Mechanical Engineering Chapter 4 Problem The System Equilibrium Youy Represent The Forces Fab And Fac Force

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subject Authors Anthony M. Bedford, Wallace Fowler

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Problem 4.146 The system is in equilibrium. If you
represent the forces FAB and FAC by a force Facting at
Aand a couple M, what are Fand M?
A
A
BC
60°40°
y
FAB FAC
Problem 4.147 Three forces act on a beam.
(a) Represent the system by a force Facting at the
origin Oand a couple M.
(b) Represent the system by a single force. Where does
the line of action of the force intersect the xaxis?
y
5 m
30 N
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Problem 4.148 The tension in cable AB is 400 N, and
the tension in cable CD is 600 N.
(a) If you represent the forces exerted on the left post
by the cables by a force Facting at the origin O
and a couple M, what are Fand M?
(b) If you represent the forces exerted on the left post
by the cables by the force Falone, where does its
line of action intersect the yaxis?
400 mm
A
C
B
y
The tension in CD is
The sum of the moments acting on the left post is the product of the
moment arm and the x-component of the tensions:
MD0.7357.77k0.3561.8kD419kN-m
Check: The position vectors at the point of application are rAB D0.7j,
and rCD D0.3j. The sum of the moments is
MDrAB ðTABCrCD ðTCD
D
ijk
00.70
357.77 178.89 0
C
ijk
00.30
561.8210.67 0
D0.7357.77k0.3561.8kD419k
from which DD419
919.6D0.456 m, Check.
240
c
2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
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Problem 4.149 Consider the system shown in Problem
4.148. The tension in each of the cables AB and CD is
400 N. If you represent the forces exerted on the right
post by the cables by a force F, what is F, and where
does its line of action intersect the yaxis?
732.19 D0.496 m on the positive yaxis.
Problem 4.150 If you represent the three forces acting
on the beam cross section by a force F, what is F, and
where does its line of action intersect the xaxis?
y
500 lb
800 lb
6 in
FYD800j.
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Problem 4.151 In Active Example 4.12, suppose that
the force FBis changed to FBD20i15jC30k(kN),
and you want to represent system 1 by an equivalent
system consisting of a force Facting at the point Pwith
coordinates (4, 3, 2) m and a couple M(system 2).
Determine Fand M.
y
FA
P
(4, 3, 2) m
System 1
FB
Solution: From Active Example 4.12 we know that
Problem 4.152 The wall bracket is subjected to the
force shown.
(a) Determine the moment exerted by the force about
the zaxis.
(b) Determine the moment exerted by the force about
the yaxis.
(c) If you represent the force by a force Facting at O
and a couple M, what are Fand M?
12 in
z
y
x
O10i – 30j + 3k (lb)
242
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Problem 4.153 A basketball player executes a “slam
dunk” shot, then hangs momentarily on the rim, exerting
the two 100-lb forces shown. The dimensions are hD
14 1
(b) The glass backboard will shatter if jMj>4000 in-
y
O
z
x
h
–100j (lb)
r
α
Solution: The equivalent force at the origin must equal the sum of
the forces applied: FEQ D200j. The position vectors of the points of
application of the forces are r1D⊲h Cr⊳i, and r2Di⊲h Crcos ˛⊳
krsin ˛. The moments about the origin are
D
ijk
2hCr⊲1Ccos ˛⊳ 0rsin ˛
For the values of h,r, and ˛given, the moment is MD822.72i
Problem 4.154 In Example 4.14, suppose that the 30-
lb upward force in system 1 is changed to a 25-lb upward
force. If you want to represent system 1 by a single force
F(system 2), where does the line of action of Fintersect
the xzplane?
y
x
System 1
(2, 0, 4) ft
(6, 0, 2) ft
(3, 0, –2) ft
x
O
20j (lb) 30j (lb)
F
Solution: The sum of the forces in system 2 must equal the sum
of the forces in system 1:
The sum of the moments about a point in system 2 must equal the sum
of the moments about the same point is system 1. We sum moments
about the origin.
⊲M2D⊲M1
Expanding the determinants results in the equations
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Problem 4.155 The normal forces exerted on the car’s
tires by the road are
NAD5104j(N),
NBD5027j(N),
NCD3613j(N),
NDD3559j(N).
If you represent these forces by a single equivalent force
N, what is N, and where does its line of action intersect
the xzplane?
1.4 m
x
z
y
0.8 m
1.4 m
DB
CA
0.8 m
Solution: We must satisfy the following three equations
Problem 4.156 Two forces act on the beam. If you
represent them by a force Facting at Cand a couple M,
what are Fand M?
Solution: The equivalent force must equal the sum of forces: FD
100jC80k. The equivalent couple is equal to the moment about C:
244
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Problem 4.157 An axial force of magnitude Pacts on
the beam. If you represent it by a force Facting at the
origin Oand a couple M, what are Fand M?
Pi
b
Problem 4.158 The brace is being used to remove a
screw.
(a) If you represent the forces acting on the brace by
a force Facting at the origin Oand a couple M,
what are Fand M?
(b) If you represent the forces acting on the brace by
a force F0acting at a point Pwith coordinates
⊲xP,y
P,z
Pand a couple M0, what are F0and M0?x
z
B
B
y
O
A
h
h
r
A
1
2
A
1
2
Solution: (a) Equivalent force at the origin Ohas the same value
as the sum of forces,
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Problem 4.159 Two forces and a couple act on the
cube. If you represent them by a force Facting at point
Pand a couple M, what are Fand M?
y
2ij (kN)
FB =
P
FA =
Solution: The equivalent force at Phas the value of the sum of
forces,
Problem 4.160 The two shafts are subjected to the
torques (couples) shown.
(a) If you represent the two couples by a force Facting
at the origin Oand a couple M, what are Fand M?
(b) What is the magnitude of the total moment exerted
by the two couples?
y
4 kN-m
6 kN-m
40°
30°
Solution: The equivalent force at the origin is zero, FD0 since
there is no resultant force on the system. Represent the couples of
4 kN-m and 6 kN-m magnitudes by the vectors M1and M2. The
246
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Problem 4.161 The two systems of forces and
moments acting on the bar are equivalent. If
FAD30iC30j20k(kN),
FBD40i20jC25k(kN),
MBD10iC40j10k(kN-m),
what are Fand M?
z
y
x
MB
FA
FB
A
B
2 m
2 m
Problem 4.162 Point Gis at the center of the block.
FBD10j10k(lb).
If you represent the two forces by a force Facting at G
and a couple M, what are Fand M?
Solution: The equivalent force is the sum of the forces:
The equivalent couple is the sum of the moments about G. The position
vectors are:
rAD15iC5jC10k(in),
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Problem 4.163 The engine above the airplane’s fuse-
lage exerts a thrust T0D16 kip, and each of the engines
under the wings exerts a thrust TUD12 kip. The dimen-
sions are hD8ft,cD12 ft, and bD16 ft. If you repre-
sent the three thrust forces by a force Facting at the
origin Oand a couple M, what are Fand M?
y
bb
z
y
2 T
U
T0
c
h
O
Solution: The equivalent thrust at the point Gis equal to the sum
of the thrusts:
The sum of the moments about the point Gis
The position vectors are r1UDCbihj,r2UDbihj, and rOD
Ccj. For hD8 ft, cD12 ft, and bD16 ft, the sum of the moments
Problem 4.164 Consider the airplane described in
by the force Falone, where does its line of action
intersect the xyplane?
Solution: The sum of the forces is now
For hD8 ft, cD12 ft, and bD16 ft, using the position vectors for
the engines given in Problem 4.163, the equivalent couple is
248
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Problem 4.165 The tension in cable AB is 100 lb, and
the tension in cable CD is 60 lb. Suppose that you want
to replace these two cables by a single cable EF so that
the force exerted on the wall at Eis equivalent to the
two forces exerted by cables AB and CD on the walls
at Aand C. What is the tension in cable EF, and what
are the coordinates of points Eand F?
x
y
z
x
y
z
B
A
D
C(4, 6, 0) ft
(7, 0, 2) ft
(3, 0, 8) ft
E
F
(0, 6, 6) ft
Solution: The position vectors of the points A,B,C, and Dare
The unit vectors parallel to the cables are obtained as follows:
eAB D0.4286i0.8571jC0.2857k.
rCD DrDrCD3i6jC2k,
Since eAB DeCD,the cables are parallel. To duplicate the force, the
single cable EF must have the same unit vector.
For the systems to be equivalent, the moments about the origin must
D788.57iC188.57j617.14k.
D788.57iC188.57j617.14k,
Thus the coordinates of point Eare E(0, 9, 2.75) ft. The coordinates
of the point Fare found as follows: Let Lbe the length of cable
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Problem 4.166 The distance sD4 m. If you represent
the force and the 200-N-m couple by a force Facting at
origin Oand a couple M, what are Fand M?
(2, 6, 0) m
y
100i 20j 20k (N)
s
Solution: The equivalent force at the origin is
The position vector of the action point from the origin is
250
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Problem 4.167 The force Fand couple Min system
1 are
FD12iC4j3k(lb),
MD4iC7jC4k(ft-lb).
Suppose you want to represent system 1 by a
wrench (system 2). Determine the couple Mpand the
coordinates xand zwhere the line of action of the force
intersects the xzplane.
y
z
x
M
F
O
y
z
x
Mp
F
O
(x, 0, z)
System 1 System 2
Solution: The component of Mthat is parallel to Fis found as
follows: The unit vector parallel to Fis
The component of Mnormal to Fis
4D1.2840 ft
Problem 4.168 A system consists of a force Facting
at the origin Oand a couple M, where
FD10i(lb),MD20j(ft-lb).
If you represent the system by a wrench consisting of
the force Fand a parallel couple Mp, what is Mp, and
where does the line of action Fintersect the yzplane?
page-pfe
Problem 4.169 A system consists of a force Facting
at the origin Oand a couple M, where
FDiC2jC5k(N),MD10iC8j4k(N-m).
If you represent it by a wrench consisting of the force F
and a parallel couple Mp, (a) determine Mp, and deter-
mine where the line of action of Fintersects (b) the xz
plane, (c) the yzplane.
Solution: The unit vector parallel to Fis
MNDMMPD9.8iC7.6j5k.
from which
D9.8iC7.6j5k,
from which
Problem 4.170 Consider the force Facting at the origin
Oand the couple Mgiven in Example 4.15. If you repre-
z
F
Solution: From Example 4.15 the force and moment are FD3iC
6jC2k(N), and MD12iC4jC6k(N-m).
The moment produced by the force must equal the normal component:
ijk
xD4.816
252
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Problem 4.171 Consider the force Facting at the ori-
gin Oand the couple Mgiven in Example 4.15. If you
represent this system by a wrench, where does the line
of action of the force intersect the plane yD3m?
Solution: From Example 4.15 (see also Problem 4.170) the force
is FD3iC6jC2k, and the normal component of the moment is
MND7.592i4.816jC3.061k.
Problem 4.172 A wrench consists of a force of magni-
tude 100 N acting at the origin Oand a couple of magni-
tude 60 N-m. The force and couple point in the direction
Solution: The vector parallel to the force is rFDiCjC2k, from
which the unit vector parallel to the force is eFD0.4082iC0.4082jC
0.8165k. The force and moment at the origin are
The force and moment are parallel. At the point (5, 3, 1) m the equiva-
40.82 40.82 81.65
Problem 4.173 System 1 consists of two forces and
a couple. Suppose that you want to represent it by a
Solution: The sum of the forces in System 1 is FD300jC
600k(N). The equivalent force in System 2 must have this value.

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