See the solution to the next problem, whenFs= 0.

figure 14-31 Problem 12 Solution.

Section 14-3: Examples of Static Equilibrium

Problem

13. Where should the child in Fig. 14-31 sit if thescale is to read (a) 100 N and (b) 300 N?

Solution

If we consider torques about the pivot point (so thatthe force exerted by the pivot does not contribute)then Equation 14-2 is sufficient to determine the

position of the child. As shown on Fig. 14-31, theweight of the tabletop (acting at its center of gravity),the weight of the child (acting a distance x from theleft end), and the scale force, Fs, produce zero torqueabout the pivot:

(1/g) (

)P = 0 =

(Fs/g)(160 cm) (60 kg)(40 cm) + (40 kg)(80 cm x).Therefore,x = 20 cm + (Fs/9.8 N)4 cm. If (a) Fs =100 N, then x = 20 cm + (400/9.8)cm = 60.8 cm, andif (b) Fs= 300 N, x= 142 cm. (Note that the child ison opposite sides of the pivot in parts (a) and (b),since without the child, Fs = 147 N.)

Problem

14. A 4.2-m-long beam is supported by a cable at itscenter. A 65-kg steelworker stands at one end ofthe beam. Where should a 190-kg bucket ofconcrete be suspended if the beam is to be instatic equilibrium?

Solution

The sum of the torques on the beam (taken about itscenter, C, so that the cables tension and beams


5/24

CHAPTER 14 211

Problem 14 Solution.

weight do not enter the equation) is equal to zero.Therefore, (190 kg)g(x) = (65 kg)g(2.1 m), orx= 71.8 cm, on the opposite side ofCfrom theworker. (Note: Since sin = sin( ) will cancel fromthe torque equation, the beam need not be horizontalto be in equilibrium; the steelworkers mentalequilibrium is greatest when the beam is horizontal.)

Problem

15. Two pulleys are mounted on a horizontal axle, asshown in Fig. 14-32. The inner pulley has adiameter of 6.0 cm, the outer a diameter of 20 cm.Cords are wrapped around both pulleys so theydont slip. In the configuration shown, with whatforce must you pull on the outer rope in order tosupport the 40-kg mass?

Solution

Since each cord is tangent to its respective pulley, thelever arms are just the radii, as shown on the figure.The two torques are equal in magnitude,R1F1 =R2F2, so that

F1 = (40 kg)(9.8 m/s2

)(6cm) = (20 cm) = 118N.


Problem

16. A 23-m-long log of irregular cross section is lyinghorizontally, supported by a wall at one end and a

cable attached 4.0 m from the other end, as shownin Fig. 14-33. The log weighs 7.5103 N, and thetension in the cable is 6.2103 N. Where is thelogs center of gravity?

Solution

The log is in equilibrium under the torques exerted bythe cable, gravity, and the wall. Calculating thetorques about the point of contact with the wall(because the other two forces are given), we find(6.2 kN)(23 4) m = (7.5kN)xCG, orxCG= 15.7 m,from the end on the wall.


Problem

17. Figure 14-34 shows a traffic signal, with massesand positions of its various members indicated.

The structure is mounted with two bolts, locatedsymmetrically about the vertical memberscenterline, as indicated. What tension force mustthe left-hand bolt be capable of withstanding?


Solution

The forces on the traffic signal structure, and theirlever arms about point 0 (on the vertical members


6/24

212 CHAPTER 14

centerline between the bolts) are shown on Fig. 14-34.The normal forces exerted by the bolts and the groundon the vertical member are designated by N andNr,measured positive upward. (Of course, the ground canonly make a positive contribution, and the bolts only anegative contribution, to these normal forces.) Thetwo conditions of static equilibrium needed todetermineN andNr are:

Fy = 0 =N+ Nr

(9.8)(320 + 170 + 65) N (the vertical component ofEquation 14-1, positive up) and (

z)0 = 0 =

(Nr N)(0.38m) (1703.5 + 658)(9.8 Nm) (theout-of-the-page-component of Equation 14-2, positiveCCW). These can be written as Nr+ N= 5.44 kN,and Nr N= 28.8 kN. Thus N= 11.7 kN, whichis downward and must be exerted by the bolt. Thereaction force on the bolt is upward and is a tensile

force. (Really, N is the difference between thedownward force exerted by the bolt and the upwardforce exerted by the ground. Tightening the boltincreases the tensile force it must withstand beyondthe minimum value calculated above, under theassumption that the ground exerts no force.)

Problem

18. Figure 14-35 shows how a scale with a capacity ofonly 250 N can be used to weigh a heavier person.The board is 3.0 m long, has a mass of 3.4 kg, andis of uniform density. It is free to pivot about theend farthest from the scale. What is the weight of

a person standing 1.2 m from the pivot end, if thescale reads 210 N? Assume that the beam remainsnearly horizontal.

Solution

Since the board is at rest, the sum of the torques(positive CCW in Fig. 14-35) is zero about the pivot(due to the weight of the person, the weight of theboard acting at its CG, and the scale force, as shown).Thus, (

z)P= 0 = (210 N)(3 m) (3.49.8 N)

(1.5 m) W(1.2 m), or W= 483 N.


Problem

19. Figure 14-36ashows an outstretched arm with amass of 4.2 kg The arm is 56 cm long, and its

center of gravity is 21 cm from the shoulder. Thehand at the end of the arm holds a 6.0-kg mass.(a) What is the torque about the shoulder due tothe weights of the arm and the 6.0-kg mass? (b) Ifthe arm is held in equilibrium by the deltoidmuscle, whose force on the arm acts 5.0 belowthe horizontal at a point 18 cm from the shoulder

joint (Fig. 14-36b), what is the force exerted bythe muscle?

Solution

(a) The magnitude of the (external) torque on thearm is 0 = [(4.2 kg)(0.21 m) + (6 kg)(0.56 m)]

(9.8 m/s2)sin105 = 40.2 Nm. The direction isclockwise (into the page) about the shoulder joint.(b) The deltoid muscle exerts a counterclockwisetorque of magnitudeF r sin = F(0.18m)sin170,which, under equilibrium conditions, equals themagnitude of the torque in part (a). Thus, F=40.2 Nm/(0.18 m)sin 170 = 1.28 kN, underscoringthe comment at the end of Example 14-5. Theskeleto-muscular structure of the human extremitiesevolved for speed and range of motion, not mechanicaladvantage.


Problem

20. Figure 14-37 shows a portable infant seat that issupported by the edge of a table. The mass of theseat is 1.5 kg, and its center of mass is located16 cm from the table edge. A 12-kg baby is sittingin the seat with her center of mass over the seats


7/24

CHAPTER 14 213

center of mass. Find the forces FA and FB thatthe seat exerts on the table.

Solution

The infant seat is in equilibrium under the reactionforces to FA andFB, and the weight of the seat andinfant. Thus,

Fy = 0

=FBFA (1.5 kg + 12 kg)(9.8 m/s2)(positive up)

(

)A = 0

= (22 cm)FB (16 + 22) cm(13.5)(9.8) N(positive CCW)

Solving for FB from the torque equation andFAfrom the force equation, we find FB = 229N andFA= 96.2 N, in the directions shown in the figure.

940


Problem

21. A 15.0-kg door measures 2.00 m high by 75.0 cmwide. It hangs from hinges mounted 18.0 cm fromtop and bottom. Assuming that each hinge carrieshalf the doors weight, determine the horizontaland vertical forces that the door exerts on eachhinge.

Solution

If the door is properly hung, all the forcesonthe door

are coplanar. We assume that the CM is at thegeometrical center of the door. The conditions forequilibrium are:

Fx = 0 = FAx+ FBx,Fy = 0 = FAy+ FBy M g,

(

)B = 0 = rA FA+ rcm Mg= (164 cm) (FAx + FAy )

+(37.5 cm + 82 cm) (Mg),where we chose to calculate torques about the lowerhinge atB . Expanding the cross products, 0 =

164FAx(k) 37.5Mg(k), we find FAx= (37.5)(15 kg)(9.8 m/s

2)/(164) = 33.6 N.From the

xequation, FBx = FAx= 33.6 N, and by assumption,FAy =FBy =

1

2

Mg= 73.5 N. Of course, the forcesexertedbythe door on the hinges (by Newtons thirdlaw) are the reactions to the forces, FA and FB, justcalculated.


Problem

22. Figure 14-38 shows a popular system for mountingbookshelves. An aluminum bracket is mounted on

a vertical aluminum support by small tabs insertedinto vertical slots. If each bracket in a shelf systemsupports 32 kg of books, with the center of gravity12 cm out from the vertical support, what is thehorizontal component of the force exerted on theupper of the two bracket tabs? Assume contactbetween the bracket and support occurs only atthe upper tab and at the bottom of the bracket,4.5 cm below the upper tab.

Solution

The forces on the bracket are shown superposed onFig. 14-38, assuming only a normal force on the

bottom. Equilibrium implies

Fx = 0 = FAx+ FB ,


8/24

214 CHAPTER 14

and (

z)A= 0 = (4.5 cm)FB (12 cm)W.(Wechose to evaluate the zcomponent (out of the page) ofthe torques about the upper tab, at A, and note that

Fy = 0 is not needed.) The solution for FAx isimmediate: FAx= FB = (12/4.5)W= (2.67)(32 kg)(9.8 m/s

2) = 836 N. The negative

xcomponent means that a tensile force is exerted onthe upper tab, as expected. (Note that the upper tabmust also support the weight of the books.)


Problem

23. Figure 14-39 shows a house designed to have highcathedral ceilings. Following a heavy snow, thetotal mass supported by each diagonal roof rafteris 170 kg, including building materials as well assnow. Under these conditions, what is the force inthe horizontal tie beam near the roof peak? Is thisforce a compression or a tension? Neglect anyhorizontal component of force due to the verticalwalls below the roof. Ignore the widths of thevarious structural components, treating contactforces as though they were concentrated at theroof peak and at the outside edge of therafter/wall junction.

Solution

The forces on one of the diagonal rafters are drawn onFig. 14-39. If the rafters are symmetrical (withoutinternal stress), and we neglect the weight of the tiebeam, FB and FCwill be horizontal. Wand FA arevertical, the latter by assumption, and we supposeWacts at the center of the rafter. The equilibriumconditions needed to find FB are:

Fx= 0 =FB

FC, and (

z)A = 0, or 0 = (4 m)FC (3.2 m)FB(2.4 m)W.(Point A was chosen to eliminate FA fromthe equation.) The solution is FB = 2.4W/(4 3.2) =

5.00 kN. The force on the tie beam (the reaction toFB is a tension. (The function of the tie beam isprecisely to relieve any horizontal force that the roofmay exert on the walls.)


Problem

24. Repeat Example 14-4, now assuming that thecoefficient of friction at the floor is 1 and that atthe wall 2. Show that the minimum angle atwhich the board will not slip is now given by

= tan1

1 1221

.

figure14-11(b) Problem 24 Solution.

Solution

The addition of a frictional force on the board where itcontacts the wall is shown in the sketch (seeFig. 14-11b). We assume that f2 = 2N2 is themaximum frictional force and that it acts to opposethe fall of the board. (Iff2 is not proportional to N2,there is insufficient data to solve this Problem.) Theequations of static equilibrium in Example 14-4


9/24

CHAPTER 14 215

become: f1N2 = 0 (horizontal), N1 + 2N2 mg= 0(vertical), andN2L sin + 2N2L cos 12

mgL cos = 0 (torque about bottom of board). Theminimum angle can be found from the horizontalequation and the requirement that N2 = f11N1. N1 can be eliminated from the verticalequation,N2 1(mg 2N2) or (1 + 12)N2 1mg.N2 can be eliminated from the torqueequation,N2(tan + 2) =

12

mg, so this conditionbecomes 1

2mg(1 + 12)/(2+ tan ) 1mg, or

tan (1 12)/21.

Problem

25. A uniform sphere of radius R is supported by arope attached to a vertical wall, as shown inFig. 14-40. The point where the rope is attached

to the sphere is located so a continuation of therope would intersect a horizontal line through thespheres center a distance R/2 beyond the center,as shown. What is the smallest possible value forthe coefficient of friction between wall and sphere?

Solution

In equilibrium, the sum of the torques about thecenter of the sphere must be zero, so the frictionalforce is up, as shown. The lever arm of the tension inthe rope is 1

2R cos30 =

3R= 4, and the weight and

normal force exert no torque about the center. Thus,f R= T

3R/4.The sum of the horizontal components

of the forces is zero also, so 0 = N Tsin30, orT = 2N. Thereforef= 1

2

3 N. Since f sN, this

impliess f /N= 12

3 = 0.87.


Problem

26. Show that if the wall in the previous problem isfrictionless, then a continuation of the rope line

must pass through the center of the sphere.

Solution

From the preceding solution, it is clear that, in theabsence of friction, the condition that the sum of thetorques about the center of the sphere is zero impliesthat the lever arm of the tension is zero, or that therope must be in line with the center.

Problem

27. A garden cart loaded with firewood is beingpushed horizontally when it encounters a step 8.0cm high, as shown in Fig. 14-41. The mass of the

cart and its load is 55 kg, and the cart is balancedso that its center of mass is directly over the axle.The wheel diameter is 60 cm. What is theminimum horizontal force that will get the cart upthe step?

8.0 cm60 cm

F

figure 14-41 Problem 27.


Solution

We assume that a horizontal push on the cart resultsin a horizontal force exerted on the wheels by the axle,as shown. (We also suppose both wheels share theforces equally, so they can be treated together.) Alsoshown are the weight of the cart and the normal forceof the ground, both acting through the center of thewheels, and the force of the step, Fs. If we considerthe sum of the torques (positive CCW) about the step,


10/24

216 CHAPTER 14

the latter does not contribute, and the wheels (andcart) will remain stationary as long as (

)step=

M gR sin N R sin F R cos = 0. When N= 0,however, the wheels begin to lose contact with theground and go over the step. This occurs whenF =M g tan . From the geometry of the situation,R(1 cos ) = h, the height of the step, so =cos1(1 h/R) = cos1(1 8/30). Then F= (559.8 N) tan(cos1(11/15)) = 500 N is the minimumforce.

Problem

28. Figure 14-42 shows the foot and lower leg of aperson standing on the ball of one foot. Threeforces act on the foot to maintain this equilibrium:the tension forceT in the Achilles tendon, contact

force FCat the ankle joint, and the normal forceN of the ground that supports the personsweight. The persons mass is 70 kg, and the forceapplication points are as indicated in Fig. 14-42.Find the magnitude of (a) the tension in theAchilles tendon and (b) the contact force at theankle joint.


Solution

If we approximate the bones in the foot as a massless,planar, rigid body, the equilibrium conditions for thesituation depicted in Fig. 14-42 are:

Fx=

Tsin25

FC,x= 0,Fy =Tcos25 + N FC,y= 0,and ( )ankle joint= N(12 cm) T(7 cm)sin 115 =

0. Then with N= 70 9.8 N, one finds (a) T =N(12 cm)/(7 cm) sin 115 = 1.30 kN, and (b)FC,x=Tsin25 = 0.548 kN, FC,y=Tcos25 + N= 1.86 kN,

and FC=

F2C,x+ F2C,y = 1.94 kN.

Problem

29. The leaning Tower of Pisa (Fig. 14-43) currentlyleans at a 4.7 angle to the vertical. Treating thetower as a solid cylinder, what is the maximumangle at which it can lean before falling over?

Treat the tower as a uniform cylinder 7.0 m indiameter and 55 m high, and assume the groundsupports the towers weight but does not provideany torque.

Solution

The center of mass of the tower must be somewhereover its footprint on the ground, or it will topple. Thiswill be so, for a uniform cylindrical model of thetower, if the angle of tilt is less than the angle that a diagonal makes with the length, as indicated onFig. 14-43. Therefore = tan1 (7m/55m) =7.25.


Problem

30. A uniform 5.0-kg ladder is leaning against africtionless vertical wall, with which it makes a15 angle. The coefficient of friction betweenladder and ground is 0.26. Can a 65-kg personclimb to the top of the ladder without it slipping?If not, how high can the person climb? If so, howmassive a person would make the ladder slip?

Solution

It is shown in the solution to Problem 45, that thecondition for a person of mass mto climb up afraction of length of ladder without the ladder slippingis scot + (m/m)(scot 12 ), wheres is thecoefficient of friction with the floor, the angle withthe vertical frictionless wall, and m the mass of theladder. For the situation in this problem, scot =0.26 cot15 = 0.970 and 0.970 + (5/65)(0.470) =1.01. Therefore (since 1, by definition) a 65-kgperson can climb all the way to the top. However, theright-hand side of the condition is less than 1 for m >m(scot 12 )/(1scot ) = 5 kg(0.470)/(0.0297) =79.3 kg, so a person with mass greater than 79.3 kgcauses the ladder to slip before reaching the top.


11/24

CHAPTER 14 217

Problem

31. The boom in the crane of Fig. 14-44 is free topivot about point Pand is supported by the cable

that joins halfway along its 18-m total length. Thecable passes over a pulley and is anchored at theback of the crane. The boom has mass 1700 kg,distributed uniformly along its length, and themass hanging from the end of the boom is2200 kg. The boom makes a 50 angle with thehorizontal. What is the tension in the cable thatsupports the boom?


Solution

The forces on the boom are shown superposed on the

figure. By assumption, Tis horizontal and acts at theCM of the boom. To find T, we compute the torquesaboutP, (

)P = 0, obtaining:

T12

sin50 mbg 12 cos50 mg cos50 = 0,or

T = (2m + mb)g cot50 = (4400 + 1700)(9.8 N)cot50

= 50.2 kN.

Problem

32. A uniform board of length and weight Wissuspended between two vertical walls by ropes of

length/2 each. When a weight wis placed on theleft end of the board, it assumes the configurationshown in Fig. 14-45. Find the weight win termsof the board weight W.

Solution

The conditions for equilibrium (about the origindrawn on the figure) are:

Fx = 0 = T2 sin60 T1 sin 35,

Fy = 0 = T2 cos60 + T1cos35

w W,(

z)0 = 0 = T2 sin20.8

W12

sin99.2.

Therefore:

T2 = 0.5Wsin99.2/ sin20.8 = 1.39W,

T1 = T2sin 60/ sin35 = 2.10W,

and w= T1cos 35 + T2cos 60 W = 1.41W.

figure

14-45 Problem 32 Solution.

Problem

33. Figure 14-46 shows a 1250-kg car that has slippedover the edge of an embankment. A group ofpeople are trying to hold the car in place bypulling on a horizontal rope, as shown. The carsbottom is pivoted on the edge of the embankment,and its center of mass lies further back, as shown.If the car makes a 34 angle with the horizontal,what force must the group apply to hold it in

place?Solution

Three forces act on the car, as shown added toFig. 14-46. The unknown force, FP, exerted by theedge of the embankment, does not contribute toEquation 14-2 (positive torques CCW) if evaluatedabout point P, so the tension necessary to keep the carin equilibrium can be found directly. (

)P = 0 =

Mg(2.4 m 1.8 m)cos34 T(1.8 m)sin34, orT= (12509.8 N)/3tan34 = 6.05 kN.


Problem

34. A uniform board of length is dangling over africtionlessedge, secured by a horizontalrope, as


12/24

218 CHAPTER 14

shown in Fig. 14-47. Show that the angle it makeswith the horizontal must be = sin1

2d/,

where dis the distance from the edge to the centerof the board.

Solution

See the solution to the next Problem.

Problem

35. Figure 14-47 shows a uniform board dangling overa frictionlessedge, secured by a horizontalrope. Ifthe angle in Fig. 14-47 were 30, what fractionwould the distance dshown in the figure be of theboard length ?

figure14-47 Problems 34, 35.

Solution

The three forces acting on the board are in the sameconfiguration as those in Problem 33, so Equation 14-2about the edge gives M gd cos = T( 1

2 d)sin . If

the edge is frictionless, then FPis perpendicular tothe board, so Equation 14-1 requires T =FPsin andM g= FPcos . Substituting above, we findFPd cos

2 = FP(12

d)sin2 , or d(cos2 + sin2 ) =12

sin2 and = sin1

2d/.For = 30,d/=12sin

2 30 = 18 .

Section 14-4: Stability of Equilibria

Problem

36. A portion of a roller coaster track is described byh= 0.94x 0.010x2, where hand xare the heightand horizontal position in meters. (a) Find a pointwhere the roller coaster car could be in staticequilibrium on this track. (b) Is the equilibriumstable or unstable? (c) What is the height of thetrack at the equilibrium point?

Solution

The potential energy of the roller coaster car, in theequivalent one-dimensional problem, is U(x) = mgh =

mg(0.94x 0.01x2). (a) Equation 14-3, the conditionfor equilibrium, gives dU/dx= 0 = mg(0.94 0.02x),or x = 47 m (b) Since d2U/dx2 = 0.02mg < 0,Equation 14-5 implies this is an unstable equilibrium.(c)h(47) = 0.94(47) 0.01(47)2 = 22.1 m.

(a) (b)

figure 14-48 Problem 37.

Problem

37. A roly-poly toy clown is made from part of asphere topped by a cone. The sphere is truncatedat just the right point so that there is nodiscontinuity in angle as the surface changes fromsphere to cone (Fig. 14-48a). If the clown alwaysreturns to an upright position, what is the

maximum possible height for its center of mass?Would your answer change if the continuity-of-angle condition were not met, as in Fig. 14-48b?

Solution

Suppose the roly-poly rests on a flat horizontalsurface. Its spherical surface is always tangent to thehorizontal if the continuity-of-angle condition holds, asin sketch (a) (except when upside-down on the pointof the cone). Gravity will always exert a restoringtorque if the CM lies to the left of the vertical throughthe left-most point of contact, B, as shown. Since thisvertical passes through O, the center of the sphere, theCM should lie below O, as measured from thebottom point, A, on the axis.

If the continuity-of-angle condition is not met, thecone, in general, intersects the sphere in one of twocircles (through BB orC C, as in sketch (b)). IfC C

is the actual boundary, the reasoning in the firstparagraph still applies. However, ifBB is theboundary, the maximum distance of the CM frompoint A should be < AQ(considerably lowerthanO).


13/24

CHAPTER 14 219


Problem

38. A uniform rectangular block is twice as long as itis wide. Letting be the angle that the longdimension makes with the horizontal, determinethe angular positions of any static equilibria, andcomment on their stability.

Solution

Consideration of the block, tilted in a planeperpendicular to its thickness (the unspecifieddimension in this question) reveals that = 0 is astable equilibrium position, = 90 a metastable one,and + = 90 an unstable one ( is the angle adiagonal makes with the longer side, as shown). Sincetan = /2= 1

2, the unstable equilibrium is at

= 90 tan1( 12

) = 63.4.


Problem

39. The potential energy as a function of position fora certain particle is given by

U(x) = U0

x3x30

+ a x2

x20+ 4 x

x0

,

whereU0, x0, and aare constants. For what valuesofawill there be two static equilibria? Commenton the stability of these equilibria.

Solution

The equilibrium condition, dU/dx= 0 (Equa-tion 14-3), requires 3(x/x0)2 + 2a(x/x0) + 4 = 0. Thisquadratic has two real roots if the discriminant ispositive, i.e., a2 12> 0, or|a| >23. The roots are(x/x0) =

13

(

a

a2

12). The second derivative of

the potential energy, evaluated at these roots, isd2U

dx2

= U0

x20

6

x

x0

+ 2a

= 2

a2 12

U0x20

.

Thus, the plus root is a position of metastableequilibrium (Equation 14-4), while the minus rootrepresents unstable equilibrium, (Equation 14-5). Aplot of the potential energy, which is a cubic, willclarify these remarks. For|a| 23, U(x)has nowiggles, as shown. (Upasses through the origin, but

its position depends on the value of a, and is notshown.)

750


Problem40. A cubical block rests on an inclined board with

two sides parallel to the direction of the incline.The coefficient of static friction between block andboard is 0.95. If the inclination angle of the boardis increased, will the block first slide or first tipover?

Solution

Reference to the solution of Problem 57 shows thatthe cube will tip over when >tan1(w/h) = 45, but


14/24

220 CHAPTER 14

will slide when >tan1 0.95 = 43.5. It thus slidesbefore tipping.

Paired Problems

Problem

41. Figure 14-49 shows a 66-kg sign hung centeredfrom a uniform rod of mass 8.2 kg and length2.3 m. At one end the rod is attached to the wallby a pivot; at the other end its supported by acable that can withstand a maximum tension of800 N. What is the minimum height habove thepivot for anchoring the cable to the wall?

Solution

Suppose that the sign is centered on the rod, so thatits CM lies under the center of the rod. Then the totalweight may be considered to act through the center ofthe rod, as shown. In equilibrium, Equation 14-2calculated about the pivot (which does not contain theunknown pivot force) yields 0 = T sin Mg 1

2, or

T =M g/2sin . But, tan = h/,so T = 12

M g1 + 2/h2 (use the identity 1 + cot2 = csc2).

Therefore, the conditionT Tmax implies 1 + 2 h2 (2Tmax/Mg)2, or h /

(2Tmax/Mg)2 1 =

2.3 m/

(1600/74.29.8)2 1 = 1.17 m.


Problem

42. A crane in a marble quarry is mounted on therock walls of the quarry and is supporting a2500-kg slab of marble as shown in Fig. 14-50.The center of mass of the 830-kg boom is locatedone-third of the way from the pivot end of its15 m length, as shown. Find the tension in the

horizontal cable that supports the boom.

Solution

As in the previous problem, the equilibrium conditionfor torques about the pivot does not contain theunknown pivot force, and thus allows the tension to bedirectly determined without use of the force equations.Thus,T sin50 =M g cos50 + mg 1

3 cos50, or

T= (M+ 13

m)g cot50 = (2500 + 13 830)(9.8 N)

cot50 = 22.8 kN.


Problem

43. A 4.2-kg plant hangs from the bracket shown in

Fig. 14-51. The bracket has a mass of 0.85 kg, andits center of mass lies 9.0 cm from the wall. Asingle screw holds the bracket to the wall, asshown. Find the horizontal tension force in thescrew. Hint: Imagine that the bracket is slightlyloose and pivoting about its bottom end. Assumethe wall is frictionless.

Solution

We assume that the screw provides the total supportfor the bracket, exerting a force with horizontalcomponentFx (the reaction to which is a tensile force

on the screw) and vertical componentFy (the reactionto which is a shearing force on the screw equal to thetotal weight) as shown. A normal contact forceexerted by the wall could be distributed along thebracket (e.g., by tightening the screw), but if we onlywish to estimate the minimum Fx, we may consider allthe normal force to act at the lowest point of contactof the bracket, point O. Then Equation 14-2 about Ogives Fx(7.2 cm) = [(4.2 kg)(28 cm) + (0.85 kg)(9.0 cm)](9.8 m/s

2) orFx= 170 N.


15/24

CHAPTER 14 221


Problem

44. A 160-kg highway sign of uniform density is 2.3 m

wide and 1.4 m high. At one side it is secured to apole with a single bolt, mounted a distance dfromthe top of the sign. The only other place wherethe sign contacts the pole is at its bottom corner.If the bolt can sustain a horizontal tension of2100 N, what is the maximum permissible valuefor the distance d ?

Solution

The forces on the sign are in a similar configuration tothose in the preceding problem, as shown. (The weightof the sign is shown acting at its center and the force

at the bottom corner could have a verticalcomponent.) To apply the limit on the tensile forceFx, we need only consider Equation 14-2 about pointO: Fx(1.4 m d) = (1609.8 N)(1.15 m). ThenFx= (1.80 kNm)/(1.4 m d)< 2.1 kN implies thatd


16/24

222 CHAPTER 14

force equation to find N.) For a person at the top ofthe ladder, = 1 and the condition for no slippingbecomesm m(scot 12 )/(1 scot ). With thedata given for the ladder (note that cot = tan 66)m

(9.5 kg)(0.42 tan66 0.5)/(1 0.42 tan 66) =74.3 kg.


Problem

46. To what vertical height on the ladder in thepreceding problem could a 95-kg person reachbefore the ladder starts to slip?

Solution

For a 95-kg person, the condition that the ladder notslip gives 0.42 tan 66 + (9/95)(0.42 tan 66 0.5) = 98.8% as the maximum fraction up along thelength of the ladder. (See previous problem.) Themaximum height above the ground is just cos =(0.988)(5 m) sin 66 = 4.51 m.

Problem

47. A uniform, solid cube of mass mand side sis instable equilibrium when sitting on a level tabletop.How much energy is required to bring it to anunstable equilibrium where its resting on its

corner?

Solution

When balancing on a corner, the CM of auniform cube (i.e., its center) is a distance

(s/2)2 + (s/2)2 + (s/2)2 =

3s/2 above the cornerresting on the tabletop. When in stable equilibrium,the CM is s/2 above the tabletop. Thus, the potentialenergy difference is U=mg ycm= mgs(

3 1)/2.

Problem

48. An isosceles triangular block of mass mand heighthis in stable equilibrium, resting on its base on a

horizontal surface. How much energy does it taketo bring it to unstable equilibrium, resting on itsapex? Hint: Consult Example 10-3.

Solution

Example 10-3 shows that the CM of an isoscelestriangle is 1

3the height from the base, or 2

3 the height

from the apex. The difference in energy between theunstable and stable equilibrium mentioned is U=mg ycm= mg(

23

h 13

h) = 13

mgh.

Supplementary Problems

Problem

49. A uniform pole of mass Mis at rest on an inclineof angle, secured by a horizontal rope as shownin Fig. 14-52. What is the minimum coefficient offriction that will keep the pole from slipping?


Solution

The forces acting on the pole are the tension in therope, gravity (acting at the CM at its center) and thecontact force of the incline (perpendicular componentNand parallel component f) as shown. Considerationof Equation 14-2 about the CM shows that a frictionalforce fmust be acting up the plane if the rod is toremain in static equilibrium. (Since the weight of therod, mg, and the normal force, N, contribute no

torques about the CM, there must be a force tooppose the torque of the tension, T.) The equationsfor static equilibrium (parallel and perpendicularcomponents of Equation 14-1, and CCW-positivecomponent of Equation 14-2) are:

F||= 0 =

f+ Tcos mg sin ,F= 0 = N Tsin mg cos , and (

)cm= 0 =T

12

cos f12

.The solutions for the forces aref= 1

2mg sin , T =

12

mg tan , and N= 12

mg(2 cos + sin2 / cos ),subject to the condition that f N. Thereforesin (2 cos + sin2 / cos ) or tan (2 + tan2 ). (By use of the identities sin 2=


17/24

CHAPTER 14 223

2sin cos , cos2= cos2 sin2 , and sin2 =1 cos2 , this may be rewritten as sin2 (3 + cos 2).)

Problem50. For what angle does the situation of Problem 49

require the greatest coefficient of friction?

Solution

The minimum coefficient of friction found in theprevious problem, min() = tan /(2 + tan

2 ), is apositive function which is zero at = 0 and 90 (thelimits of its domain). Therefore it has a maximumwhendmin/d= 0, or 2 + tan

2 2tan2 = 0, or= tan1

2 = 54.7.

Problem51. One end of a board of negligible mass is attached

to a spring of spring constant k, while its otherend rests on a frictionless surface, as shown inFig. 14-53. If a mass mis placed on the middle ofthe board, by how much will the spring compress?

Solution

If the frictionless surface is horizontal, the three forcesacting on the board are vertical, as shown. Forequilibrium,N+ Fs= mg and (

)cm= 0. The

latter implies N=Fs, so Fs = 1

2mg= k x, and the

compression of the spring is x= mg/2k.


Problem

52. A uniform ladder of mass mis leaning against africtionless vertical wall with which it makes anangle. The coefficient of static friction at thefloor is. Find an expression for the maximummass for a person who is able to climb to the topof the ladder without its slipping. Use your result

to show that anyonecan climb to the top if tan , but that no onecan if < 1

2tan .

Solution

The condition for a person of mass mto be able toreach the top of the ladder, which was found in thesolution to Problem 45, can be written as m m(2s tan )/2(tan s). Since m is positive, thiscondition cannot be fulfilled ifs 12tan , i.e., noone can climb to the top without causing the ladder toslip, whereas ifs = tan , the limit is, so anyonecan climb to the top. (Note: the original expression, scot + (m/m)(scos 12 ), shows that = 1is allowed for any m, providedscot >1.)

Problem

53. Figure 14-54 shows a wheel on a slope withinclination angle = 20, where the coefficient offriction is adequate to prevent the wheel fromslipping; however, it might still roll. The wheel isa uniform disk of mass 1.5 kg, and it is weightedat one point on the rim with an additional 0.95-kgmassm. Find the angle shown in the figure suchthat the wheel will be in static equilibrium.

Solution

The wheel doesnt slide if

F|| = 0 and it doesnt rollif (

)center= 0. (|| means parallel to the incline,and center is the center and CM of the wheel. These

are the only equilibrium conditions needed for thesolution of this problem.) With reference to the forcesshown added to Fig. 14-52, these conditions aref= (M+ m)g sin andf R= mgR cos . Together,they imply f=mg cos = (M+ m)g sin or =cos1[(1 + M/m)sin ] = cos1[(1 + 1.5/0.95)sin20] = 28.1.


Problem

54. The wheel in Fig. 14-54 has mass Mand isweighted with an additional mass m. The slopeangle is . Show that static equilibrium is possibleonly ifm > Msin /(1 sin ).


18/24

224 CHAPTER 14

Solution

Static equilibrium is possible only if the CM of theweighted wheel lies to the left of the point of contact

with the incline (or they

axis in the sketch). Thesmallest additional mass would have to be placedhorizontally to the left of the center, as shown. Then(M+ m)xcm= M R sin m(R R sin )< 0, impliesm > Msin /(1 sin ).


Problem

55. A 2.0-m-long rod has a density described by =a +bx, where is the density in kilograms permeter of length, a = 1.0 kg/m, b= 1.0 kg/m

2, andxis the distance in meters from the left end of therod. The rod rests horizontally with its ends eachsupported by a scale. What do the two scalesread?


Solution

The rod is in static equilibrium under the threevertical forces shown in the sketch, so

Fy = 0

impliesFsr+ Fsr = M g, and (

)cm= 0 impliesFsxcm= Fsr( xcm). The solution for the left andright scale forces is Fs= M g Fsr = M g

(1 xcm/). Equation 10-5 gives

xcm=

0

xdx

0

dx

=

0

(ax + bx2) dx

0

(a + bx) dx

= (a 12

2 + b 13

3)/(a + b 12

2)

=(3a + 2b)/(6a + 3b).

For the values given, xcm/= 712

and note thatM=a + 1

2b2 = 4 kg. Thus, Fsr = M gxcm/=

22.9 N andFs= 16.3 N.

Problem

56. What horizontal force applied at its highest pointis necessary to keep a wheel of mass Mfrom rolling

down a slope inclined at angle to the horizontal?

Solution

Consider the conditions for static equilibrium of thewheel, under the action of the forces shown (Fapp isthe applied horizontal force, Fc is the contact forceof the incline, normal plus friction, and we assumedthat the CM is at the center). The torques about thepoint of contact sum to zero, or FappR(1 + cos ) =MgR sin . ThereforeFapp= M g sin /(1 + cos ) =Mg tan 1

2.


Problem

57. A rectangular block twice as high as it is wide isresting on a board. The coefficient of staticfriction between board and block is 0.63. If theboard is tilted as shown in Fig. 14-55, will theblock first tip over or first begin sliding?

Solution

We suppose that the block is oriented with two sidesparallel to the direction of the incline, and that its CMis at the center. The condition for sliding is that


19/24

CHAPTER 14 225

mg sin > fmaxs =sN=smg cos , or tan > s.Fors = 0.63, this condition is >tan

1 0.63 = 32.2.The condition for tipping over is that the CM lie tothe left of the lower corner of the block (see sketch).Thus > , where = tan1(w/h) is the diagonalangle of the block. Forh = 2w, = tan1 0.5 = 26.6.Therefore, this block tips over before sliding.


Problem

58. What condition on the coefficient of friction in thepreceding problem will cause the block to slidebefore it tips?

Solution

Ifs< tan = 0.5, the block in the previous problemwill slide before tipping.

Problem

59. A uniform solid cone of height hand basediameter 1

3h is placed on the board of Fig. 14-55.

The coefficient of static friction between the coneand incline is 0.63. As the slope of the board isincreased, will the cone first tip over or beginsliding? Hint: Begin with an integration to findthe center of mass.

Solution

The analysis for Problem 57 applies to the cone, where is the angle between the symmetry axis and a linefrom the CM to the edge of the base. The integrationto find the CM is fastest when the cone is oriented likethe aircraft wing in Example 10-3, for then, theequation of the sloping side is simple, as shown in thesketch. For mass elements, take thin disks parallel tothe base, sodm = y 2dx= (3M/h3)x2 dx, where= M/ 1

3R2h is the density (assumed constant) and

Mis the mass of the cone. Then xcm= M1 x dm=

(3/h3) h0 x3 dx= 34 h, or the CM is 14 h above the base.

Since tan = ( 16

h)/( 14

h) = 23

>0.63 = s, this conewill slide before tipping over.


Problem60. In Fig. 14-56 a uniform boom of mass 350 kg is

attached to a vertical wall by a pivot, and its farend is supported by a cable as shown. If the cablecan withstand a maximum tension of 4.0 kN, whatis the maximum value for the angle ?

Solution

The situation is like that in Problem 41, but withdifferent angles. In equilibrium, the net torque aboutthe pivot point is zero, so M g

2sin40 =

T sin(40 ). (The exterior angle of a triangleequals the sum of the two alternate interior angles.)The condition T < Tmax means that sin(40 ) (M g/2 Tmax)sin40 = 0.276,for the data given, so 40 sin1(0.276) = 24.0 is required.


Problem

61. An interstellar spacecraft from an advancedcivilization is hovering above Earth, as shown in


20/24

226 CHAPTER 14

Fig. 14-57. The ship consists of two pods of massmseparated by a rigid shaft of negligible massthat is one Earth radius (RE) long. Find (a) themagnitude and direction of the net gravitationalforce on the ship and (b) the net torque about thecenter of mass. (c) Show that the ships center ofgravity is displaced approximately 0.083RE fromits center of mass.

Solution

(a) The force on each pod (pod #1 over the Northpole) can be calculated from the law of universalgravitation, Equation 9-1 (vectors in the x-yEarth-centered frame shown on Fig. 14-57):

F1 = GMEm

(2RE)2(),F2 = GMEm

(

5RE)2

1

5 2

5

.

When we replaceGMEbygR2E,the net force becomes:

F1+ F2 = mg

5

5+

1

4+

2

5

5

= mg(0.894+ 4.29) 101.

This has magnitude (4.38101)mg and is directed11.8 clockwise from the negative yaxis. (b) Thepositions of the pods, relative to their CM midwaybetween them, are r1 = 12 RE and r2 = 12 RE.Therefore, the net torque about the CM is:

r1 F1+ r2 F2=1

2RE

mg4

+

12

RE mg

5

5( 2)

= mgRE

8k mgRE

5

5k

= (3.56 102)mgREk (out of page).(c) The CG is positioned between the pods such thatthe net gravitational torque about it is zero. If thepositions of the pods relative to the CG are1REand 2RE, respectively, where 1+ 2 = 1, then

O= (1RE) mg

4

+ (2RE) mg5

5( 2)

=mgRE

14 22

5

5

k.

Solving for 1 (or2), we find 1 = 1 2 = 1 5

51/8, or 1 = (1 + 5

5/8)1 = 0.417 (and2 =0.583).Thus, the CG is (0.5 1)RE= 0.0829REcloser to pod #1 than the CM.

0202614Art: 14 57.epsfigure 14-57 Problem 61 Solution.

PART 1 CUMULATIVE PROBLEMS


21/24

CHAPTER 14 227

chapters in earlier parts, or they present special

challenges.

Problem

1. A 170-g apple sits atop a 2.8-m-high post. A 45-garrow moving horizontally at 130 m/s passeshorizontally through the apple and strikes theground 36 m from the base of the post, as shown in

Fig. 1. Where does the apple hit the ground?Neglect the effect of air resistance on either objectas well as any friction between apple and post.

SolutionWe can assume that momentum is conserved duringthe inelastic collision (in a brief interval at t = 0)between the arrow (m1) and the apple (m2). The


22/24

228 CHAPTER 14

2.8 m

v= 130 m/s

figure 1 Cumulative Problem 1.

velocities of the arrow before and after are specified tobe horizontal (in thexdirection), therefore thevelocity of the apple (which was at rest before) is alsohorizontal after the collision. Thus, m1v1i,x =m1v1f,x+ m2v2f,x . Since both are moving horizontallyafter the collision, the arrow and the apple will eachfall to the ground through the same vertical distancey(equal to the height of the post), in the same timet=

2y/g.However, they strike the ground at

different horizontal positions, which (in the absence ofair resistance) are x1 = v1f,xt and x2 = v2f,xt, relativeto the base of the post. Since x1 = 36 m, y= 2.8 m,and v1i,x= 130 m/s are given, v1f,x and v2f,x can beeliminated and a solution for x2 obtained: x2 =(m1/m2)(v1i,x v1f,x)t= (m1/m2)(v1i,x

2y/g

x1) = (45/170)[(130 m/s)

2(2.8 m)/(9.8 m/s

2)

36 m] = 16.5 m. (Alternatively, since externalhorizontal forces are neglected, the center of mass ofthe arrow/apple system moves horizontally at constantspeed until it reaches ground level, vcm,x= constant =

m1v1i,x/(m1+ m2) (its value before the collision).Then at ground level, m2x2 = (m1 + m2)xcmm1x1 =(m1+ m2)vcm,xt m1x1 = m1(v1i,xt x1), as before.)Refer to relevant material in Chapters 4, 10, and 11 ifnecessary.

Problem

2. A fire departments tanker truck has a total mass of21103 kg, including 15103 kg of water. Itsbrakes fail at the top of a long 3 slope and itbegins to roll downward, starting from rest. In anattempt to stop the truck, firefighters direct a

stream of water parallel to the slope, as shown inFig. 2, beginning as soon as the truck starts to roll.The water leaves the 6.0-cm-diameter hose nozzleat 50 m/s. Will the truck stop before it runs out ofwater? If so, when? If not, what is the minimumspeed reached?

Solution

The fire truck operates like a rocket, with the streamof water acting as expelled fuel. The rocket equation(Equation 10-10a) can be modified, as follows, toinclude the additional change in momentum due to the

3

figure2 Cumulative Problem 2.

external force of gravity.In the texts derivation, we now set P=Fgravt

(instead of P = 0), so M(v+ v) + m(v vex) (M+ m)v= Mv vexm= Fgrav t, andEquation 10-10a becomes M(v/t) = vex(Mt) + Fgrav . This is a one-dimensional equation withpositive components in the direction of the thrust,which is up the slope in this problem. Thegravitational force is opposite to the thrust, and in thelimit m 0, is just the downslope component of thefire trucks weight, orM g sin . Therefore, theequation of motion of the fire truck (which replacesEquation 10-10b) is M(dv/dt) = vex(dM/dt) Mg sin .

This equation can be integrated (aftermultiplication bydt/M) to find the speed vf as afunction of time tf,

vf vi = vfvi

dv = vex MfMi

dM

M g sin

tfti

dt

=vexlnMiMf

g sin (tf ti)(this replaces Equation 10-11). For this fire truck,Mi = 21103 kg andvi= 0, at ti= 0, and Mf(21 15)103 kg = 6103 kg, which is the trucksmass without any water. Since vex is constant, so isthe mass rate of flow of water (see Equation 18-4b),

thus dM/dt= wvexR2 = (103 kg/m3)(50 m/s)(3 cm)2 = 45 kg/s, where R = 3 cm isthe radius of the nozzle, and w the density of water.Then the mass of the fire truck (at any time until thewater runs out) isMf =Mi (45 kg/s)tf. At thestart, the thrust, vex|dM/dt| = (50 m/s)(45 kg/s) = 7.07 kN, is less than the magnitude ofthe downslope component of the weight,Mig sin = (21103 kg)(9.8 m/s2)sin3 = 10.8 kN,so the truck begins to roll downslope. When the waterruns out, attf= (15103 kg)/(45 kg/s) = 106 s,vf =vex ln(21/6) g sin (106 s) = 8.22 m/s, which ispositive, upslope. Therefore, the firefighters succeed instopping the truck (only instantaneously, unless theyalso stop the water and block the wheels) when vf =0 = vex ln[Mi/(Mi wvexR2t)] (g sin )t, or, withthe above numbers, ln[1 (6.732103 s1)t] =(1.026102 s1)t. A numerical solution for t,


23/24

CHAPTER 14 229

obtained with three iterations of Newtons method orwidely available PC software, yields t = 88.8 s.

Problem

3. A block of massm1 is attached to the axle of auniform solid cylinder of mass m2 and radius R bymassless strings. The two accelerate down a slopethat makes an angle with the horizontal, as shownin Fig. 3. The cylinder rolls without slipping andthe block slides with coefficient of kinetic friction between block and slope. The strings are attachedto the cylinders axle with frictionless loops so thatthe cylinder can roll freely without any torque fromthe string. Find an expression for the accelerationof the pair, assuming that the string remains taut.

figure 3 Cumulative Problem 3.

Cumulative Problem 3 Solution.

Solution

One must consider the forces on the block and thecylinder, exerted by gravity, the inclined surface, andthe strings, as sketched (since the strings are assumedmassless and other forces are neglected). If the strings

remain taut, then the downslope acceleration, a||, ofthe block and the cylinders center of mass are thesame. If the string tension is parallel to the slope, thenormal force on the block is N1 = m1g cos , and theparallel component of its equation of motion isFnet,||= m1g sin + T m1g cos = m1a||. Whenrolling without slipping, the point of contact, P, of thecylinder with the slope, is instantaneously at rest, sothe acceleration of its center of mass is a||= R (thisfollows from v = R and = c = cm; seeSection 12-5). Since only the string tension andgravity exert torques about point P, the equation of

motion of the cylinder is P = (m2g sin T)R=IP= (

32

m2R2)(a||/R), or m2g sin T = 32 m2a||

(see the parallel-axis theorem in Chapter 12 forIP). IfTis eliminated by adding the two equations of motion,one finds a||= 2g[(m1+ m2)sin m1cos ](2m1+ 3m2).

Problem

4. A missile is launched from pointA in Fig. 4,heading for target C. The launch angle is 45.0, andthe launch speed is 1.90 km/s. An antimissiledefense system is located at point B, 310 kmdownrange from the launch site. It fires aninterceptor rocket at a 65.0 launch angle, with theintention of hitting the attacking missile when thelatter is 270 km downrange of its launch site.

(a) What should be the interceptors launch speed?(b) How long after the launch of the attackingmissile should the interceptor be launched? (c) Atwhat altitude does the interception take place?

270 km

310 kmA B

C

figure4 Cumulative Problem 4.

Solution

It is simplest to answer each part of this questionusing a different coordinate system, so that equationsin the text can be used without modification. We aregiven enough information to specify the location of thepoint of interception; it is 270 km from point A, themissile launch site, and 310 270 = 40 km from pointB, the interceptor rocket launch site.

(c) The trajectory of the missile (Equation 4-9with origin at point A in Fig. 4) gives the altitude ofthe point of interception whenx = 270 m and other

data for the missile launch are substituted:

y = (270 km) tan 45 (0.0098 km/s2

)(270 km)2

2(1.90 km/s)2 cos2 45

= 72.1 km

The time of flight of the missile to the point ofinterception is t = x/v0x= (270 km)(190 cos 45 km/s) = 201 s.

(a) With altitude from part (c) above and theother initial data for the interceptor rocket, itstrajectory equation (Equation 4-9 with origin at pointBand xaxis leftward in Fig. 4) can be solved for the


24/24

230 CHAPTER 14

launch speed, as in Example 4-4. (The point ofinterception has coordinatesx = 40 km andy= 72.1 km relative to point B.) Thus,

v0 =

(0.0098 km/s2)2(40 tan 65 72.1) km

1/2 40 kmcos65

= 1.79 km/s.

(b) The time of flight for the interceptor rocket ist= x/v0x= (40 km)/(1.79cos65

km/s) = 52.8 s.Therefore, it should be launched 201 52.8 = 148 safter the missile.

Problem

5. A solid ball of radiusR is set spinning with angularspeed about a horizontal axis. The ball is then

lowered vertically with negligible speed until it justtouches a horizontal surface and is released (Fig. 5).If the coefficient of kinetic friction between the balland the surface is , find (a) the linear speed of theball once it achieves pure rolling motion and (b) thedistance it travels before its motion is pure rolling.

v

figure

5 Cumulative Problem 5.

Solution

(a) While there is relative motion at the point ofcontact between the ball and the horizontal surface,the force of sliding friction (f=N=mg) slows the

balls rotation and accelerates its center of mass. Theequation for the latter is f=mg = macm, or acm=g (positive to the right in the sketch and Fig. 5).The equation for the former is =

f R=

mgR=

I = (2mR2/5), where = 5g/2Ris the angularacceleration about the horizontal axis through thecenter of the ball (positive clockwise in the sketch andFig. 5). The accelerations are constant, so thevelocities are given by Equations 2-17 and 12-9 asvcm= acmt= gt and = 0+ t= 0 5gt/2R,where the ball is released at t = 0 and the initialvelocities,v0 = 0 and 0, are given. The acceleratedmotion continues until the point of contact isinstantaneously at rest (no more sliding friction). Theball rolls without slipping thereafter, at a constantvelocity given by vcm= R. This occurs at a time t,

when (0 5gt/2R)R=gt, or t = 20R/7g. Thus,the final velocity isvcm= gt = 20R/7. (b) Thedistance traveled during this time is x= 1

2acmt

2 =12

(g)(20R/7g)2 = 220R

2/49g (or x= v2cm/2acmwith the same result).

Documents

Chapter 14 Static Equilibrium