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Solutions Manual

Engineering Mechanics: Statics1st Edition

Michael E. PleshaUniversity of WisconsinMadison

Gary L. GrayThe Pennsylvania State University

Francesco CostanzoThe Pennsylvania State University

With the assistance of:

Chris Punshon

Andrew J. Miller

Justin High

Chris OBrien

Chandan Kumar

Joseph Wyne

Jonathan Fleischmann

Version: August 24, 2009

The McGraw-Hill Companies, Inc.

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Copyright 20022010

Michael E. Plesha, Gary L. Gray, and Francesco Costanzo

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Statics 1e 3

Important Information about

this Solutions Manual

We encourage you to occasionally visit

http://www.mhhe.com/pgc

to obtain the most up-to-date version of this solutions manual.

Contact the Authors

If you find any errors and/or have questions concerning a solution, please do not hesitate to contact the

authors and editors via email at:

stat_solns@email.esm.psu.edu

We welcome your input.

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4 Solutions Manual

Accuracy of Numbers in Calculations

Throughout this solutions manual, we will generally assume that the data given for problems is accurate to

3significant digits. When calculations are performed, all intermediate numerical results are reported to4significant digits. Final answers are usually reported with 3 significant digits. If you verify the calculations in

this solutions manual using the rounded intermediate numerical results that are reported, you should obtain

the final answers that are reported to3 significant digits.

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270 Solutions Manual

Chapter 4 SolutionsProblem 4.1

Compute the moment of force Fabout pointB , using the following procedures.

(a) Determine the moment armdand then evaluateMBD F d.(b) Resolve forceF intox and y components at pointA and use the principle of

moments.

(c) Use the principle of moments withF positioned at pointC.

(d) Use the principle of moments withF positioned at pointD .

(e) Use a vector approach.

Solution

Part (a) We first use geometry and trigonometry to find that

tan1. 3=4/D 36:9, tan1. 4=3/D 53:1, and 180 60 53:1 D 66:9 asshown in the figure to the right. We then use the law of sines with triangleABD

to find that12 in:

sin 53:1D b

sin 66:9) bD 13:8 in: (1)

Through trigonometry, it follows that the moment arm d is

dD b sin 53:1 D 11:0 in: (2)

Using Eq. (4.1) on page 182, and noting that positive moment is counterclockwise,MB is then given by

MBD F dD .25 lb/.11:0 in:/ D 276 in.lb: (3)

Part (b) Using the figure to the right, and noting that positive moment is counter-

clockwise, Eq. (4.1) provides

MBD 25 lb

3

5

!.12 in: sin 60/ C 25 lb

4

5

!.12 in: cos 60/ D 276 in. lb:

(4)

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Statics 1e 271

Part (c) Using triangleAB C shown at the right, the law of sines provides

12 in:

sin 36:9D h

sin 113:1) h D 18:4 in: (5)

In the figure to the right, we see that the vertical force at Cproduces no moment aboutB . By using Eq. (4.1), and noting that positive moment is counterclockwise,MB is

given by

MBD 25 lb 35

.18:4 in:/ D 276 in.lb: (6)

Part (d) In the figure to the right, we see that the horizontal force on D

produces no moment aboutB . Using Eq. (4.1), and noting that positive moment

is counterclockwise,MB is given by

MBD 25 lb 45

.13:8 in:/ D 276 in.lb: (7)

Part (e) We find that

ErBAD 12 in: cos 60 O{ C sin 60 O| (8)

EFD 25 lb3

5O{ 4

5O|

: (9)

Using Eq. (4.2) on page 183, EMB is given by

EMBD ErBA EFD .12 in:/.25 lb/ O{ O| Ok cos 60 sin 60 0

35

45

0

D 276Okin. lb: (10)

Note that instead of Eq. (8), we could have used ErBC or ErBD instead.

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272 Solutions Manual

Problem 4.2

Compute the moment of forceF about pointB , using the following procedures.

(a) Determine the moment armdand then evaluateMBD F d.

(b) Resolve forceF intox and y components at pointA and use the principleof moments.

(c) Use the principle of moments withF positioned at pointC.

(d) Use the principle of moments withF positioned at pointD .

(e) Use a vector approach.

Solution

Part (a) Using the 30 and 60 angles given in the problem state-

ment, we identify the additional angles shown in the figure at the right.

We then use the law of sines for triangle ABD to determine

12 in:

sin 30D b

sin 30) bD 12 in: (1)

Through trigonometry, it follows that the moment arm d is

dD b sin 30 D 6 in: (2)

Using Eq. (4.1) on page 182, and noting that positive moment is counterclockwise,MB

is then given by

MBD F dD 25 lb.6 in:/ D 150 in.lb: (3)

Part (b) Using the figure to the right, and noting that positive moment is coun-

terclockwise, Eq. (4.1) provides

MBD 25 lb cos 30.12 in: sin 60/ C 25 lb sin 30.12 in: cos 60/D 150 in.lb:

(4)

(5)

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Statics 1e 273

Part (c) Using triangleAB C shown at the right, the law of sines provides

12 in:

sin 120D h

sin 30) h D 6:93 in: (6)

In the figure to the right, we see that the vertical force atC produces no moment aboutB . Using Eq. (4.1), and noting that positive moment is counterclockwise,MB is given

by

MBD 25 lb cos 30.6:93 in:/ D 150 in.lb: (7)

Part (d) In the figure to the right, we see that the horizontal force atD

produces no moment about B . Using Eq. (4.1), and noting that positive

moment is counterclockwise,MB is given by

MBD 25 lbsin 30.12 in:/ D 150 in.lb: (8)

Part (e) We find that

ErBAD 12 in: cos 60 O{ C sin 60 O| (9)

EFD 25 lb cos 30 O{ sin 30 O| : (10)Using Eq. (4.2) on page 183, EMB is given by

EMBD ErBA EFD .12 in:/.25 lb/

O{ O| Ok

cos 60 sin 60 0cos 30

sin 30 0

D 150Okin. lb: (11)

Rather than ErBA, we could have used ErBC or ErBD instead.

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274 Solutions Manual

Problem 4.3

The cover of a computer mouse is hinged at point B so that it may be clicked.

Repeat Prob. 4.1 to determine the moment about point B .

Solution

Part (a) Using trigonometry and the figure to the right, we determine that the

moment armd is

dD 60 mmsin 45 D 42:4 mm (1)Using Eq. (4.1) on page 182, and noting that positive moment is counterclockwise,

MB is given by

MBD F dD 3 N.42:4 mm/ D 127 Nmm: (2)

Part (b) We first use geometry to determinehD20 mmas shown in the figureto the right. Using Eq. (4.1), and noting that positive moment is counterclockwise,

we determine that

MBD 3 Ncos 45.20 mm/ 3 Nsin 45.40 mm/

D 127 Nmm:

(3)

(4)

Part (c) In the figure to the right, we see that the vertical force atCproduces no moment

aboutB . Using Eq. (4.1), and noting that positive moment is counterclockwise,MB is given

by

MBD 3 Ncos 45.60 mm/ D 127 Nmm: (5)

Part (d) In the figure to the right, we see that the horizontal force at D produces no

moment aboutB . Using Eq. (4.1), and noting that positive moment is counterclockwise,

MB is given by

MBD 3 Nsin 45.60 mm/ D 127 Nmm (6)

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Statics 1e 275

Part (e) Using the value ofh found in Part (b), we find that

ErBAD .40 O{ C 20O| / mm (7)EFD 3 N

sin 45 O{ cos 45 O|

: (8)

Using Eq. (4.2) on page 183, EMB is given by

EMBD ErBA EFD 3 Nmm O{ O| Ok40 20 0

sin 45 cos 45 0

D 127OkN mm (9)

Rather than ErBA, we could have used ErBC or ErBD instead.

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276 Solutions Manual

Problem 4.4

The cover of a computer mouse is hinged at point B so that it may be clicked.

Repeat Prob. 4.1 to determine the moment about point B .

Solution

Part (a) We use some elementary geometry to determine the dimensions shown

in the figure to the right. The moment armdis then

dD 20 mm cos 45

D 14:1 mm (1)Using Eq. (4.1) on page 182, and noting that positive moment is counterclock-

wise,MB is given by

MBD F dD 3 N.14:1 mm/ D 42:4 Nmm: (2)

Part (b) Using Eq. (4.1), and noting that positive moment is counterclockwise,

we determine

MBD 3 Ncos 45.20 mm/ 3 Nsin 45.40 mm/D 42:4 Nmm:

(3)

(4)

Part (c) In the figure to the right, we see that the vertical force at C produces no

moment aboutB . Using Eq. (4.1) again,MB is given by

MBD 3 Ncos 45.20 mm/ D 42:4 Nmm: (5)

Part (d) We see that the horizontal force on Dproduces no moment about Bin the figure

to the right, then by using Eq. (4.1), and noting that positive moment is counterclockwise,MB is given by

MBD 3 Nsin 45.20 mm/ D 42:4 Nmm: (6)

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Statics 1e 277

Part (e) We find that

ErBAD .40 O{ C 20O| / mm (7)EFD 3 N

cos 45 O{ sin 45 O|

: (8)

Using Eq. (4.2) on page 183, EMB is given by

EMBD ErBA EFD 3 Nmm O{ O| Ok40 20 0

cos 45 sin 45 0

D 42:4OkNmm (9)

Rather than ErBA, we could have used ErBC or ErBD instead.

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278 Solutions Manual

Problem 4.5

An atomic force microscope (AFM) is a state-of-the-art device used to study

the mechanical and topological properties of surfaces on length scales as small

as the size of individual atoms. The device uses a flexible cantilever beamAB

with a very sharp, stiff tipB Cthat is brought into contact with the surface tobe studied. Due to contact forces atC, the cantilever beam deflects. If the tip

of theAFM is subjected to the forces shown, determine the resultant moment of

both forces about pointA. Use both scalar and vector approaches.

Solution

For the scalar approach we use Eq. (4.1) on page 182. Noting that positive moment is counterclockwise, MAis given by

MAD

.13 nN/.12m/C

.5 nN/.1:5m/D

149 nNm109N

nN!106m

m!

D 1:491013Nm:

(1)

(2)

For the vector approach we use Eq. (4.2) on page 183. It follows that

ErACD .12 O{ 1:5O| / m (3)EFD .5 O{ C 13O| / nN (4)

and EMAis given by

EMAD ErACEFD O{

O| Ok

12 1:5 05 13 0

nNm

D 149OknN mD 1:491013OkN m:

(5)

(6)

(7)

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Statics 1e 279

Problem 4.6

The door of an oven has 22 lbweight which acts vertically through point D. The

door is supported by a hinge at pointA and two springs that are symmetrically

located on each side of the door. For the positions specified below, determine

the force needed in each of the two springs if the resultant moment about pointAof the weight and spring forces is to be zero.

(a) D 45.(b) D 90.(c) Based on your answers to Parts (a) and (b), determine an appropriate

stiffness for the springs.

Solution

Part (a) We first determine some needed dimensions using trigonometry as shown in the figure at the left

below. The vertical and horizontal distances between pointsA andC are.5 in:/ sin 45 D .5 in:/ cos 45 D3:54 in. Then, using the Pythagorean theorem we determine the distance between pointsB andC as

LBCDq

.10 in: C 3:54 in:/2 C .8 in: 3:54 in:/2 D 14:25 in: (1)

LetFbe the force in one spring. The forces in the springs act along lineB C, therefore the total horizontal

and vertical forces atCfrom the springs are, respectively,

FhorizontalD 2F.13:54 in:=14:25 in:/; F verticalD 2F.4:46 in:=14:25 in:/; (2)

and these forces are shown in the figure at the right below. The moment arm for the 22 lb weight is

.12 in:/ cos 45 D 8:49 in:as shown in the figure to the right below.

With positive moment being counterclockwise, the resultant moment of the two spring forces and the weight

of the door about pointA is

MAD .22 lb/.8:49 in:/ C 2F13:54in:

14:25 in:.3:54 in:/ C 2F 4:46 in:

14:25 in:.3:54 in:/: (3)

RequiringMAD 0, as stated in the problem statement, we set Eq. (3) equal to zero to solve for F as

FD 21:0 lb: (4)

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280 Solutions Manual

Part (b) When the door is horizontal (fully open), pointsA,B , andChave the positions shown in the

figure at the left below. The Pythagorean theorem is used to determine the distance between points B andC

as

LBCDq

.10 in: C 5 in:/2 C .8 in:/2 D 17 in: (5)LetFbe the force in one spring. The forces in the springs act along lineB C, therefore the total horizontal

and vertical forces atCfrom the springs are, respectively

FhorizontalD 2F.15 in:=17 in:/; F verticalD 2F.8 in:=17 in:/: (6)

and these forces are shown in the figure to the right below.

With positive moment being counterclockwise, and noting that the horizontal force at C produces no moment

about pointA, the resultant moment of the two spring forces and the weight of the door about pointA is

MAD .22 lb/.12 in:/ C 2F 8in:

17:0 in:.5 in:/: (7)

RequiringMAD 0, as stated in the problem statement, we set Eq. (7) equal to zero to solve for F as

FD 56:1 lb: (8)

Part (c) We use the spring law to determine the necessary spring stiffness so that the springs will produce

the forces determined in Parts (a) and (b) forD 45

andD 90

. Thus

FD k ) kD F

) kD change in forcechange in length

D 56:1 lb 21:0 lb17:0 in: 14:3 in:D 13:0 lb=in: (9)

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Statics 1e 281

Problem 4.7

The ball of a trailer hitch is subjected to a forceF. If failure occurs when the

moment ofFabout pointA reaches 10,000 in.lb, determine the largest valueF may have and specify the value of for which the moment about A is the

largest. Note that the value ofFyou determine must not produce a momentaboutA that exceeds 10,000 in.lb for any possible value of.

Solution

The maximum moment for a given value of forceF occurs when the moment arm

is at its maximum. Using the Pythagorean theorem in the figure to the right, the

maximum value of the moment armdis given by

dDq

.9 in:/2 C .6 in:/2 D 10:8 in: (1)

We then use Eq. (4.1) on page 182 to determine the maximum force F, given that the moment of this forceabout pointA may not exceed10;000 in.lb. Thus,

MAD F dD F.10:8 in:/ D 10;000 in.lb ) FD 926 lb: (2)

The angle at which this maximum force acts is then found to be

dcos D 9 in: ) D 33:6 (3)

D 180 .180 90/ D 123:6: (4)

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282 Solutions Manual

Problem 4.8

Repeat Prob. 4.7 if the moment at pointB may not exceed 5000 in.lb.

Solution

The maximum moment for a given value of forceF occurs when the moment arm

is at its maximum. Using the Pythagorean theorem in the figure to the right, the

maximum value of the moment armdis given by

dDq

.5 in:/2 C .6 in:/2 D 7:81 in: (1)

We then use Eq. (4.1) on page 182 to determine the maximum force F, given that the moment of this force

about pointB may not exceed5000 in.lb. Thus,

MBD F dD F.7:81 in:/ D 5000 in. lb ) FD 640 lb: (2)

The angle at which this maximum force acts is then found to be

dcos D 5 in: ) D 50:2 (3)

D 180 .180 90/ D 140: (4)

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Statics 1e 283

Problem 4.9

The port hull of a catamaran (top view shown) has cleats at pointsA,B , andC.

A rope having100 Nforce is to be attached to one of these cleats. If the force

is to produce the largest possible counterclockwise moment about point O,

determine the cleat to which the rope should be attached and the direction therope should be pulled (measured positive counterclockwise from the positivex

direction). Also, determine the value ofMO produced. Assume all cleats, point

O, and the rope lie in the same plane.

Solution

CleatCoffers the largest possible moment arm, therefore

Attaching the rope to cleatC will produce the largest moment. (1)

As shown to the right, the Pythagorean theorem is used to determine the

moment armd as

dDq

.1:5 m/2 C .3 m/2 D 3:35 m: (2)

The angle at which the force acts is given by

dcos D 3 m ) D 26:4 (3)

D 90 C D 116:4: (4)The moment of the100 N force aboutO is given by Eq. (4.1) on page 182 as

MOD 100 N.3:35 m/ D 335 Nm: (5)

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284 Solutions Manual

Problem 4.10

Frame AB C has a frictionless pulley at C around which a cable is wrapped.

Determine the resultant moment about pointA produced by the cable forces if

WD 5 kN and(a) D 0.(b) D 90.(c) D 30.

Solution

Part (a) TD WD 5 kNfor the pulley to be in equilibrium. We may determine the moment of the cableforces about pointAby considering the forces supported by the cables, or by considering the forces the pulley

applies to the structure at point C. Both of these options are illustrated in the figures below, to the left and

right, respectively.

Use of either of the force systems shown above is convenient, and we will use the force system shown at the

left. Noting that positive moment is counterclockwise, we then use Eq. (4.1) on page 182 to determine themoment about pointA as

MAD 5 kN.3 m C 0:5 m/ C 5 kN.3 m 0:5 m/D 30:0 kNm:

(1)

(2)

Part (b) We use the same procedure as for Part (a), and use either of the following force systems.

Use of either of the force systems shown above is convenient, and we will use the force system shown at the

left. Noting that positive moment is counterclockwise, we then use Eq. (4.1) on page 182 to determine the

moment about pointA as

MAD 5 kN.4 m C 0:5 m/ C 5 kN.3 m 0:5 m/D 35:0 kNm:

(3)

(4)

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Statics 1e 285

Part (c) In principle, either of the force systems shown below may be used to determine the moment about

pointA.

However, if the force system at the left is used, it will be tedious to determine the moment arm for the cable

force with the 30 orientation. Thus, it will be considerably more convenient to use the force system shown

at the right, which provides

MAD 5 kN.3 m/ C 5 kN.cos 30/3 m C 5 kN.sin 30/4 mD 38:0 kNm:

(5)

(6)

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286 Solutions Manual

Problem 4.11

Frame AB C has a frictionless pulley at C around which a cable is wrapped.

If the resultant moment about point A produced by the cable forces is not to

exceed20 kNm, determine the largest weightWthat may be supported and thecorresponding value of.

Solution

First we determine the angle that provides the largest moment arm for force

T. Using the figure shown at the right, it follows that

tan D 4 m

3 m ) D D 53:1

: (1)

Since the pulley is in equilibrium, TD W . The Pythagorean theorem providesdD

p.4 m/2 C .3 m/2 D 5 m. We then use Eq. (4.1) on page 182, to evaluate the

moment of the cable forces about pointA, noting that positive moment is counterclockwise

MAD W .3 m/ C W cos 53:1.3 m/ C W sin 53:1.4 m/: (2)

As given in the problem statement,MAis not to exceed20 kNm. Thus we set Eq. (2) equal to 20 kNmandsolve forW to obtain

W

D2:5 kN: (3)

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Statics 1e 287

Problem 4.12

The load carrying capacity of the frame of Prob. 4.11 can be increased by

placing a counterweightQ at pointD as shown. The resultant moment about

pointA due to the cable forces andQ is not to exceed 20 kNm.(a) Determine the largest value of counterweightQ. (Hint: LetWD 0 and

determineQ so thatMAD 20 kNm.)(b) With the value ofQ determined in Part (a), determine the largest weight

W that may be supported and the corresponding value of.

Solution

Part (a) We begin by settingWD 0, and then use Eq. (4.1) on page 182to evaluate the moment ofQ about pointA, noting that this moment may not

exceed20 kNm. With positive moment being counterclockwise, it follows that

MAD Q.2 m/ D 20 kNm ) Q D 10 kN: (1)

Part (b) WithW 0, we know that TD W, since the pulley is in equilibrium. As determined in thesolution to Prob. 4.11, the moment ofT about pointA is largest whenD 53:1 (the moment ofT aboutpointA is largest when its line of action is perpendicular to the line connecting pointsA and C, as shown

above). Using the figure above, we evaluate the moment about point A, taking positive moment to be

counterclockwise, and we require this moment to not exceed20 kN

m. Thus,

MAD 10 kN.2 m/ C W .3 m/ C W cos 53:1.3 m/ C W sin 53:1.4 m/ D 20 kNmWD 5 kN:

(2)

(3)

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288 Solutions Manual

Problem 4.13

A flat rectangular plate is subjected to the forces shown, where all forces are

parallel to the x or y axis. IfFD 200 N and PD 300 N, determine theresultant moment of all forces about the

(a) axis.

(b) aaxis, which is parallel to the axis.

Solution

Part (a) We use Eq. (4.1) on page 182 to find

M D F.300 mm/ F.500 mm/ C P.200 mm/ P.600 mm/

D 1:6

105N

mm

D 160 N

m:

(1)

(2)

In writing the above expression, positive moment is taken to be in the positive direction, according to the

right-hand rule.

Part (b) We use Eq. (4.1) again to find

MaD F.300 mm/ C F.200 mm/ P.500 mm/D 1:7105Nmm D 170 Nm:

(3)

(4)

In writing the above expression, positive moment is taken to be in the positive a direction, according to the

right-hand rule.

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Statics 1e 289

Problem 4.14

A flat rectangular plate is subjected to the forces shown, where all forces are

parallel to the x or y axis. IfPD 300 N, determine F when the resultantmoment of all forces about the axis isM D 100 Nm.

Solution

Use Eq. (4.1) on page 182 to evaluate the moment ofF andPD 300 N about the axis

M D F.300 mm/ F.500 mm/ C 300 N.200 mm/ 300 N.600 mm/: (1)

In writing the above expression, positive moment is taken to be in the positive direction, according to the

right-hand rule. As given in the problem statement, M D 100 Nm. Thus, Eq. (1) becomes

100 Nm 103 mmm

D F.300 mm/ F.500 mm/ C 300 N.200 mm/ 300 N.600 mm/; (2)which may be solved for

FD 100 N: (3)

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290 Solutions Manual

Problem 4.15

Repeat Prob. 4.14 if the resultant moment of all forces about the a axis is

MaD 100 Nm.

Solution

Use Eq. (4.1) on page 182 to evaluate the moment ofF andPD 300 N about thea axis

MaD F.300 mm/ C F.200 mm/ 300 N.500 mm/: (1)

In writing the above expression, positive moment is taken to be in the positive a direction, according to the

right-hand rule. As given in the problem statement, MaD 100 Nm. Thus, Eq. (1) becomes

100 Nm 103 mmm

D F.300 mm/ C F.200 mm/ 300 N.500 mm/; (2)which may be solved for

FD 500 N: (3)

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Statics 1e 291

Problem 4.16

StructureOBCDis built in at point O and supports a 50 lb cable force at point

Cand 100 and 200 lb vertical forces at points B andD, respectively. Using a

vector approach, determine the moment of these forces about

(a) pointB .

(b) pointO .

Solution

From the figure in the problem statement, the following vector expressions may be written

ErBCD 20Okin. ErBDD 15 O{ C 20Ok

in. ErOCD 36Okin. ErODD

15 O{ C 36Ok

in.

EFD 200O|lb EPD 100O|lb ETD 50 lb 12 O{ C 18O| 36Ok42

:

Part (a) Observing that EPproduces no moment about B , we use Eq. (4.2) on page 183 to evaluate theresultant moment ofET andEFabout pointB as

EMBD ErBCETC ErBD EF (1)

D .20 in:/.50 lb/42

O{ O| Ok0 0 1

12 18 36

C 200 in.lb

O{ O| Ok15 0 20

0 1 0

(2)

D

1000 in. lb42

hO{.

18/

O| .

12/

COk.0/iC

200 in.

lb hO

{.20/

O|.0/

COk.

15/i (3)

EMBD

3570 O{ C 286O| 3000Ok

in.lb: (4)

Part (b) We use Eq. (4.2) on page 183 to evaluate the resultant moment ofET,EF, andEP about pointO asEMOD ErOCETC ErOD EFC ErOBEP (5)

D .36 in:/.50 lb/42

O{ O| Ok0 0 1

12 18 36

C 200 in. lb

O{ O| Ok15 0 36

0 1 0

C .100 lb/.16 in:/ O

{ O

| O

k

0 0 1

0 1 0

(6)

D 1800 in.lb42

hO{.18/ O| .12/ COk.0/

iC 200 in.lb

hO{.36/ O|.0/ COk.15/

iC 1600 O{in. lb

(7)

EMOD

8030 O{ C 514O| 3000Ok

in.lb: (8)

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292 Solutions Manual

Problem 4.17

Repeat Prob. 4.16, using a scalar approach.

Solution

The vector expression forET is

ET

D50 lb

12 O{ C 18O| 36Ok42

: (1)

Using this expression, the figure at the right is drawn, where

the forceETis shown in terms of its components. We then useEq. (4.1) on page 182 to determine the components of EMB ineach of the coordinate directions, taking positive moment com-

ponents to be in the positive coordinate directions.

Part (a) It follows that

.MB/xD 200 lb.20 in:/ 50 lb 1842

.20 in:/ D 3570 in.lb

.MB/yD 50 lb 1242

.20 in:/ D 286 in.lb.MB/D 200 lb.15 in:/ D 3000 in.lb:

(2)

(3)

(4)

Part (b) It follows that

.MO/xD 200 lb.36 in:/ 50 lb 1842

.36 in:/ C 100 lb.16 in:/ D 8030 in.lb

.MO/yD 50 lb 1242

.36 in:/ D 514 in.lb.MO/D 200 lb.15 in:/ D 3000 in.lb:

(5)

(6)

(7)

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Statics 1e 293

Problem 4.18

Structure OAB is built in at point O and supports forces from two cables.

Cable CAD passes through a frictionless ring at point A, and cable DBE

passes through a frictionless ring at pointB . If the force in cableCAD is 250 N

and the force in cableDBE is 100 N, use a vector approach to determine

(a) the moment of all cable forces about pointA.

(b) the moment of all cable forces about pointO .

Solution

From the figure provided in the problem statement, the following

vector expressions may be written

EFBDD 100 N 150 O{ 200Ok250

ErABD 120Okmm

EFBED 100 N 150O| 200Ok

250 ErOBD 200Okmm

EFACD 250 N60O| 80Ok

100 ErOAD 80Okmm

EFADD 250 N 150O{ 80Ok170

:

These force vectors are shown in the figure at the right.

Part (a) Noting that EFAC and EFAD produce no moment about pointA, we use Eq. (4.2) on page 183 todetermine

EMAD ErABEFBD CEFBE

D .120 mm/.100 N/

250

O{ O| Ok0 0 1

150 150 400

D 48 Nmm O{.150/ O| .150/D .7200 O{ C 7200O| / Nmm:

(1)

(2)

(3)

Part (b) We use Eq. (4.2) to determine

EMOD ErOBEFBD CEFBE

C ErOA

EFACCEFAD

D .200 mm/.100 N/250

O{ O| Ok0 0 1

150 150 400

C .80 mm/.250 N/

O{ O| Ok0 0 1150170

60100

80100

80170

D 80 Nmm O{.150/ O| .150/ C 20;000 Nmm O{.0:6/ O| .0:882/D .29;600O| / N mm:

(4)

(5)

(6)

(7)

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294 Solutions Manual

Problem 4.19

Repeat Prob. 4.18, using a scalar approach.

Solution

We begin by adding the two cable forces at point A to obtain EFA, and adding the two cable forces at pointBto obtainEFB , as follows

EFADEFACCEFADD 250 N60O| 80Ok100

C 250 N 150O| 80Ok170

D

221 O{ 150O| 318Ok

N; (1)

EFBDEFBD CEFBED 100 N 150 O{ 200Ok

250 C 100 N 150O| 200

Ok250

D

60 O{ C 60O| 160Ok

N: (2)

The components ofEFAand EFB are shown in the figure below.

Part (a) Taking positive moment components to be in the positive coor-

dinate directions, we use Eq. (4.1) on page 182 to write

.MA/x

D 60 N.120 mm/

D 7200 N

mm

.MA/yD 60 N.120 mm/ D 7200 Nmm

.MA/D 0:

(3)

(4)

(5)

Part (b) Taking positive moment components to be in the positive coor-

dinate directions, we use Eq. (4.1) on page 182 to write

.MO/xD 60 N.200 mm/ C 150 N.80 mm/ D 0 Nmm

.MO/yD 60 N.200 mm/ C 221 N.80 mm/ D 29;700 Nmm

.MO/D 0:

(6)

(7)

(8)

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Statics 1e 295

Problem 4.20

Structure OABCis built in at pointO and supports forces from two cables.

Cable EAD passes through a frictionless ring at point A, and cable OC G

passes through a frictionless ring at point C. If the force in cableEAD is

800 lb and the force in cable O C G is 400 lb, determine

(a) the moment of forces from cableO C G about pointB .

(b) the moment of all cable forces about pointA.

(c) the moment of all cable forces about pointO .

Solution

From the figure provided in the problem statement, the following

vector expressions may be written

EFCGD 400 lb 20 O{ 9O| 12Ok

25 ErBCD 9O| ft

EFCOD 400 lb9O| 12Ok

15 ErACD

9O|C 6Ok

ft

EFAED 800 lb 8O{ 6Ok

10 ErOCD

9O|C 12Ok

ft

EFADD 800 lb 8O| 6Ok

10 ErOAD 6Okft:

These force vectors are shown in the figure at the right.

Part (a) We use Eq. (4.2) on page 183 to determine the moment about pointB due to the forces in cable

OC G as

EMBD ErBCEFCGCEFCO

D .400 ftlb/

O{ O| Ok0 9 02025

925

915

1225

1215

D 400 ftlbhO{.11:5/ O|.0/ COk.7:20/

iD4600 O{ 2880Ok

ftlb:

(1)

(2)

Part (b) Noting that

EFAE and

EFAD produce no moment about pointA, we use Eq. (4.2) on page 183 to

determine

EMAD ErACEFCGCEFCO

D .400 ftlb/

O{ O| Ok0 9 62025

925

915

1225

1215

D 400 ftlbhO{.5:76/ O| .4:80/ COk.7:20/

iD2304 O{ C 1920O| 2880Ok

ftlb:

(3)

(4)

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296 Solutions Manual

Part (c) We use Eq. (4.2) on page 183 to determine

EMOD ErOCEFCGCEFCO

C ErOA

EFAECEFAD

D .400 ftlb/ O{ O| Ok0 9 122025

925

915

1225

1215

C .6 ft/.800 lb/10

O{ O| Ok0 0 18 8 12

D 400 ftlbh O| .9:60/ COk.7:20/

iC 480 ftlb O{.8/ O| .8/

D3840 O{ C 7680O| 2880Ok

ftlb:

(5)

(6)

(7)

(8)

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Statics 1e 297

Problem 4.21

Repeat Prob. 4.20, using a scalar approach.

Solution

We begin by adding the two cable forces at pointA to obtainEFA, and adding the two cable forces at pointCto obtainEFC, as follows

EFADEFAECEFADD 800 lb 8 O{ 6Ok10

C 800 lb 8O| 6Ok10

D 640 O{ C 640O| 960Ok lb; (1)EFCD EFCGCEFCOD 400 lb 20

O{ 9O| 12Ok25

C 400 lb9O| 12Ok

15 D

320 O{ 384O| 512Ok

lb: (2)

The components ofEFAand EFCare shown in the figure below.

Part (a) Taking positive moment components to be in the positive coordinate

directions, we use Eq. (4.1) on page 182 to determine the moment about point B

due to the forces in cableO C G as

.MB/x

D 512 lb.9 ft/

D 4610 ft

lb

.MB/yD 0

.MB/D 320 lb.9 ft/ D 2880 ftlb:

(3)

(4)

(5)

Part (b) Noting that the forces at pointA produce no moment about point A,

and taking positive moment components to be in the positive coordinate directions,

we use Eq. (4.1) on page 182 to write

.MA/xD 384 lb.6 ft/ 512 lb.9 ft/ D 2300 ftlb

.MA/yD 320 lb.6 ft/ D 1920 ftlb

.MA/D 320 lb.9 ft/ D 2880 ftlb:

(6)

(7)

(8)

Part (c) Taking positive moment components to be in the positive coordinate directions, we use Eq. (4.1)

on page 182 to write

.MO/xD 384 lb.12 ft/ 512 lb.9 ft/ 640 lb.6 ft/ D 3840 ftlb

.MO/yD 320 lb.12 ft/ C 640 lb.6 ft/ D 7680 ftlb

.MO/D 320 lb.9 ft/ D 2880 ftlb:

(9)

(10)

(11)

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298 Solutions Manual

Problem 4.22

StructureOAB is built in at point O and supports two forces of magnitude F

parallel to the y and axes. If the magnitude of the moment about point O

cannot exceed1:0 kNm, determine the largest valueFmay have.

Solution

Either the vector approach or the scalar approach may be conveniently used to determine the moment of the

the two forces about pointO , and we will use the scalar approach. With the dimensions provided, we use

Eq. (4.1) on page 182 to determine

.MO/xD F.90 mm/ (1)

.MO/yD F.400 mm/ (2)

.MO/D F.400 mm/ (3)The magnitude of the moment at pointO may not exceed1:0 kNm, thus

.MO/2xC .MO/2yC .MO/2

1=2 D 1:0 kNm

103mm

m

!

F

.90 mm/2 C .400 mm/2 C .400 mm/21=2 D 1000 kNmm(4)

(5)

which may be solved for

F

D1:75 kN: (6)

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Statics 1e 299

Problem 4.23

Repeat Prob. 4.22 if thex component of the moment (torsional component) at

pointO may not exceed0:5 kNmand the resultant of the y and components(bending components) may not exceed 0:8 kNm (i.e., qM2yC M2 0:8 kNm).

Solution

Either the vector approach or the scalar approach may be conveniently used to determine the moment of the

the two forces about pointO , and we will use the scalar approach. Taking positive moment components to be

in the positive coordinate directions, we use Eq. (4.1) on page 182 to determine

.MO/xD F.90 mm/ (1)

.MO/yD F.400 mm/ (2)

.MO/D F.400 mm/: (3)The absolute value of the torsional component of the moment,.MO/x , may not exceed0:50 kNm, thus

.MO/x

D F.90 mm/ 0:50 kNm

103mm

m

! ) F 5:56 kN: (4)

The bending component of the moment,q

.MO/2yC .MO/2, may not exceed0:80 kNm, thusq

.MO/2yC .MO/2 0:8 kNm (5)

Fq

.400 mm/2 C .400 mm/2 0:8 kNm103mmm

! ) F 1:41 kN: (6)Both Eqs. (4), and (6) are satisfied by

F 1:41 kN: (7)

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300 Solutions Manual

Problem 4.24

Forces of 3 kN and 200 N are exerted at points B andCof the main rotor of a

helicopter, and forceFis exerted at pointD on the tail rotor. The 3 kN forces

are parallel to the axis, the 200 N forces are perpendicular to the main rotor

and are parallel to the xy plane,Fis parallel to they axis, andD 45

.

(a) Determine the value ofFso that the component of the moment about

pointO of all rotor forces is zero.

(b) Using the value ofFfound in Part (a), determine the resultant moment of

all rotor forces about pointO .

(c) If is different than 45, do your answers to Parts (a) and (b) change?

Explain.

Solution

Part (a) We use Eq. (4.1) on page 182 to evaluate the moment of the rotor forces about the axis, and as

instructed in the problem statement, this moment is required to be zero. Thus, we obtain

.MO/D 200 N.3 m/ 200 N.3 m/ C F .5 m/ D 0 ) FD 240 N: (1)

Part (b) Using the geometry provided in the problem statement, with the aid of the sketch below to

determinex and y components of the main rotor forces, the following vector expressions may be written

EFB

D200 N sin 45

O{

cos 45

O|C 3000 NOk (2)EFCD 200 N sin 45 O{ C cos 45 O|C 3000 NOk (3)

EFDD F . O| / (4)ErOCD 3 m

cos 45 O{ C sin 45 O|C 1:8Okm (5)ErOBD 3 m

cos 45 O{ sin 45 O|C 1:8Okm (6)

ErODD 5 O{m C 1:2Okm (7)Using Eq. (4.2) on page 183, the moment about pointO of all rotor forces is

EMOD ErOBEFBC ErOCEFCC ErOD EFD (8)

D O{ O| Ok3 cos 45 3 sin 45 1:8

200 sin 45 200 cos 45 3000

Nm

C O{ O| Ok3 cos 45 3 sin 45 1:8

200 sin 45 200 cos 45 3000

Nm C

O{ O| Ok5 0 1:2

0 240 0

Nm

(9)

DhO{.6109/ O|.6618/ COk.600/

iNm C

hO{.6109/ O| .6618/ COk.600/

iNm

ChO{.288/ O|.0/ COk.1200/

iNm; (10)

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Statics 1e 301

which simplifies to

EMOD .288 O{/ N m: (11)

Part (c) If 45, the answer to Part (a) does not change, since the moment produced by the main rotorforces about theaxis is not a function of. The answer to Part (b) does not change, for the following reason.

The main rotor forces produce moments about thex and y axes that are functions of , but the moment arms

for components of force at pointsB andC are always opposite (negative of one another) so that the netx

andy components are zero regardless of.

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302 Solutions Manual

Problem 4.25

The moment of force EF about point A can be computed using EMA;1D ErABEFor using EMA;2D ErAC EF, whereB andCare points on the line of action of EF.Noting that

ErAC

D ErAB

C ErBC, show that

EMA;1

D EMA;2.

Solution

As specified in the problem statement,

MA;2D ErACEF ; and (1)ErACD ErABC ErBC: (2)

Equations (1) and (2) may be combined to write

EMA;2DErABC ErBC EF : (3)

Using the distributive property of the cross product, Eq. (3) may be expanded to write

EMA;2D ErABEFC ErBCEF : (4)

Because ErBC andEFare parallel, ErBCEFDE0. Hence, Eq. (4) becomes

MA;2D ErABEFD MA;1: (5)

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Statics 1e 303

Problem 4.26

The steering wheel of a Ferrari sports car has circular shape with 190 mmradius,

and it lies in a plane that is perpendicular to the steering column AB . PointC,

where the drivers hand applies force EF to the steering wheel, lies on they axiswithy coordinateyCD 190 mm. PointA is at the origin of the coordinatesystem, and pointB has the coordinatesB .120; 0;50/ mm. Determine themoment ofEF about lineAB of the steering column if(a) EF has10 N magnitude and lies in the plane of the steering wheel and has

orientation such that its moment arm to lineAB is 190 mm. Also determine

the vector expression for this force.

(b) EFD .10 N/ 18 O{ 3O|C 14Ok

23 .

(c) EFD

.10 N/12 O{ 5Ok

13

.

Solution

Part (a) The moment of forceEFabout lineAB can be evaluated using the scalar approach as

MABD .10 N/.190 mm/ D 1900 Nmm: (1)

To determine the vector expression for this force, note thatEF is perpendicular to both ErAB and ErAC. Thus,EFis parallel to the directionEnwhere

En D ErAB ErACD .120 O{ 50Ok/ mm .190O| / mmD O{ O| Ok120 0 50

0 190 0

mm2

D O{.9500/ O|.0/ COk.22800/mm: (2)Using Eq. (2), a unit vectorOnin the direction ofEnis

On D En

jEnjD 0:3846 O{ C 0:9231Ok: (3)

Note that rather than using Eqs. (2) and (3), it is possible to write an expression for the unit normal vector by

inspection asOn D .50 O{ C 120Ok/=130.Since the force EFis parallel toOn, the vector expression for the force is

EFD .10 N/.0:3846 O{ C 0:9231Ok/ D .3:85 O{ C 9:23Ok/ N: (4)

Note that it is possible that the force EF could be in the direction opposite to On, thus the answer EFD.3:85 O{ 9:23Ok/ N is also acceptable.This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission

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304 Solutions Manual

Part (b) Our strategy will be to determine the moment of force EFabout point A, and then use the dotproduct to determine the component of this moment that acts in the direction of line AB . The moment of

forceEF about pointA is

EMAD ErACEFD O{ O| Ok0 190 018 3 14

10 Nmm

23

D O{.2660/ O|.0/ COk.3420/10 Nmm23

: (5)

The component of EMAin the direction of line AB is

MABD EMA ErABjErAB j

D .2660/.120/ C .0/.0/ C .3420/.50/10 Nmm

23

1

130

!

D 472 Nmm: (6)

Part (c) Our strategy is the same as that used in Part (b). Thus, the moment of forceEFabout pointA is

EMAD ErACEFD O{ O| Ok0 190 0

12 0 5

10 Nmm

13

D O{.950/ O|.0/ COk.2280/10 Nmm13

: (7)

The component of

EMAin the direction of line AB is

MABD EMA ErABjErAB j

D .950/.120/ C .0/.0/ C .2280/.50/10 Nmm13

1

130

!

D 0: (8)

The reason that the forceEF produces no moment about lineAB is because EFis parallel to lineAB .

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Statics 1e 305

Problem 4.27

A rectangular piece of sheet metal is clamped along edge AB in a machine

called a brake. The sheet is to be bent along lineAB by applying a y direction

forceF. Determine the moment about lineAB ifFD 200 lb. Use both vectorand scalar approaches.

Solution

For the vector solution, we use the data given in the problem statement to write the following expressions

ErACD 36 O{in.; (1)EFD 200O| lb; (2)

ErABD

20 O{ 15Ok

in. (3)

We then use Eq. (4.2) on page 183 to determine the moment ofEF about pointA as

EMAD ErACEFD O{ O| Ok36 0 0

0 200 0

in.lb D O{.0/ O|.0/ COk.7200 in.lb/ D 7200Okin. lb: (4)

The component of EMAthat acts in the direction of line AB is given by

MABD EMA ErABrAB

D .0 in.lb/.20 in:/ C .0 in.lb/.0 in:/ C .7200 in.lb/.15 in:/25 in:

D 4320 in.lb;(5)

(6)

whererAB is the magnitude ofErAB . The negative answer in Eq. (6) means that the moment ofFabout lineAB is in the direction fromB to A. Note: We could have also used the scalar triple product to obtain Eq. (6).

For the scalar solution, we first determine the moment arm das shown in

the figure to the right. Through geometry and trigonometry, we find that

D tan11520

D 36:9; (7)dD 36 in: sin D 21:6 in: (8)

Then by applying Eq. (4.1) on page 182, taking positive moment to be in the

direction fromA to B according to the right-hand rule, it follows that

MABD 200 lb.21:6 in:/ D 4320 in.lb: (9)

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306 Solutions Manual

Problem 4.28

In Prob. 4.27 determineF if the moment about lineAB is to be2000 in.lb.

Solution

Using the data given in the problem statement, we write the following expressions

ErACD 36 O{in.; (1)EFD FO| ; (2)

ErABD

20 O{ 15Ok

in. (3)

Then by using the scalar triple product, we determine the moment of forceEFabout lineAB as

MABD ErACEF ErABrAB

D 125

20 0 1536 0 0

0 F 0

in. (4)

D 125

.15/.36/F in. (5)

WithMABD 2000 in. lb, we solve Eq. (5) for Fto obtain

FD 25 2000 in.lb.

15/.36/ in.

D 92:6 lb ) FD 92:6 lb: (6)

In writing our final answer forF, the negative sign is unimportant and is thus omitted.

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Statics 1e 307

Problem 4.29

In the pipe assembly shown, pointsB andClie in the xy plane and force F is

parallel to the axis. IfFD 150 N, determine the moment ofF about linesOAand AB . Use both vector and scalar approaches.

Solution

Using the figure to the right, we write the following expressions

ErOAD 350O| mm; (1)ErABD 350 mm

sin 30 O{ C cos 30 O| D .175 O{ C 303O| / mm; (2)

ErBCD 350 mm cos 30 O{ sin 30 O| D .260 O{ 150O| / mm; (3)EFD 150OkN: (4)

For the vector approach, we use Eq. (4.2) on page 183 to determine the moment of the force about point A,

namely EMA. Then, EMA ErOA=rOAproduces the moment of the force about line OA, and EMA ErAB=rABproduces the moment of the force about lineAB . Thus

EMAD ErACEFDErABC ErBC EFD

O{ O| Ok0:435 0:153 0

0 0 150

Nm (5)

DhO{.23:0/ O| .65:2/ COk.0/

iNm D .23:0 O{ C 65:2O| / Nm: (6)

MOAD EMA ErOArOA

D .65:2 Nm/.0:350 m/0:350 m

D 65:2 Nm;

MABD EMA ErABrAB

D .23:0 Nm/.0:175 m/ C .65:2 Nm/.0:303 m/0:350 m

D 44:9 Nm:

(7)

(8)

For the scalar approach, we use Eq. (4.1) on page 182 and the figure shown above, taking positive moment

MOAto be in the direction from O to A, and positive moment MAB to be in the direction fromA to B ,

according to the right-hand rule, to determine

MOAD F d1D 150 N .0:350 m/ sin 30 C .0:300 m/ cos 30 D 65:2 Nm;MABD F d2D 150 N .0:300 m/ D 45:0 Nm:

(9)

(10)

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308 Solutions Manual

Problem 4.30

In the pipe assembly shown, pointsB andClie in the xy plane and force F is

parallel to the axis. If a twisting moment (torque) of50 Nmwill cause a pipeto begin twisting in the flange fitting atO or at either end of the elbow fitting at

A, determine the first fitting that twists and the value ofF that causes it.

Solution

We use Eq. (4.1) on page 182, and the figure shown at the right for the determi-

nation of the moment armsd1 and d2, to determine the moments of force F about

linesOAand AB as

MOAD F d1D F

.0:350 m/ sin 30 C .0:300 m/ cos 30

; (1)

MABD

F d2D

F.0:300 m/: (2)

According to the failure criteria given in the problem statement, pipe OA will twist at O and A ifMOAreaches50 Nm, and pipeAB will twist at A and B ifMAB reaches50 Nm. Thus,

IfMOAD 50 Nm; then FD 50Nm

.0:350 msin 30 C 0:300 mcos 30/D 115 N; (3)

IfMABD 50 Nm; then FD 50Nm

.0:300 m/D 167 N: (4)

The smallest value ofF among Eqs. (3) and (4) is the value of the force that will cause twisting to occur.

Thus, the largest force F that may be applied is

FD 115 N; and twisting occurs at both threads on pipeOA. (5)

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Statics 1e 309

Problem 4.31

Determine the moment of force F about lineAB as follows.

(a) Determine the moment ofF about point A, EMA, and then determine thecomponent of this moment in the direction of line AB .

(b) Determine the moment ofF about pointB , EMB , and then determine thecomponent of this moment in the direction of line AB .

(c) Comment on differences and/or agreement between EMA, EMB , and themoment about line AB found in Parts (a) and (b). Also comment on the

meaning of the sign (positive or negative) found for the moment about line

AB .

Solution

With the information provided in the problem statement, the following position vectors may be written

ErABD .15 O{ 20O| / in.; ErACD20O|C 18Ok

in.; ErBCD

15 O{ C 18Ok

in. (1)

Part (a) We use Eq. (4.2) on page 183 to determine the moment of the force about pointA as

EMAD ErACEFD O{ O| Ok0 20 18

10 80 40

in.lb; (2)

EMAD hO{.640/ O| .180/ COk.200/i in.lb D 640 O{ C 180O|C 200

Ok in.lb: (3)The component of EMAin the direction of line AB is given by

MABD EMA ErABrAB

D .640/.15/ C .180/.20/ C .200/.0/25

in.lb D 528 in.lb: (4)

Part (b) We use Eq. (4.2) on page 183 to determine the moment of the force about pointB as

EMBD ErBCEFD

O{ O| Ok15 0 1810 80 40

in.lb (5)

(6)

EMBDhO{.1440/ O|.420/ COk.1200/

iin.lb D

1440 O{ 420O| 1200Ok

in.lb: (7)

The component of EMB in the direction of lineAB is given by

MABD EMB ErABrAB

D .1440/.15/ C .420/.20/ C .1200/.0/25

in.lb D 528 in.lb: (8)

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310 Solutions Manual

Part (c)

1. EMAand EMB are different, both in magnitude and direction.2. MAB is the same regardless if EMAor EMB is used.

3. The negative result forMAB means that the twisting action ofF is directed fromB toA.

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Statics 1e 311

Problem 4.32

Determine the moment of force F about lineAB as follows.

(a) Determine the moment ofF about point A, EMA, and then determine thecomponent of this moment in the direction of line AB .

(b) Determine the moment ofF about pointB , EMB , and then determine thecomponent of this moment in the direction of line AB .

(c) Comment on differences and/or agreement between EMA, EMB , and themoment about line AB found in Parts (a) and (b). Also comment on the

meaning of the sign (positive or negative) found for the moment about line

AB .

Solution

With the information provided in the problem statement, the following position vectors may be written

ErABD4 O{ 7O|C 4Ok

mm; ErACD

4 O{ C 4Ok

mm; ErBCD 7O| mm: (1)

Part (a) We use Eq. (4.2) on page 183 to determine the moment of the force about pointA as

EMAD ErACEFD 27

O{ O| Ok4 0 4

3 2 6

Nmm; (2)

EMA

D 2

7hO{.8/ O| .36/ COk.8/iNmm: (3)

The component of EMAin the direction of line AB is given by

MABD EMA ErABrAB

D 27

Nmm .8/.4/ C .36/.7/ C .8/.4/19D 8:00 Nmm: (4)

Part (b) We use Eq. (4.2) on page 183 to determine the moment of the force about pointB as

EMBD ErBCEFD 27

O{ O| Ok0 7 0

3 2 6

Nmm; (5)

EMBD 27

hO{.42/ O|.0/ COk.21/

iNmm: (6)

The component of EMB in the direction of lineAB is given by

MABD EMB ErABrAB

D 27

Nmm .42/.4/ C .0/.7/ C .21/.4/19D 8:00 Nmm: (7)

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312 Solutions Manual

Part (c)

1. EMAand EMB are different, both in magnitude and direction.2. MAB is the same regardless if EMAor EMB is used.

3. The negative result forMAB means that the twisting action ofF is directed fromB toA.

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Statics 1e 313

Problem 4.33

The moment of force EFabout lineAB can be computed usingMAB;1 or MAB;2where

MAB;1D .ErACEF / ErABjErABj ; MAB;2D .ErBCEF / E

rABjErABj ;

whereCis a point on the line of action ofEF. Noting thatErACD ErABC ErBC,show thatMAB;1D MAB;2.

Solution

We summarize the information provided in the problem statement by writing the following expressions

MAB;1D ErACEF

ErABrAB

; (1)

MAB;2DErBCEF

ErABrAB

; (2)

ErACD ErABC ErBC: (3)

Substituting Eq. (3) into Eq. (1) provides

MAB;1DErABC ErBC EF ErAB

rAB(4)

DErABEFC ErBCEF

ErABrAB

(5)

D ErABEF ErABrAB C ErBCEF ErAB

rAB : (6)

In Eq. (6), the result ofErABEF

is a vector that is orthogonal toErAB , therefore,

ErABEF

ErABD 0.

Thus, Eq. (6) reduces to

EMAB;1D 0 CErBCEF

ErABrAB

: (7)

Equation (7) is identical to Eq. (2), thus

EMAB;1D EMAB;2: (8)

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314 Solutions Manual

Problem 4.34

An automobile windshield wiper is actuated by a force EFD .120 O{ 50O| / N.Determine the moment of this force about shaftOB .

Solution

We use Eq. (4.2) on page 183 to determine the moment of forceEFabout pointO as

EMOD ErOA EFD

O{ O| Ok0 60 80

120

50 0

Nmm (1)

DhO{.4000/ O| .9600/ COk.7200/

iNmm: (2)

The component of EMO about lineOB is

MOBD EMO ErOBrOB

D .4000/.20/ C .9600/.90/ C .7200/.60/110

NmmD 12;500 Nmm D 12:5 Nm:

(3)

(4)

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Statics 1e 315

Problem 4.35

In the windshield wiper of Prob. 4.34, ifEF has130 N magnitude, determine thedirection in which it should be applied so that its moment about shaft OB is as

large as possible, and determine the value of this moment.

Solution

To maximize the moment ofFabout lineOB ,EFshould be perpendicular to the plane containing lines OAandOB . Thus, we determine a vectorEn, and unit vectorOn, that are perpendicular to this plane as

En

D ErOB

ErOA

D O{ O| Ok

20 90 60

0 60 80 mm2

D hO{.10800/ O|.1600/ COk.1200/imm2; (1)

On DEnnD 0:983 O{ 0:146O| 0:109Ok: (2)

UsingOn, we constructEF as

EFD 130 N On D 130 N

0:983 O{ 0:146O| 0:109Ok

D

128 O{ 19:0O| 14:2Ok

N: (3)

We then use the scalar approach to determine the moment of this force about line OB as

MOBD F dD F rOAD .130 N/q

.60 mm/2 C .80 mm/2 D 13;000 Nmm D 13 Nm: (4)

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316 Solutions Manual

Problem 4.36

Wrench AB is used to twist a nut on a threaded shaft. Point A is located

120 mmfrom pointO , and linea has x,y , and direction angles of36,60,

and72, respectively. PointB is located at.0; 300; 100/ mm, and the force is

EFD .80O|C 20Ok/ N.(a) Determine the moment ofEFabout pointA, EMA, and then find the compo-

nent of this moment about line a,Ma.

(b) Determine the moment ofEFabout pointO , EMO , and then find the compo-nent of this moment about line a,Ma.

Solution

Part (a) Using the coordinates and direction angles provided, the following position vectors may be written

ErOAD 120 mm

cos 36 O{ C cos 60 O|C cos 72 Ok

D

97:1 O{ C 60O|C 37:1Ok

mm; (1)

ErOBDhO{.0 0/ C O|.300 0/ COk.100 0/

imm D

300O|C 100Ok

mm: (2)

It then follows that

ErABD ErAOC ErOBD ErOAC ErOB (3)D97:1 O{ C 240O|C 62:9Ok

mm: (4)

We use Eq. (4.2) on page 183 to determine the moment of forceEFabout pointA as

EMAD ErABEFD O{ O| Ok97:1 240 62:9

0 80 20

Nmm

DhO{.9830/ O| .1940/ COk.7770/

iNmm D

9830 O{ C 1940O|C 7770Ok

Nmm:

(5)

(6)

The component of EMAabout linea is given by

MaD EMA OrOAD .9830/.cos 36/ C .1940/.cos 60/ C .7770/.cos 72/

Nmm

D 11;300 Nmm D 11:3 Nm:

(7)

(8)

Part (b) We use Eq. (4.2) to determine the moment of forceEF about pointO as

EMOD ErOBEFD O{ O| Ok0 300 100

0 80 20

Nmm

DhO{.14000/ O|.0/ COk.0/

iNmm D 14000 O{N mm:

(9)

(10)

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Statics 1e 317

The component of EMO about linea is given by

MaD EMO OrOAD

.14000/.cos 36/ C .0/.cos 60/ C .0/.cos 72/NmmD 11;300 Nmm D 11:3 Nm:

(11)

(12)

Observe that EMAand EMO are different, yetMa is the same.

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318 Solutions Manual

Problem 4.37

A trailer has a rectangular door DEGH hinged about edge GH. If EFD.2 O{ C 5O|C 14Ok/ lb, determine the moment ofFabout edgeGH.

Solution

With the coordinates provided in the problem statement, the following position vector may be written

ErHDD

16O| 32Ok

in. (1)

We then use Eq. (4.2) on page 183 to determine the moment of force EF about pointH as

EMHD ErHD EFD O{ O| Ok0 16 32

2 5 14

in. lb; (2)

Dh

384 O{ .64/O|C 32Oki

in.lb: (3)

The component of EMHin the direction of lineGHis simply the negative of the x component of EMH (or,evaluate the dot productMGHD EMH .O{/ to obtain the following result)

MGHD 384 in.lb: (4)

The negative sign for MGHmeans that the twisting action of the force is in the direction from H to G ,

according to the right-hand rule.

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Statics 1e 319

Problem 4.38

In the trailer of Prob. 4.37, ifFD 15 lb, determine the direction in which itshould be applied so that its moment about edge GHof the rectangular door is

as large as possible, and determine the value of this moment.

Solution

To maximize the moment ofEFabout lineGH,EFshould be applied perpendicular (normal) to the door. Fromthe information provided in the problem statement, the following position vectors may be written

OrHGD O{; (1)ErHDD hO{.40 C 40/ C O|.24 8/ C

Ok.20 52/i in. D 16O| 32Ok in. (2)

We then determine a unit vector normal to these two vectors as

En D OrHG ErHDD O{ O| Ok1 0 0

0 16 32

in. D h O| .32/ COk.16/i in. (3)

On DEnnD

32O|C 16Ok

in.p

.32 in:/2 C .16 in:/2D 0:894O|C 0:447Ok: (4)

We then construct EF in this direction as

EFD 15 lb On D 15 lb 0:894O|C 0:447Ok D 13:4O|C 6:71Ok lb: (5)Using the scalar approach, the moment ofEFabout lineGH is

MGHD F rHDD .15 lb/q

.16 in:/2 C .32 in:/2 D 537 in.lb: (6)

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320 Solutions Manual

Problem 4.39

A trailer has a triangular door AB C hinged about edgeB C. IfEQ D . O{ C4O| C8Ok/ lb, determine the moment ofQ about edgeB C.

Solution

With the coordinates provided in the problem statement, the following position vectors may be written

ErBADhO{.0 24/ C O|.28 4/ COk.12 12/

iin. D .24 O{ C 24O| / in.; (1)

ErBCDhO{.0 24/ C O| .4 4/ COk.60 12/

iin. D

24 O{ C 48Ok

in. (2)

We then use Eq. (4.2) on page 183 to determine the moment of force EQabout pointB as

EMBD ErBA EQ D O{ O| Ok24 24 0

1 4 8

in.lb; (3)

Dh

192 O{ .192/O|C .120/Oki

in.lb: (4)

The component of EMB in the direction of lineB C is

MBCD EMB ErBC

rBC D .192/.

24/

C.192/.0/

C.

120/.48/ in.2

lbp

.24 in:/2 C .48 in:/2D 193 in.lb:

(5)

(6)

The negative sign for MBC means that the twisting action of the force is in the direction from C to B,

according to the right-hand rule.

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Statics 1e 321

Problem 4.40

In the trailer of Prob. 4.39, ifQD 9 lb, determine the direction in which itshould be applied so that its moment about edgeB C of the triangular door is

as large as possible, and determine the value of this moment.

Solution

To maximize the moment ofEQabout lineB C, EQshould be applied perpendicular (normal) to the door. Fromthe information provided in the problem statement, the following position vectors may be written

ErBAD .24 O{ C 24O| / in. D . O{ C O|/24 in:; (1)ErBCD .24 O{ C 48O| / in. D O{ C 2

Ok 24 in: (2)We then determine a unit vector normal to these two vectors as

En D ErBA ErBCD O{ O| Ok1 1 0

1 0 2

.24 in:/2 D hO{.2/ O| .2/ COk.1/i .24 in:/2 (3)

On DEnnD

.24 in:/2

2 O{ C 2O|COk

.24 in:/2p

.2/2 C .2/2 C .1/2D 0:667 O{ C 0:667O|C 0:333Ok: (4)

We then construct EQin this direction as

EQ D 9 lb

0:667 O{ C 0:667O|C 0:333Ok

D

6 O{ C 6O|C 3Ok

lb: (5)

Using Eq. (4.2) on page 183, the moment of EQabout pointB is

EMBD ErBA EQ D O{ O| Ok1 1 0

6 6 3

24 in.lb (6)

DhO{.3/ O| .3/ COk.12/

i24 in.lb D

3 O{ C 3O| 12Ok

24 in.lb: (7)

The component of EMB about lineB C is

MBCD EMB ErBCrBC

D 1.3/ C 0.3/ C 2.12/ .24 in.lb/.24 in:/p.1/2 C .2/2.24 in:/

D 290 in.lb: (8)

The negative sign for MBC means that the twisting action of the force is in the direction from C to B,

according to the right-hand rule.

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322 Solutions Manual

Problem 4.41

A poorly leveled crane rotates about linea, which lies in the xy plane and has

ay direction angle of5. LineAB liesin thexplane. If the crane supports a

weightWD 5000lb andD 45, determine the moment ofWabout linea.

Solution

From the information provided in the problem statement, the following vector expressions may be written

EWD 5000O|lb; (1)ErABD 40 ft

cos 45 O{ C sin 45 Ok

: (2)

Using Eq. (4.2) on page 183, the moment of

EWabout pointA is

EMAD ErABEWD O{ O| Ok cos 45 0 sin 45

0 1 0

.40 ft/.5000 lb/ (3)

D 2105 ftlbhO{.0:707/ O|.0/ COk.0:707/

iD 1:41105 ftlb

O{ COk

: (4)

Noting that thea direction lies in thexy plane and has a y direction angle of 5, the unit vectorOuin theadirection is

Ou D sin 5 O{ C cos 5 O| : (5)The component of EMAin thea direction is given by

MaD EMA Ou D EMA sin 5 O{ C cos 5 O|

D 1:41105 ftlb .1/. sin 5/ C .0/.cos 5/ C .1/.0/D 12;300 ftlb:

(6)

(7)

(8)

The negative sign forMa means that the twisting action of the weight is in the adirection, according to theright-hand rule.

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Statics 1e 323

Problem 4.42

If the poorly leveled crane of Prob. 4.41 can have any position , where0 135, determine the largest moment of the weight WD 5000 lb aboutline a and the position that produces this moment. AssumehD 40 ft forany position (this requires the operator to slightly change the booms heightand/or length as the crane rotates about linea).

Solution

From the information provided in the problem statement, the following vector expressions may be written

EWD 5000O|lb; (1)ErABD 40 ft

cos O{ C sin Ok

: (2)

Using Eq. (4.2) on page 183, the moment of

EWabout pointA is

EMAD ErABEWD O{ O| Ok cos 0 sin

0 1 0

.40 ft/.5000 lb/ (3)

D 2105 ftlbh sin O{ cos Ok

i: (4)

Noting that thea direction lies in thexy plane and has a y direction angle of 5, the unit vectorOuin theadirection is

Ou D sin 5 O{ C cos 5 O| : (5)The component of EMAin thea direction is given by

MaD EMA Ou D EMA sin 5 O{ C cos 5 O| (6)

D 2105 ftlb . sin /. sin 5/ C .0/.cos 5/ C . cos /.0/ (7)D 17;400 ftlb.sin /: (8)

From this we see that Ma is maximum when sin D 1, which occurs when D 90. Therefore themaximum moment about linea is

MaD 17;400 ftlb at D 90: (9)

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324 Solutions Manual

Problem 4.43

Three-wheel and four-wheel all-terrain vehicles (ATVs) are shown.

For both ATVs, the combined weight of the driver and vehicle is

W D 2600 N. During a hard turn, the tendency for the ATVto tip is modeled by the force F which acts in the negativey direction. Determine the value of Fsuch that the resultant moment of

F and the weight about line AB is zero. Comment on whichATV is more

prone to tip during hard turns. Note: One of these models ofATVare no longer

manufactured because of their poor safety record.

Solution

We first consider the three-wheel ATV. With pointO being the origin of the coordinate system, the following

vector expressions may be written

ErAOD0:9 O{ C 0:8Ok

m; (1)

EQ DEFCEWD FO| 2600 NOk: (2)

Note that the lines of action of bothEF and EW intersect at pointO . Using Eq. (4.2) on page 183, the momentof force EQabout pointA is

EMAD ErAOEQ D O{ O| Ok

0:9 m 0 0:8 m0 F 2600 N D hO{.0:8 mF / O|.2340 Nm/ COk.0:9 mF /i : (3)

Noting that the position vector from pointA to pointB is

ErABD .1:6 O{ C 0:6O| / m; (4)the component of EMAabout lineAB is given by

MABD EMA ErABrAB

(5)

D .0:8 mF /.1:6 m/ C .2340 Nm/.0:6 m/ C .0:9 mF/.0 m/p.1:6 m/2

C.0:6 m/2

: (6)

As instructed in the problem statement,MABD 0, hence we set Eq. (6) equal to zero and solve for F toobtain

FD 1404 Nm1:28 m D 1096 N ) FD 1096 N: (7)

In Eq. (7), the negative sign for Fis irrelevant and is thus ignored.

For the four-wheelATV, we use the scalar approach (Eq. (4.1) on page 182) to determine the moment of

the forces about lineAB as

MABD F.0:8 m/ 2600 N.0:6 m/: (8)This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission

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Statics 1e 325

As instructed in the problem statement,MABD 0, hence we set Eq. (8) equal to zero and solve for F toobtain

FD 1950 N: (9)Comparing Eqs. (7) and (9), we see that the force required to tip the four-wheelATVis 78% larger than that

for the three-wheel ATV. Note that three-wheelATVs are no longer manufactured because of their poor safety

record.

Photo credit: Richard Hamilton Smith/CORBIS.

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