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Problem 4.14 The moment exerted about point E bythe weight is 299 in-lb. What moment does the weightexert about point S?

S

30°

E 40°12 in.

13 in.

Solution: The key is the geometry

From trigonometry,

cos 40◦ =d2

13 in, cos 30◦ =

d1

12 in

Thus d1 = (12 in) cos 30◦

d1 = 10.39′′

and d2 = (13 in) cos 40◦

d2 = 9.96′′

We are given that

299 in-lb = d2W = 9.96 W

W = 30.0 lb

Now,

Ms = (d1 + d2)W

Ms = (20.35)(30.0)

Ms = 611 in-lb clockwise

S

E

30°

40°12 in.

13 in.

12 in

30°

40°

S

d1

d2E

13 inW

Problem 4.15 Three forces act on the square plate.Determine the sum of the moments of the forces (a) aboutA, (b) about B, (c) about C.

200 N

200 N

200 N3 m

3 m

A

C

B

Solution: Determine the perpendicular distance between the points and thelines of action. Determine sign, and calculate moment. (a) The distances frompoint A to the lines of action is zero, hence the moment about A is MA = 0.(b) The perpendicular distances of the lines of action from B are: 3 m for theforce through A, with a positive action, and for the force through C, DC =( 1

2

) √32 + 32 = 2.12 m with a negative action. The moment about B is

MB = (3)(200) − 2.12(200) = 175.74 N-m (c) The distance of the forcethrough A from C is 3 m, with a positive action, and the distance of the forcethrough B from C is 3 m, with a positive action. The moment about C isMC = 2(3)(200) = 1200 N-m.

200 N

200 N

200 N

3 m

3 m

A

C

FB

Problem 4.16 Determine the sum of moments of thethree forces about (a) point A, (b) point B, (c) point C.

100 lb 200 lb 100 lb

2 ft 2 ft 2 ft 2 ft

A B C

Solution:

(a) The sum of the moments about A:

∑MA = −(2)(100) + (4)(200) − (6)(100) = 0.

(b) The sum of the moments about B:

∑MB = +(2)(100) − (2)(100) = 0

(c) The sum of the moments about C:

∑MC = +(6)(100) − (4)(200) + (2)(100) = 0.

100 lb

200 lb

100 lb

2 ft 2 ft 2 ft 2 ft

A B C

Problem 4.17 Determine the sum of the moments ofthe five forces acting on the Howe truss about point A.

4 ft 4 ft 4 ft 4 ft 4 ft 4 ft

800 lb

8 ft

A

B

C

G

F

E

D

H I J K L

600 lb600 lb

400 lb400 lb

Solution: All of the moments about A are clockwise (negative).The equation for the sum of the moments about A in units of ft-lb isgiven by:

∑MA = −4(400) − 8(600) − 12(800) − 16(600) − 20(400)

or∑

MA = −33,600 ft-lb.

4 ft 4 ft 4 ft 4 ft 4 ft 4 ft

800 lb

8 ft

A

B

C

G

F

ED

H I J K L

600 lb600 lb

400 lb400 lb

Problem 4.22 The vector sum of the three forces iszero, and the sum of the moments of the three forcesabout A is zero.(a) What are FA and FB?(b) What is the sum of the moments of the three forces

about B?

A

FA FB

B

900 mm 400 mm

80 N

Solution: The forces are:

FA = |FA|(0i + 1j),FB = |FB |(0i + 1j),

and F = 80(0i − 1j).

The sum of the forces is:

∑F = FA + FB + F = 0,

from which

∑FY = (|FA| + |FB | − 80)j = 0.

The sum of the moments:

∑MA = −(0.9)(80) + (1.3)(|FB |) = 0.

(a) Solve these two equations to obtain: |FB | = 55.4 N, and |FA| =24.6 N (b) The moments about B:

∑MB = (80)(0.4) − (1.3)|FA| = 0

Problem 4.23 The weights (in ounces) of fish A, B,and C are 2.7, 8.1, and 2.1, respectively. The sum of themoments due to the weights of the fish about the pointwhere the mobile is attached to the ceiling is zero. Whatis the weight of fish D?

12 in 3 in

2 in6 in

2 in7 in

A

CD

B

Solution:∑MO = (12)(2.7) − 3(10.2 + D)

Solving D = 0.6 oz12 in 3 in

2 in6 in

2 in7 in

A

CD

B

12 3

2,7

0

(8.1+2.1+D) = (10.2+D)

Problem 4.28 Five forces act on a link in the gear-shifting mechanism of a lawn mower. The vector sumof the five forces on the bar is zero. The sum of theirmoments about the point where the forces Ax and Ay

act is zero.(a) Determine the forces Ax, Ay , and B.(b) Determine the sum of the moments of the forces

about the point where the force B acts.

650 mm

650 mm 350 mm

450 mm30 kN

45°

20°

B

Ax

Ay

25 kN

Solution: The strategy is to resolve the forces into x- andy-components, determine the perpendicular distances from B to theline of action, determine the sign of the action, and compute the mo-ments.

The angles are measured counterclockwise from the x-axis. Theforces are

F2 = 30(i cos 135◦ + j sin 135◦) = −21.21i + 21.21j

F1 = 25(i cos 20◦ + j sin 20◦) = 23.50i + 8.55j.

(a) The sum of the forces is

∑F = A + B + F1 + F2 = 0.

Substituting:

∑FX = (AX + BX + 23.5 − 21.2)i = 0,

and∑

FY = (AY + 21.2 + 8.55)j = 0.

Solve the second equation: AY = −29.76 kN. The distancesof the forces from A are: the triangle has equal base and altitude,hence the angle is 45◦, so that the line of action of F1 passesthrough A. The distance to the line of action of B is 0.65 m,with a positive action. The distance to the line of action of they-component of F2 is (0.650 + 0.350) = 1 m, and the actionis positive. The distance to the line of action of the x-componentof F2 is (0.650−0.450) = 0.200 m, and the action is positive.The moment about A is

∑MA = (8.55)(1) + (23.5)(0.2) + (BX)(0.65) = 0.

650 mm

650 mm 350 mm

450 mm

F2 = 30 kN

45°

20°

B

Ax

Ay

F1 = 25 kN

Solve: BX = −20.38 kN. Substitute into the force equation to obtainAX = 18.09 kN

(b) The distance from B to the line of action of the y-component of F1 is0.350 m, and the action is negative. The distance from B to the line ofaction of AX is 0.650 m and the action is negative. The distance from Bto the line of action of AY is 1 m and the action is positive. The distancefrom B to the line of action of the x-component of F2 is 0.450 m and theaction is negative. The sum of the moments about B:

∑MB = −(0.350)(21.21) − (0.650)(18.09)

+ (1)(29.76) − (0.450)(23.5) = 0

Problem 4.29 Five forces act on a model truss builtby a civil engineering student as part of a design project.The dimensions are b = 300 mm and h = 400 mm;F = 100 N. The sum of the moments of the forcesabout the point where Ax and Ay act is zero. If theweight of the truss is negligible, what is the force B?

b b b b b b

h

60°F

60°F

Ax

Ay B

Solution: The x- and y-components of the force F are

F = −|F|(i cos 60◦ + j sin 60◦) = −|F|(0.5i + 0.866j).

The distance from A to the x-component is h and the action is positive.The distances to the y-component are 3b and 5b. The distance to B is6b. The sum of the moments about A is

∑MA = 2|F|(0.5)(h) − 3b|F|(0.866) − 5b|F|(0.866) + 6bB = 0.

Substitute and solve: B = 1.6784|F|1.8 = 93.2 N

b b b b b b

h

60°F 60°F

Ax

Ay B

Problem 4.36 The cable from B to A (the sailboat’sforestay) exerts a 230-N force at B. The cable from B toC (the backstay) exerts a 660-N force at B. The bottomof the sailboat’s mast is located at x = 4 m, y = 0.What is the sum of the moments about the bottom of themast due to the forces exerted at B by the forestay andbackstay?

y

x

B (4,13) m

C(9,1) m

A(0,1.2) m

Solution: Triangle ABP

tan α =4

11.8, α = 18.73◦

Triangle BCQ

tan β =512

, β = 22.62◦

� +MO = (13)(230) sin α − (13)(660) sin β

� +MO = −2340 N-m

230 N 660 N

A (0,1.2)

O (4,0)

C (9,1)

B (4,13)

αβ

P

Q

y

x

B (4,13) m

C(9,1) mA

(0,1.2) m

α

α

β

13 m

O

230 sin β660 sin

Problem 4.60 The direction cosines of the force F arecos θx = 0.818, cos θy = 0.182, and cos θz = −0.545.The support of the beam at O will fail if the magnitude ofthe moment of F about O exceeds 100 kN-m. Determinethe magnitude of the largest force F that can safely beapplied to the beam.

z

y

O

x

F

3 mSolution: The strategy is to determine the perpendicular dis-tance from O to the action line of F, and to calculate the largestmagnitude of F from MO = D|F|. The position vector fromO to the point of application of F is rOF = 3i (m). Resolvethe position vector into components parallel and normal to F. Thecomponent parallel to F is rP = (rOF · eF )eF , where theunit vector eF parallel to F is eF = i cos θX + j cos θY +k cos θZ = 0.818i + 0.182j − 0.545k. The dot product isrOF · eF = 2.454. The parallel component is rP =2.007i + 0.4466j − 1.3374k. The component normal to Fis rN = rOF − rP = (3 − 2)i − 0.4466j + 1.3374k.The magnitude of the normal component is the perpendiculardistance: D =

√12 + 0.44662 + 1.3372 = 1.7283 m.

The maximum moment allowed is MO = 1.7283|F| = 100 kN-m,from which

|F| =100 kN-m

1.7283 m= 57.86 ∼= 58 kN

z

y

O

x

F

3 m

Problem 4.61 The force F exerted on the grip of theexercise machine points in the direction of the unit vectore = 2

3 i− 23 j+

13k and its magnitude is 120 N. Determine

the magnitude of the moment of F about the origin O.

x

y

z

O

F

150 mm

200 mm

250 mm

Solution: The vector from O to the point of application of theforce is

r = 0.25i + 0.2j − 0.15k m

and the force is F = |F|e

or

F = 80i − 80j + 40k N.

The moment of F about O is

MO = r × F =

∣∣∣∣∣i j k

0.25 0.2 −0.1580 −80 40

∣∣∣∣∣ N-m

or

MO = −4i − 22j − 36k N-m

and

|MO| =√

42 + 222 + 362 N-m

|MO| = 42.4 N-m

y

z

x

F

150 mm

200 mm

250 mm

O

Problem 4.62 The force F in Problem 4.61 points inthe direction of the unit vector e = 2

3 i − 23 j + 1

3k. Thesupport at O will safely support a moment of 560 N-mmagnitude.(a) Based on this criterion, what is the largest safe mag-

nitude of F?(b) If the force F may be exerted in any direction, what

is its largest safe magnitude?

Solution: See the figure of Problem 4.61.

The moment in Problem 4.61 can be written as

MO =

∣∣∣∣∣i j k

0.25 0.2 −0.1523F − 2

3F + 13F

∣∣∣∣∣ where F = |F|

MO = (−0.0333i − 0.1833j − 0.3k)F

And the magnitude of MO is

|MO| = (√

0.03332 + 0.18332 + 0.32)F

|MO| = 0.353 F

If we set |MO| = 560 N-m, we can solve for |Fmax|

560 = 0.353|Fmax||Fmax| = 1586 N

(b) If F can be in any direction, then the worst case is when r ⊥ F. Themoment in this case is |MO| = |r||Fworst|

|r| =√

0.252 + 0.22 + 0.152 = 0.3536 m

560 = (0.3536)|FWORST||Fworst| = 1584 N

Problem 4.63 An engineer estimates that under themost adverse expected weather conditions, the total forceon the highway sign will beF = ±1.4i−2.0j (kN). Whatmoment does this force exert about the base O?

TUCCON

y

O

F

z

x

8 m

8 m

Solution: The coordinates of the point of application of the forceare: (0, 8, 8). The position vector is rOF = 8j + 8k. The crossproduct is

rOF × F =

∣∣∣∣∣i j k0 8 8

±1.4 −2 0

∣∣∣∣∣ = 16i − (∓1.4)(8)j + (∓1.4)(8)k

MO = 16i ± 11.2j ∓ 11.2k (N-m)

Check: Use perpendicular distances to forces:

MX = 8(2) = 16,

MY = 8(±1.4) = ±11.2,

MZ = −8(±1.4) = ∓11.2 .

O

F

z

x

8 m

8 m

Problem 4.70 Consider the 70-m tower in Prob-lem 4.69. Suppose that the tension in cable AB is 4 kN,and you want to adjust the tensions in cables AC andAD so that the sum of the moments about the origin Odue to the forces exerted by the cables at point A is zero.Determine the tensions.

Solution: From Varignon’s theorem, the moment is zero only ifthe resultant of the forces normal to the vector rOA is zero. FromProblem 4.69 the unit vectors are:

eAD =rAD

|rAD| =−3585.73

i − 7085.73

j − 3585.73

= −0.4082i − 0.8165j − 0.4082k

eAC =rAC

|rAC | = −4090

i − 7090

j +4090

k

= −0.4444i − 0.7778j + 0.4444k

eAB =rAB

|rAB | =40

80.6i − 70

80.6j + 0k = 0.4963i − 0.8685j + 0k

The tensions are TAB = 4eAB , TAC = |TAC |eAC , and TAD =|TAD|eAD . The components normal to rOA are

∑FX = (−0.4082|TAD| − 0.4444|TAC | + 1.9846)i = 0∑FZ = (−0.4082|TAD| + 0.4444|TAC |)k = 0.

The HP-28S calculator was used to solve these equations:

|TAC | = 2.23 kN, |TAD| = 2.43 kN

Problem 4.71 The tension in cable AB is 150 N. Thetension in cable AC is 100 N. Determine the sum of themoments about D due to the forces exerted on the wallby the cables.

C

y

x

D

zA

B

5 m

5 m

8 m

8 m

4 m

Solution: The coordinates of the points A, B, C are A (8, 0, 0),B (0, 4, −5), C (0, 8, 5), D(0, 0, 5). The point A is the intersectionof the lines of action of the forces. The position vector DA is

rDA = 8i + 0j − 5k.

The position vectors AB and AC are

rAB = −8i + 4j − 5k, rAB =√

82 + 42 + 52 = 10.247 m.

rAC = −8i + 8j + 5k, rAC =√

82 + 82 + 52 = 12.369 m.

The unit vectors parallel to the cables are:

eAB = −0.7807i + 0.3904j − 0.4879k,

eAC = −0.6468i + 0.6468j − 0.4042k.

The tensions are

TAB = 150eAB = −117.11i + 58.56j − 73.19k,

TAC = 100eAC = −64.68i + 64.68j − 40.42k.

The sum of the forces exerted by the wall on A is

TA = −181.79i + 123.24j − 32.77k.

The force exerted on the wall by the cables is −TA. The momentabout D is MD = −rDA × TA,

MD =

∣∣∣∣∣i j k8 0 −5

181.79 −123.24 +32.77

∣∣∣∣∣ = (−123.24)(5)i

−((8)(+32.77) − (−5)(181.79))j + (8)(−123.24)k

MD = −616.2i − 117.11j − 985.9k (N-m)

(Note: An alternate method of solution is to express the moment in terms of thesum: MD = (rDC × TC + (rDB × TB).)

C

y

xD Fz

A

B

5 m

5 m

8 m8 m

4 m

Problem 4.98 The tension in cable AB is 80 lb. Whatis the moment about the line CD due to the force exertedby the cable on the wall at B?

y

x

3 ft

8 ft

6 ft

B

C

D

A (6, 0, 10) ftz

Solution: The strategy is to find the moment about the point Cexerted by the force at B, and then to find the component of thatmoment acting along the line CD. The coordinates of the points B,C, D are B (8, 6, 0), C (3, 6, 0), D(3, 0, 0). The position vectorsare: rOB = 8i + 6j, rOC = 3i + 6j, rOD = 3i. The vectorparallel to CD is rCD = rOD − rOC = −6j. The unit vectorparallel to CD is eCD = −1j. The vector from point C to B isrCB = rOB − rOC = 5i.

The position vector of A is rOA = 6i + 10k. The vector parallel toBA is rBA = rOA − rOB = −2i − 6j + 10k. The magnitude is|rBA| = 11.832 ft. The unit vector parallel to BA is

eBA = −0.1690i − 0.5071j + 0.8452k.

The tension acting at B is

TBA = 80eBA = −13.52i − 40.57j + 67.62k.

The magnitude of the moment about CD due to the tension acting atB is

|MCD| = eCD · (rCB × TBA) =

∣∣∣∣∣0 −1 05 0 0

−13.52 −40.57 67.62

∣∣∣∣∣= 338.1 (ft lb).

The moment about CD is MCD = 338.1eCD = −338.1j (ft lb).The sense of the moment is along the curled fingers of the right handwhen the thumb is parallel to CD, pointing toward D.

y

z A

x

BC

D

(6, 0, 10)

6 ft

8 ft

3 ft

Problem 4.119 Four forces and a couple act on thebeam. The vector sum of the forces is zero, and the sumof the moments about the left end of the beam is zero.What are the forces Ax, Ay , and B?

4 m 4 m 3 m

Ax

Ay

200 N-m800 N

B

x

y

Solution: The sum of the forces about the y-axis is

∑FX = AY + B − 800 = 0.

The sum of the forces about the x-axis is

∑FX = AX = 0.

The sum of the moments about the left end of the beam is

∑ML = 11B − 8(800) − 200 = 0.

From the moments:

B =660011

= 600 N.

Substitute into the forces balance equation to obtain:

AY = 800 − 600 = 200 N4 m 4 m 3 m

Ax

Ay

200 N-m

800 N

B

x

y

Problem 4.120 The force F = 40i + 24j + 12k (N).(a) What is the moment of the couple?(b) Determine the perpendicular distance between the

lines of action of the two forces.F

–F

(10, 0, 1) m

x

(6, 3, 2) m

y

z

Solution:(a) The moment of the couple is given

MC = rAB × F

MC = (−4i + 3j + 1k) × (40i + 24j + 12k)

MC = 12i + 88j − 216k (N-m)

(b) |MC | = |d||F| sin 90◦

|F| =√

F 2x + F 2

y + F 2z = 48.2 N

|MC | =√

M2x + M2

y + M2z = 233.5 N

|d| = perpendicular distance

|d| = |MC |/|F||d| = 4.85 m

y

z

F

−F

x

B

A

(10, 0, 1) m

(6, 3, 2) m

Problem 4.121 Determine the sum of the momentsexerted on the plate by the three couples. (The 80-lbforces are contained in the x-z plane.)

z 60° 80 lb 60° 80 lb

40 lb

40 lb

8 ft

x

y

3 ft20 lb 20 lb

3 ft

Solution: The moments of two of the couples can be determinedfrom inspection:

M1 = −(3)(20)k = −60k ft lb.

M2 = (8)(40)j = 320j ft lb

The forces in the 3rd couple are resolved:

F = (80)(i sin 60◦ + k cos 60◦) = 69.282i + 40k

The two forces in the third couple are separated by the vector

r3 = (6i + 8k) − (8k) = 6i

The moment is

M3 = r3 × F3 =

∣∣∣∣∣i j k6 0 0

69.282 0 40

∣∣∣∣∣ = −240j.

z60°

80 lb

60°80 lb

40 lb

40 lb

8 ft

x

y3 ft

20 lb 20 lb

3 ft

The sum of the moments due to the couples:

∑M = −60k + 320j − 240j = 80j − 60k ft lb

Problem 4.122 What is the magnitude of the sum ofthe moments exerted on the T -shaped structure by thetwo couples?

z

y

x

50i + 20j – 10k (lb)

–50i – 20j + 10k (lb)

50j (lb)

–50j (lb)

3 ft3 ft

3 ft

3 ft

Solution: The moment of the 50 lb couple can be determined byinspection:

M1 = −(50)(3)k = −150k ft lb.

The vector separating the other two force is r = 6k. The moment is

M2 = r × F =

∣∣∣∣∣i j k0 0 650 20 −10

∣∣∣∣∣ = −120i + 300j.

The sum of the moments is

∑M = −120i + 300j − 150k.

The magnitude is

|M| =√

1202 + 3002 + 1502 = 356.23 ft lb

z

y

x

50 j (lb)

–50 j (lb)

3 ft3 ft

3 ft

3 ft

F

–F

Problem 4.126 In Problem 4.125, the forces

FB = 2i + 6j + 3k (kN),

FC = i − 2j + 2k (kN),

and the couple

MC = MCyj + MCzk (kN-m).

Determine the values for MCy and MCz , so that the sumof the moments of the two forces and the couple aboutA is zero.

Solution: From the solution to Problem 4.125, the sum of the moments ofthe two forces about A is

MForces = 0i − 7j + 2k (kN-m).

The required moment, MC , must be the negative of this sum.

Thus MCy = 7 (kN-m), and MCz = −2 (kN-m).

Problem 4.127 Two wrenches are used to tighten anelbow fitting. The force F = 10k (lb) on the rightwrench is applied at (6, −5, −3) in., and the force −Fon the left wrench is applied at (4, −5, 3) in.(a) Determine the moment about the x axis due to the

force exerted on the right wrench.(b) Determine the moment of the couple formed by the

forces exerted on the two wrenches.(c) Based on the results of (a) and (b), explain why two

wrenches are used.

x

y

F

–F

z

Solution: The position vector of the force on the right wrench isrR = 6i − 5j − 3k. The magnitude of the moment about the x-axisis

|MR| = eX · (rR × F) =

∣∣∣∣∣1 0 06 −5 −30 0 10

∣∣∣∣∣ = −50 in lb

(a) The moment about the x-axis is

MR = |MR|eX = −50i (in lb).

(b) The moment of the couple is

MC = (rR − rL) × FR =

∣∣∣∣∣i j k2 0 −60 0 10

∣∣∣∣∣ = −20j in lb

(c) The objective is to apply a moment to the elbow relative to con-necting pipe, and zero resultant moment to the pipe itself. Aresultant moment about the x-axis will affect the joint at theorigin. However the use of two wrenches results in a net zeromoment about the x-axis the moment is absorbed at the junctureof the elbow and the pipe. This is demonstrated by calculatingthe moment about the x-axis due to the left wrench:

|MX | = eX · (rL × FL) =

∣∣∣∣∣1 0 04 −5 30 0 −10

∣∣∣∣∣ = 50 in lb

from which MXL = 50i in lb, which is opposite in direction and equal inmagnitude to the moment exerted on the x-axis by the right wrench. Theleft wrench force is applied 2 in nearer the origin than the right wrenchforce, hence the moment must be absorbed by the space between, whereit is wanted.

x

y

F–F

z