P-1: Ԧ𝐴=3Ԧ𝑖 +2Ԧ𝑗 -5𝑘 and 𝐵 =2Ԧ𝑖 - 7Ԧ𝑗 -6𝑘. Determine;

?

?

?

?

?

?

-

AB

BA

AB

BA

BA

BA

7

P-2: The screw eye is subjected to two forces, Ԧ𝐹1 and Ԧ𝐹2 .

Determine the magnitude and direction of the resultant force.

2

P-3: The vector 𝑉 lies in the plane defined by the intersecting lines LA and LB. Its magnitude is 400

units. Suppose that you want to resolve 𝑉 into vector components parallel to LA and LB. Determine the

magnitudes of the vector components. Also determine the projections Pa and Pb of 𝑉 onto the lines LA

and LB.

80°

60°

V

LA

LB

3

P-4: Express forces as the vector form in terms of the unit vectors Ԧ𝑖 and Ԧ𝑗 .

(a) (b) (c)

Fx

Fy

jiF

jiF

jFiFF yx

39.32102.383

40sin50040cos500

-

-

Fx

Fy

Fx

Fy

40o

30o

jiF

jiF

jFiFF yx

2004.346

30sin40030cos400

-

-

jiF

F

F

jFiFF

y

x

yx

28.4

512

52.5

512

122.5

13

22

13

22

--

-

-

P-5: Determine the magnitude of the resultant 𝑉=𝑉1+𝑉2 and the angle q which 𝑉 makes with the

positive x–axis.

V1 = 9 units

V2 = 7 units

30°

y

x

4

3

5

P-6:

P-7: Determine the x-y components of the tension T which is applied to point A of the bar

OA. Neglect the effects of the small pulley at B. Assume that r and q are known. Also

determine the n-t components of the tension T for T=100 N and q=35o.

q

b

b

r + rcos q

r -

rsin

q

qqqq sin2cos23cossin22

-- rrrrrAB

qq

qb

qq

qb

sincos

sinTsinTT

sincos

cosTcosTT

y

x

223

1

223

1

-

--

-

x-y coordinates

n-t coordinates (for q=35o and T=100 N)

N..sinsinTT

N..coscosTT

.cos

sinarctan

t

n

o

5474191335100

6766191335100

1913351

351

-

bq

bq

b

q

b

T

T

qq

q

qq

qb

qq

q

qq

qb

sin2cos23

sin1

sin2cos23

sinsin

sin2cos23

cos1

sin2cos23

coscos

-

-

-

-

-

-

r

rr

r

rr

P-8: In the design of the robot to insert the small cylindrical part into a close-fitting circular hole, the robot

arm must exert a 90 N force P on the part parallel to the axis of the hole as shown. Determine the

components of the force which the part exerts on the robot along axes (a) parallel and perpendicular to the

arm AB, and (b) parallel and perpendicular to the arm BC.

(a) parallel and perpendicular to the arm AB(b) parallel and perpendicular to the arm BC.

force which the part exerts on the robot

vertical

P=90 N

vertical

horizontal

//ABP=90 N

30o

AB

45o P//AB

P AB

P//AB = PAB=90 cos45=63.64 N

vertical

horizontal

//BC

P=90 N

45o

BC

30oP//BC P BC

P//BC =90 cos30=77.94 N

P BC=90 sin30=45 N

45o

60o

15o

45o

15o

the robot arm must exert a 90 N force P on the part

P-9: The unstretched length of the spring is r.

When pin P is in an arbitrary position q,

determine the x- and y-components of the force

which the spring exerts on the pin. Evaluate your

answer for r=400 mm, k=1.4 kN/m and q =40°.

P-10: Three forces act on the bracket. Determine the magnitude and direction q of F2 so that

the resultant force is directed along the positive u axis and has a magnitude of 50 N.

F3=52 N

F1=80 N

if R=50 N q =? F2 =?

F3=52 N

F1=80 N

R

jiRjRiRR

25sin5025cos5025sin25cos --

Resultant

jiR

13.21315.45 -

iF

801

jFiFF

25sin25cos 222 - qq

jijiF

482013

1252

13

5523

jFiFRF yx

685.5425cos

315.452025cos80

2

2

-

q

q

F

FFx

13.6925sin

13.214825sin

2

2

--q

q

F

FFy

1

2

2

1

685.54

13.69

25cos

25sin

2

2

-

q

q

F

F

NF 14.8835.103

264.125tan

2

-

q

q

P-11: Determine the magnitude and direction

angles of the resultant force acting on the

bracket.

kjiF

xyxyFF

45sin45030cos45cos45030sin45cos450

11

1 -

21 FFFR

kjiF

2.31857.2751.1591 -

Resultant

kjiF

120cos60060cos60045cos6002

kjiF

30030026.4242 -

Direction angles for 𝑭 𝟐 906045 zyx qqq

1coscoscos 222 zyx qqq

1222 nml

Direction cosines

1cos60cos45cos 222 zq

5.0cos25.0cos2 zz qq

1205.0cos90 - zzz qqq

kjiF

30030026.4242 -

Direction angles for 𝑹

029.097.633

2.18cos907.0

97.633

57.575cos418.0

97.633

16.265cos

coscoscos

zyx

zz

y

yx

xR

R

R

R

R

R

qqq

qqq

Direction Cosines of Resultant Force

3.88)029.0arccos(9.24)907.0arccos(3.65)418.0arccos( zyx qqq

kjiF

2.31857.2751.1591 -

Resultant 21 FFFR

kjiR

3002.31830057.27526.4241.159 --

kjiR

2.1857.57516.265

Magnitude of Resultant Force

NRR 97.6332.1857.57516.265 222

P-12: The turnbuckle T is tightened until the tension in cable OA is 5 kN. Express the force

Ԧ𝐹 acting on point O as a vector. Determine the projection of Ԧ𝐹 onto the y-axis and onto line

OB. Note that OB and OC lies in the x-y plane.

𝑭 acting on point O

𝑭𝑭𝒛

𝑭𝒙𝒚

𝑷𝒓𝒐𝒋𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒏𝒕𝒐 𝒍𝒊𝒏𝒆 𝑶𝑩

𝒙

𝒚𝑶

𝑩30o

𝒏𝑶𝑩𝑼𝒏𝒊𝒕 𝒗𝒆𝒄𝒕𝒐𝒓 𝒐𝒇 𝒍𝒊𝒏𝒆 𝑶𝑩

jinOB

60cos30cos

kNF

F

jikjinFF

OB

OB

OBOB

63.2

5.091.2866.0358.1

5.0866.083.391.2358.1

F𝒛=5sin50=3.83 kN

F𝒙𝒚=5cos50=3.214 kN

𝒙𝒚 𝒑𝒍𝒂𝒏𝒆

𝒙

𝒚𝑶

𝑪

65o

𝑭𝒙𝒚

F𝑦= F𝑥𝑦sin65=3.214sin65=2.91 kN

F𝑥= F𝑥𝑦cos65=3.214cos65=1.358 kN

kjiF

83.391.2358.1

𝑷𝒓𝒐𝒋𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒏𝒕𝒐 𝒚 − 𝒂𝒙𝒊𝒔

kNjFFy 91.2

19

P-13: The frame shown is subjected to a

horizontal force Ԧ𝐹 = 300Ԧ𝑗. Determine themagnitude of the components of this forceparallel and perpendicular to member AB. Ԧ𝐹 = 300Ԧ𝑗

𝑻𝒉𝒆 𝒄𝒐𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔 𝒐𝒇 𝒑𝒐𝒊𝒏𝒕𝒔 𝑩 𝒂𝒏𝒅 𝑪 𝒂𝒓𝒆 𝐵 (1.6; −0.8𝑠𝑖𝑛30; 0.8𝑐𝑜𝑠30) 𝐵 (1.6;−0.4; 0.693) , 𝐶 (0; 0.7; 1.2)

𝑻𝒉𝒆 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 𝒗𝒆𝒄𝒕𝒐𝒓 𝑩𝑪 𝒊𝒔 mBCrkjir BCBC 2507.01.16.1507.01.16.1 222

// -

𝑻𝒉𝒆 𝒖𝒏𝒊𝒕 𝒗𝒆𝒄𝒕𝒐𝒓 𝒐𝒇 𝑻 𝒓𝑩𝑪 𝒊𝒔 kjikji

nnn TBCBC

253.0548.0797.0

2

507.01.16.1/ -

-

kjikjinTT BC

75.18941175.597253.0548.0797.0750 --

Tension 𝑻 acting on point B in vector form

𝑻𝑻 acting on point B BC

BCBC

r

rTnTT

/

/

P-14: The cable BC carries a tension of 750 N. Write this

tension as a force 𝑇 acting on point B in terms of the unit

vectors Ԧ𝑖, Ԧ𝑗 and 𝑘. The elbow at A forms a right angle.

P-15: The spring of constant k = 2.6 kN/m is attached to the disk at point A and to the end fitting at point B

as shown. The spring is unstretched when qA and qB are both zero. If the disk is rotated 15° clockwise and

the end fitting is rotated 30° counterclockwise, determine a vector expression for the force which the spring

exerts at point A.

P-16: The rectangular plate is supported by hinges along its side BC and by the cable AE. If

the cable tension is 300 N, determine the projection onto line BC of the force exerted on the

plate by the cable. Note that E is the midpoint of the horizontal upper edge of the structural

support.

If T=300 N, determine the projection onto line BC of the force exerted on the plate by the cable.

x

y

z

𝑪𝒐𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔 𝒐𝒇 𝒑𝒐𝒊𝒏𝒕𝒔 𝑨,𝑩, 𝑪 𝒂𝒏𝒅 𝑬 𝒘𝒊𝒕𝒉 𝒓𝒆𝒔𝒑𝒆𝒄𝒕 𝒕𝒐 𝒄𝒐𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆 𝒔𝒚𝒔𝒕𝒆𝒎

𝑨 400, 0, 0 B (0, 0, 0)

𝑪 0, 1200𝑠𝑖𝑛25,−1200𝑐𝑜𝑠25C (0, 507.14, -1087.57)

𝑬 0, 1200𝑠𝑖𝑛25,−600𝑐𝑜𝑠25E (0, 507.14, -543.78)

kjiT

kjinTT T

21.19319.18012.142

33.844

78.54314.507400300

--

--

kjkjinTT BCBC

906.0423.021.19319.18012.142 ---

Unit vector of 𝒍𝒊𝒏𝒆 𝑩𝑪

z

y

25o

25o

BCn

kjjknBC

906.0423.025sin25cos --

Projection of T onto line BC

NTBC 26.251

Consider the straight lines OA and OB.

a) Determine the components of a unit vector that is

perpendicular to both OA and OB.

b) What is the minimum distance from point A to the line OB?

A (10, -2, 3) m

y

xq

z

B (6, 6, -3) m

O

P-17:

24

a) Determine the components of a unit vector that is perpendicular to both OA and OB.

A (10, -2, 3) m

q

B (6, 6, -3) m

O

b) What is the minimum distance from point A to the line OB?

OBr /

OAr /

kjirkjir OAOB

3210366 // --

md

drrrr

rd

OBOA

r

OBOA

OA

71.9

36636.87sin

sin

9

222

////

/

q

q

Vector normal to both OA and OB is

kjir

ijikjkr

kjikjirrr OBOA

724812

18186123060

3663210//

-

-

--

Unit vector normal to both OA and OB is

kjinkji

r

rn

824.0549.0137.0724812

724812

36.87

222-

-

25

Consider the triangle ABC.

a) What is the surface area of ABC?

b) Determine the unit vector of the outer normal of surface ABC.

c) What is the angle between AC and AB?

A (12, 0, 0) m

y

x

z

B (0, 16, 0) m

C (0, 0, 5) m

P-18:

26

a)What is the surface area of ABC?

A (12, 0, 0) m

y

x

z

B (0, 16, 0) m

C (0, 0, 5) mB

A

C

c) What is the angle between AC and AB?

kiACjiAB

5121612 --

AC

AB

ikjACAB

kijiACAB

8019260

5121612

--

2222

24.1082

1926080mABCs

b) Determine the unit vector of the outer normal of surface ABC.

kjikji

ACAB

ACABn

887.0277.0369.0

48.216

1926080

368.56cos1320144

cos51216125121612

cos

2222

--

kiji

ACABACAB

27

P-19: The door is held opened by two chains. Forces in AB and CD are FA=300 N and FC=250 N, respectively.

a) Write these forces in vector form.

b) Determine the resultant of these forces and the direction angles of the resultant force.

c) Determine the projection of force FA on line CD.

28

P-20: Consider force Ԧ𝐹 = 12Ԧ𝑖 + 24Ԧ𝑗 + 24𝑘 𝑁. Resolve Ԧ𝐹

into two components, one parallel to line AB and one

perpendicular to line AB.

A(10, -2, 14) Position vector

F

kjir

kjir

AB

AB

884

146261014

/

/

-

----

B(14, 6, 6)

Unit vector

n

kjikji

r

rn

AB

AB

3

2

3

2

3

1

884

884

12

222/

/ -

-

Parallel component

kjikjinFF

NkjikjinFF

3

8

3

8

3

4

3

2

3

2

3

14

43

224

3

224

3

112

3

2

3

2

3

1242412

////

//

-

-

-

-

(in vector form)

Normal component

kjikjikjiFFF

67.2633.2167.103

8

3

8

3

4242412//

---

//F

F

29

P-21: Observers on Earth at points A and B measure direction cosines of position vectors to the

space shuttle at C as

For Ԧ𝑟𝐶/𝐴: cos qx = 0.360, cos qy = 0.480 and cos qz <90°.

For Ԧ𝑟𝐶/𝐵: cos qx = –0.515, cos qy = – 0.606 and cos qz <90°.

Determine the x, y, z coordinates of the space shuttle located at point C and also the shortest distance

from point C to direction AB.

A

B

C

x

y

z

(520, 640, 0) km 30

P-22: A house with 98000 kg mass is built on a steep slope defined by points A, B, and C.

To help assess the possibility of slope failure (mud slide), it is necessary to

a) Determine the components of the weight in directions normal and parallel (tangent) to

the slope.

b) Determine the component vectors of the weight in directions normal and parallel

(tangent) to the slope.

c) Determine the smallest distance from point O to the slope.

(g=9.81 m/s2)

31

a-b) Determine the components of the weight in directions normal and parallel (tangent) to the slope.

W

Wn

Wt

W: Weight of house

Wn: Normal component of the weight to slope

Wt : Parallel component of the weight to slope

kjiN

jikjACABN

21600720010800

18012060180

--

jiAC

kjAB

180120

60180

-

-

kjikji

N

Nn

7

6

7

2

7

3

21600720010800

21600720010800

25200

222

kkW

96138081.998000 --

AB

AC

B

A

C

NkjiknWWn 8240407

6

7

2

7

3961380 -

-

kjikjinWW nn

706320235440353160

7

6

7

2

7

3824040 ---

-

(Vector normal to the slope)

(Unit vector normal to the slope)

(its magnitude)

(in vector form)

32

c) Determine the smallest distance from point O to the slope.

W

Wn

Wt

kjiW

kjikWWW

t

nt

255060235440353160

706320235440353160961380

-

------

(its magnitude)

(in vector form)

NW

W

t

t

4.495186

255060235440353160 222

O

B

C

n

mkjiinOCd 43.517

6

7

2

7

3120

mkjijnOAd

mkjiknOBd

43.517

6

7

2

7

3180

43.517

6

7

2

7

360

CThe smallest distance can also be determined as follows

33

P-23: An overhead crane is used to reposition the boxcar within a railroad car-repair shop.

If the boxcar begins to move along the rails when the x-component of the cable tension

reaches 3 kN, calculate the necessary tension T in the cable. Determine the angle qxy

between the cable and the vertical x-y plane.

Tx=3 kN, calculate tension T, the angle qxy between the cable and the vertical x-y plane.

x-component of 𝑻

kjin

kjin

T

T

154.0617.077.0

145

45

222

kjiTT

154.0617.077.0

kNTT 896.3377.0 (magnitude of 𝑻)

kNTkNT zy 6.0896.3154.04.2896.3617.0

y and z-components of 𝑻

kjiT

6.04.23

Unit vector of 𝑻

𝑻 in vector form

kNTTT yxxy 84.34.23 2222

598.9896.089.3

84.3cos xy

xy

xyT

Tqq

P-24: In opening a door which is equipped with a heavy-duty return

mechanism, a person exerts a force P of magnitude 40 N as shown.

Force P and the normal n to the face of the door lie in a vertical plane.

Express P as a vector and determine the angles qx, qy and qz which the

line of action of P makes with the positive x-, y- and z-axes.

P𝒛=40sin30=20 N P𝒙𝒚=40cos30=34.64 N

P𝑥= P𝑥𝑦cos20=34.64cos20=32.55 N

kjiP

20848.1155.32

𝒙𝒚 𝒑𝒍𝒂𝒏𝒆

𝒙

𝒚

20o

𝒏

//𝒙

//𝒚

P𝒙𝒚

P𝑦= P𝑥𝑦sin20=34.64cos20=11.848 N

angles qx, qy and qz which the line of action of P makes with the positive x-, y- and z-axes

6040

20cos77.72

40

848.11cos536.35

40

55.32cos

aaa zyx qqq

P-25: The y and z scalar components of a force are 100 N and 200 N, respectively. If the

direction cosine l=cosqx of the line of action of the force is -0.5, write Ԧ𝐹 as a vector.

5.0200100 - lNFNF zy

75.05.011 22222222 - nmnmnml

yzzy FNFF 61.22322

FnFFFmFF zzyy qq coscos

NFF 2.25861.22375.0

NFF xx 1.129cos - q

kjiF

2001001.129 -