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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
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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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