Mechanical Engineering Chapter 4 Problem The Cable Exerts Force Ona The Building Cranes Boom The Cable

Problem 4.37 The cable AB exerts a 290-kN force on

the building crane’s boom at B. The cable AC exerts a

148-kN force on the boom at C. Determine the sum of

the moments about Pdue to the forces the cables exert

on the boom.

8 m

16 m

38 m

56 m

P

G

A

BC

Boom

Solution:

p3200 ⊲290 kN⊳⊲56 m⊳8

p320 ⊲148 kN⊳⊲16 m⊳

D3.36 MNm

MPD3.36 MN-m CW

290 kN

148 kN

88

56

A

Problem 4.38 The mass of the building crane’s boom

in Problem 4.37 is 9000 kg. Its weight acts at G. The

sum of the moments about Pdue to the boom’s weight,

the force exerted at Bby the cable AB, and the force

exerted at Cby the cable AC is zero. Assume that the

tensions in cables AB and AC are equal. Determine the

tension in the cables.

Solution:

A

56

Problem 4.39 The mass of the luggage carrier and the

suitcase combined is 12 kg. Their weight acts at A. The

F

0.28 m 0.14 m

A

C

y

a

Solution: Ois the origin of the coordinate system

MODF⊲1.2 m cos ˛⊳

1.2 m cos ˛

Problem 4.40 The hydraulic cylinder BC exerts a

300-kN force on the boom of the crane at C. The force

is parallel to the cylinder. What is the moment of the

force about A?

Solution: The strategy is to resolve the force exerted by the hydra-

ulic cylinder into the normal component about the crane; determine the

distance; determine the sign of the action, and compute the moment.

1.2D63.43°.

617.15 kN-m Check: The force exerted by the actuator can be resolved

α

3 m

2.4 m

1.2 m

180

Problem 4.41 The hydraulic piston AB exerts a 400-lb

force on the ladder at Bin the direction parallel to the

piston. The sum of the moments about Cdue to the force

exerted on the ladder by the piston and the weight Wof

the ladder is zero. What is the weight of the ladder?

6 ft

W

Problem 4.42 The hydraulic cylinder exerts an 8-kN

force at Bthat is parallel to the cylinder and points from

Ctoward B. Determine the moments of the force about

points Aand D.

1 m

0.6 m Scoop

AB

D

C

0.15 m

0.6 m

1 m

Hydraulic

cylinder

Solution: Use x,ycoords with origin A. We need the unit vector

from Cto B,eCB. From the geometry,

eCB D0.780i0.625j

5.00 kN

6.25 kN

Problem 4.43 The structure shown in the diagram is

one of the two identical structures that support the scoop

of the excavator. The bar BC exerts a 700-N force at C

that points from Ctoward B. What is the moment of this

force about K?

H

LK

100

mm

160

mm

Shaft

260

mm

320

mm

380

mm

C

D

1040

mm 1120

mm

260

mm

Scoop

B

180

mm

J

Solution:

MKD353 Nm CW

520 mm

320

K

Problem 4.44 In the structure shown in Problem 4.43,

the bar BC exerts a force at Cthat points from C

toward B. The hydraulic cylinder DH exerts a 1550-N

force at Dthat points from Dtoward H. The sum of the

moments of these two forces about Kis zero. What is

the magnitude of the force that bar BC exerts at C?

Solving we ﬁnd

FD796 N

1120

100

1550 N

K

182

c

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Problem 4.45 In Active Example 4.4, what is the

moment of Fabout the origin of the coordinate system?

(0, 6, 5) ft

A

C (7, 7, 0) ft

y

F

Solution: The vector from the origin to point Bis

Problem 4.46 Use Eq. (4.2) to determine the moment

of the 80-N force about the origin Oletting rbe the

vector (a) from Oto A; (b) from Oto B.

Solution:

(a) MODrOA ðF

Problem 4.47 A bioengineer studying an injury sus-

tained in throwing the javelin estimates that the magni-

Solution: The magnitude of the moment is jFj⊲0.55 m⊳D⊲360 N⊳

⊲0.55 m⊳D198 N-m. The moment vector is perpendicular to the xy

Problem 4.48 Use Eq. (4.2) to determine the moment

of the 100-kN force (a) about A, (b) about B.

A

y

100j (kN)

Solution: (a) The coordinates of Aare (0,6,0). The coordinates of

the point of application of the force are (8,0,0). The position vector

Problem 4.49 The cable AB exerts a 200-N force on

the support at Athat points from Atoward B. Use

Eq. (4.2) to determine the moment of this force about

point Pin two ways: (a) letting rbe the vector from P

to A; (b) letting rbe the vector from Pto B.

y

(0.3, 0.5) m

(0.9, 0.8) m

P

A

Solution: First we express the force as a vector. The force points

in the same direction as the position vector AB.

184

Problem 4.50 The line of action of Fis contained in

the xyplane. The moment of Fabout Ois 140k(N-

m), and the moment of Fabout Ais 280k(N-m). What

O

x

(5, 3, 0) m

Solution: The strategy is to ﬁnd the moments in terms of the

components of Fand solve the resulting simultaneous equations. The

position vector from Oto the point of application is rOF D5iC3j.

Take the dot product of both sides with kto eliminate k. The simul-

taneous equations are:

O

Problem 4.51 Use Eq. (4.2) to determine the sum of

the moments of the three forces (a) about A, (b) about B.

6 kN

y

Problem 4.52 Three forces are applied to the plate.

Use Eq. (4.2) to determine the sum of the moments of

the three forces about the origin O.

3 ft

200 lb

y

Solution: The position vectors from Oto the points of applica-

Problem 4.53 Three forces act on the plate. Use

Eq. (4.2) to determine the sum of the moments of the

three forces about point P.

3 kN

4 kN

30⬚

45⬚

y

Solution:

Problem 4.54 (a) Determine the magnitude of the

moment of the 150-N force about Aby calculating the

perpendicular distance from Ato the line of action of

the force.

(b) Use Eq. (4.2) to determine the magnitude of the

moment of the 150-N force about A.

A

y

x

150k (N)

(0, 6, 0) m

(6, 0, 0) m

Solution:

(a) The perpendicular from Ato the line of action of the force lies

186

Problem 4.55 (a) Determine the magnitude of the

moment of the 600-N force about Aby calculating the

perpendicular distance from Ato the line of action of

the force.

(b) Use Eq. (4.2) to determine the magnitude of the

moment of the 600-N force about A.

y

x

(0.6, 0.5, 0.4) m

0.8 m

A

600i (N)

Solution:

(a) Choose some point P⊲x, 0,0.8m⊳. on the line of action of the

force. The distance from Ato Pis then

Problem 4.56 what is the magnitude of the moment of

Fabout point B?

y

A

(4, 4, 2) ft

F ⫽ 20i ⫹ 10j ⫺ 10k (lb)

Solution: The position vector from Bto Ais

Problem 4.57 In Example 4.5, suppose that the attach-

ment point Cis moved to the location (8,2,0) m and the

tension in cable AC changes to 25 kN. What is the sum

of the moments about Odue to the forces excerted on

the attachment point Aby the two cables?

y

x

(6, 3, 0) m

(0, 4, 8) m O

B

C

Solution: The position vector from Ato Cis

The moment about Ois

ij k

Problem 4.58 The rope exerts a force of magnitude

jFjD200 lb on the top of the pole at B. Determine the

magnitude of the moment of Fabout A.

yB (5, 6, 1) ft

F

Solution: The position vector from Bto Cis

188

Problem 4.59 The force FD30iC20j10k(N).

(a) Determine the magnitude of the moment of F

about A.

(b) Suppose that you can change the direction of Fwhile

keeping its magnitude constant, and you want to

choose a direction that maximizes the moment of

Fabout A. What is the magnitude of the resulting

maximum moment?

y

x

z

(4, 3, 3) m

A(8, 2, – 4) m

F

Solution: The vector from Ato the point of application of Fis

rD4i1j7km

and

(b) The maximum moment occurs when r?F. In this case

Problem 4.60 The direction cosines of the force Fare

cos xD0.818, cos yD0.182, and cos zD0.545.

The support of the beam at Owill fail if the magnitude of

the moment of Fabout Oexceeds 100 kN-m. Determine

the magnitude of the largest force Fthat can safely be

applied to the beam. z

y

O

x

F

3 m

Solution: The strategy is to determine the perpendicular distance

Problem 4.61 The force Fexerted on the grip of the

exercise machine points in the direction of the unit vector

eD2

3i2

3jC1

3kand its magnitude is 120 N. Determine

the magnitude of the moment of Fabout the origin O.

y

O

F

150 mm

200 mm

Solution: The vector from Oto the point of application of the

force is

or

Problem 4.62 The force Fin Problem 4.61 points in

the direction of the unit vector eD2

3i2

3jC1

3k. The

support at Owill safely support a moment of 560 N-m

magnitude.

(a) Based on this criterion, what is the largest safe

magnitude of F?

(b) If the force Fmay be exerted in any direction, what

is its largest safe magnitude?

Solution: See the ﬁgure of Problem 4.61.

If we set jMOjD560 N-m, we can solve for jFmaxj

190

Problem 4.63 A civil engineer in Boulder, Colorado

estimates that under the severest expected Chinook

winds, the total force on the highway sign will be

FD2.8i1.8j(kN). Let MObe the moment due to F

about the base Oof the cylindrical column supporting

the sign. The ycomponent of MOis called the torsion

exerted on the cylindrical column at the base, and the

component of MOparallel to the xzplane is called

the bending moment. Determine the magnitudes of the

torsion and bending moment.

y

O

F

x

8 m

Solution: The total moment is

Problem 4.64 The weights of the arms OA and AB of

the robotic manipulator act at their midpoints. The direc-

y

600 mm

B

Solution: By deﬁnition, the direction cosines are the scalar compo-

nents of the unit vectors. Thus the unit vectors are e1D0.5iC0.866j,

The sum of moments is

MDr1ðW1Cr2ðW2

ijk

ij k

Problem 4.65 The tension in cable AB is 100 lb. If

you want the magnitude of the moment about the base

Oof the tree due to the forces exerted on the tree by the

two ropes to be 1500 ft-lb, what is the necessary tension

in rope AC ?

B

O

C

z

(0, 0, 10) ft (14, 0, 14) ft

Solution: We have the forces

F1D100 lb

p164 ⊲8jC10k⊳, F2DTAC

p456 ⊲14i8jC14k⊳

Thus the total moment is

Problem 4.66* A force Facts at the top end Aof the

pole. Its magnitude is jFjD6 kN and its xcomponent

is FxD4 kN. The coordinates of point Aare shown.

Determine the components of Fso that the magnitude

Solution: The force is given by FD⊲4kNiCFyjCFzk⊳.

Since the magnitude is constrained we must have

192

Problem 4.67 The force FD5i(kN) acts on the ring

Awhere the cables AB,AC, and AD are joined. What is

the sum of the moments about point Ddue to the force

Fand the three forces exerted on the ring by the cables?

(0, 6, 0) m

y

D

AF

Problem 4.68 In Problem 4.67, determine the mo-

ment about point Ddue to the force exerted on the ring

Aby the cable AB.

Solution: We need to write the forces as magnitudes times the

D(0, 6, 0)

194

Problem 4.69 The tower is 70 m tall. The tensions

in cables AB,AC, and AD are 4 kN, 2 kN, and 2 kN,

respectively. Determine the sum of the moments about

the origin Odue to the forces exerted by the cables at

point A.

35 m

40 m 40 m

40 m

A

y

x

z

B

O

C

D

Solution: The coordinates of the points are A(0, 70, 0), B(40, 0,

0), C(40, 0, 40) D(35, 0, 35). The position vectors corresponding

to the cables are:

The sum of the forces acting at Aare

TAD0.2792i6.6615jC0.07239k(kN-m)

The position vector of Ais rOA D70j. The moment about Ois MD

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 and

AD so that the sum of the moments about the origin O

due to the forces exerted by the cables at point Ais zero.

Determine the tensions.

Solution: From Varignon’s theorem, the moment is zero only if

the resultant of the forces normal to the vector rOA is zero. From

Problem 4.69 the unit vectors are:

eAD DrAD

jrADjD35

85.73 i70

85.73 j35

85.73

The tensions are TAB D4eAB,TAC DjTACjeAC, and TAD DjTADjeAD.

The components normal to rOA are

FXD⊲0.4082jTADj0.4444jTACjC1.9846⊳iD0

Problem 4.71 The tension in cable AB is 150 N. The

tension in cable AC is 100 N. Determine the sum of the

moments about Ddue to the forces exerted on the wall

by the cables.

C

y

x

D

z

A

B

5 m

8 m

4 m

Solution: The coordinates of the points A,B,Care A(8, 0, 0),

B(0, 4, 5), C(0, 8, 5), D(0, 0, 5). The point Ais the intersection of

the lines of action of the forces. The position vector DA is

rDA D8iC0j5k.

The position vectors AB and AC are

The unit vectors parallel to the cables are:

ijk

⊲⊲8⊳⊲C32.77⊳⊲5⊳⊲181.79⊳⊳jC⊲8⊳⊲123.24⊳k

196

Problem 4.72 Consider the wall shown in Prob-

lem 4.71. The total force exerted by the two cables in

the direction perpendicular to the wall is 2 kN. The

magnitude of the sum of the moments about Ddue to

the forces exerted on the wall by the cables is 18 kN-m.

What are the tensions in the cables?

Problem 4.73 The tension in the cable BD is 1 kN. As

a result, cable BD exerts a 1-kN force on the “ball” at

z

Solution: We have the force and position vectors

Problem 4.74* Suppose that the mass of the sus-

pended object Ein Problem 4.73 is 100 kg and the mass

of the bar AB is 20 kg. Assume that the weight of the

bar acts at its midpoint. By using the fact that the sum

Solution: We have the following forces applied at point B.

F1D⊲100 kg⊳⊲9.81 m/s2⊳j,F2DTBC

p33 ⊲4iCj4k⊳,

MyD2.089TBC 3.333TBD D0

198