PROBLEM 4.54
A small winch is used to raise a 120–lb load.
Find (a) the magnitude of the vertical force P
that should be applied at C to maintain
equilibrium in the position shown, (b) the
reactions at A and B, assuming that the bearing at
B does not exert any axial thrust.
SOLUTION
Dimensions in in.
(32 in.) (10 in.)cos30 (10 in.)sin30
32 8.6603 5
C=+−
=+−
ri j k
i jk
0: (10 4 ) ( 120 ) (20 ) ( ) (32 8.6603 5 ) ( ) 0
1200 480 20 20 32 5 0
A yz
yz
BB P
B B PP
Σ = + ×− + × + + + − ×− =
− + + − − −=
M ik j i j k jk j
i
k i k j ki
Equating the coefficients of the unit vectors to zero,
: 480 5 ) 0 96.0 lb;PP−= =i
( ) 96.0 lbaP
=
: 20 0
z
B=j
0
z
B=
: 1200 20 32(96.0) 0
y
B−+ − =k
213.6 lb
y
B=
0:
x
FΣ=
0
x
A=
0:
y
FΣ=
120 213.6 96.0 0
y
A−+ − =
2.40 lb
y
A=
0:
z
FΣ=
0
zz
AB+=
0
zz
AB=−=
(b)
(2.40 lb) ; (214 lb)= =A jB j
PROBLEM 4.55
A 200–mm lever and a 240–mm–diameter pulley are
welded to the axle BE that is supported by bearings at C
and D. If a 720–N vertical load is applied at A when the
lever is horizontal, determine (a) the tension in the cord,
(b) the reactions at C and D. Assume that the bearing
at D does not exert any axial thrust.
SOLUTION
Dimensions in mm
We have six unknowns and six equations of equilibrium. —OK
0: ( 120 ) ( ) (120 160 ) (80 200 ) ( 720 ) 0
C xy
DD TΣ = − × + + − × + − ×− =M k i j j ki k i j
..
Equating to zero the coefficients of the unit vectors:
:k
3
120 144 10 0T− +×=
(a)
1200 NT=
:i
3
120 57.6 10 0 480 N
yy
DD+ ×= =−
: 120 160(1200 N) 0
x
D−− =j
1600 N
x
D= −
0:
x
FΣ=
0
xx
CDT+ +=
1600 1200 400 N
x
C=−=
0:
y
FΣ=
720 0
yy
CD+− =
480 720 1200 N
y
C=+=
0:
z
FΣ=
0
z
C=
(b)
(400 N) (1200 N) ; (1600 N) (480 N)=+ =−−C i jD i j
PROBLEM 4.56
Solve Problem 4.55, assuming that the axle has been
rotated clockwise in its bearings by 30° and that the
720-N load remains vertical.
PROBLEM 4.55 A 200–mm lever and a 240–mm–
diameter pulley are welded to the axle BE that is
supported by bearings at C and D. If a 720–N vertical
load is applied at A when the lever is horizontal,
determine (a) the tension in the cord, (b) the reactions at
C and D. Assume that the bearing at D does not exert
any axial thrust.
SOLUTION
Dimensions in mm
We have six unknowns and six equations of equilibrium.
(a)
0: ( 120 ) ( ) (120 160 ) (80 173.21 ) ( 720 ) 0
C xy
DD TΣ = − × + + − × + − ×− =M k i j j ki k i j
33
120 120 120 160 57.6 10 124.71 10 0
xy
D DTT− + − − +×+ × =j ikj i k
Equating to zero the coefficients of the unit vectors,
3
: 120 124.71 10 0 1039.2 NTT− + ×= =k
1039 NT=
3
: 120 57.6 10 0 480 N
yy
DD+ ×= =−i
: 120 160(1039.2)
x
D−−
j
1385.6 N
x
D= −
0:
x
FΣ=
0
xx
CDT+ +=
1385.6 1039.2 346.4
x
C=−=
0:
y
FΣ=
720 0
yy
CD+− =
480 720 1200 N
y
C=+=
0:
z
FΣ=
0
z
C=
(b)
(346 N) (1200 N) (1386 N) (480 N)=+ =−−C i jD i j
PROBLEM 4.57
The rectangular plate shown weighs 80 lb and is supported by three
vertical wires. Determine the tension in each wire.
SOLUTION
Free–Body Diagram:
///
0: ( 80 lb) 0
B ABA CBC GB
T TrΣ = × + × + ×− =M r jr j j
(60 in.) [(60 in.) (15 in.) ] [(30 in.) (30 in.) ] ( 80 lb) 0
AC
TT× + + × + + ×− =kj i k j i k j
60 60 15 2400 2400 0
AC C
TT T−+ −− + =
ikik i
Equating to zero the coefficients of the unit vectors:
:i
60 15(40) 2400 0
A
T− +=
30.0 lb
A
T=
:k
60 2400 0
C
T−=
40.0 lb
C
T=
0:
y
FΣ=
80 lb 0
ABC
TTT++− =
30 lb 40 lb 80 lb 0
B
T++ − =
10.00 lb
B
T=
PROBLEM 4.58
The rectangular plate shown weighs 80 lb and is supported by three
vertical wires. Determine the weight and location of the lightest block
that should be placed on the plate if the tensions in the three wires are
to be equal.
SOLUTION
Free–Body Diagram:
Let
b
W−j
be the weight of the block and x and z the block’s coordinates.
SOLUTION Continued
Solving (4) and (5) for
/
b
WW
and recalling of
0 60 in.,x≤≤
0 90 in.,z≤≤
(4):
55
0.125
40 40 0
b
W
Wz
=≥=
−−
(5):
10 10 0.5
20 20 0
b
W
Wx
=≥=
−−
Thus,
min
( ) 0.5 0.5(80) 40 lb
b
WW
= = =
min
( ) 40.0 lb
b
W=
Making
0.5
b
WW=
in (4) and (5):
5 ( 40)(0.5 ) 0Wz W+− =
30.0 in.z=
10 ( 20)(0.5 ) 0Wx W− −− =
0 in.x=
PROBLEM 4.59
An opening in a floor is covered by a
1 1.2-m×
sheet of
plywood of mass 18 kg. The sheet is hinged at A and B and
is maintained in a position slightly above the floor by a
small block C. Determine the vertical component of the
reaction (a) at A, (b) at B, (c) at C.
SOLUTION
/
/
/
0.6
0.8 1.05
0.3 0.6
BA
CA
GA
=
= +
= +
ri
r ik
r ik
(18 kg)9.81
176.58 N
W mg
W
= =
=
///
0: ( ) 0
A BA CA GA
M BC WΣ = × + × + ×− =
r jr jr j
(0.6 ) (0.8 1.05 ) (0.3 0.6 ) ( ) 0BC W× + + × + + ×− =ij i kj i k j
0.6 0.8 1.05 0.3 0.6 0B C CW W+ − − +=kk iki
Equate coefficients of unit vectors to zero:
0.6
: 1.05 0.6 0 176.58 N 100.90 N
1.05
CW C
+== =
i
: 0.6 0.8 0.3 0BCW
+− =k
0.6 0.8(100.90 N) 0.3(176.58 N) 0 46.24 NBB+−==−
0: 0
y
F ABCWΣ = ++− =
46.24 N 100.90 N 176.58 N 0 121.92 NAA−+ + ==
( ) 121.9 N ( ) 46.2 N ( ) 100.9 NaA bB cC= =−=
consent of McGraw–Hill Education.
PROBLEM 4.60
Solve Problem 4.59, assuming that the small block C is
moved and placed under edge DE at a point 0.15 m from
corner E.
PROBLEM 4.59 An opening in a floor is covered by a
1 1.2-m×
sheet of plywood of mass 18 kg. The sheet is
hinged at A and B and is maintained in a position slightly
above the floor by a small block C. Determine the vertical
component of the reaction (a) at A, (b) at B, (c) at C.
SOLUTION
/
/
/
0.6
0.65 1.2
0.3 0.6
BA
CA
GA
=
= +
= +
ri
r ik
r ik
2
(18 kg) 9.81m/s
176.58 N
W mg
W
= =
=
///
0: ( ) 0
A BA CA GA
M BC WΣ = × + × + ×− =
r jr jr j
0.6 (0.65 1.2 ) (0.3 0.6 ) ( ) 0BC W× + + × + + ×− =ij i k j i k j
0.6 0.65 1.2 0.3 0.6 0B C CW W+ −− + =k kik i
Equate coefficients of unit vectors to zero:
0.6
: 1.2 0.6 0 176.58 N 88.29 N
1.2
CW C
−+ = = =
i
: 0.6 0.65 0.3 0B CW+ −=k
0.6 0.65(88.29 N) 0.3(176.58 N) 0 7.36 NBB+−==−
0: 0
y
F ABCWΣ = ++− =
7.36 N 88.29 N 176.58 N 0 95.648 NAA−+ − = =
( ) 95.6 N ( ) 7.36 N ( ) 88.3 NaA bB cC= =−=
consent of McGraw–Hill Education.
PROBLEM 4.61
A 48-in. boom is held by a ball-and–socket joint at C and by
two cables BF and DAE; cable DAE passes around a
frictionless pulley at A. For the loading shown, determine
the tension in each cable and the reaction at C.
SOLUTION
Free–Body Diagram:
Five unknowns and six equations of equilibrium, but equilibrium is
maintained
( 0).
AC
MΣ=
PROBLEM 4.54
A small winch is used to raise a 120–lb load.
Find (a) the magnitude of the vertical force P
that should be applied at C to maintain
equilibrium in the position shown, (b) the
reactions at A and B, assuming that the bearing at
B does not exert any axial thrust.
SOLUTION
Dimensions in in.
(32 in.) (10 in.)cos30 (10 in.)sin30
32 8.6603 5
C=+−
=+−
ri j k
i jk
0: (10 4 ) ( 120 ) (20 ) ( ) (32 8.6603 5 ) ( ) 0
1200 480 20 20 32 5 0
A yz
yz
BB P
B B PP
Σ = + ×− + × + + + − ×− =
− + + − − −=
M ik j i j k jk j
i
k i k j ki
Equating the coefficients of the unit vectors to zero,
: 480 5 ) 0 96.0 lb;PP−= =i
( ) 96.0 lbaP
=
: 20 0
z
B=j
0
z
B=
: 1200 20 32(96.0) 0
y
B−+ − =k
213.6 lb
y
B=
0:
x
FΣ=
0
x
A=
0:
y
FΣ=
120 213.6 96.0 0
y
A−+ − =
2.40 lb
y
A=
0:
z
FΣ=
0
zz
AB+=
0
zz
AB=−=
(b)
(2.40 lb) ; (214 lb)= =A jB j
PROBLEM 4.55
A 200–mm lever and a 240–mm–diameter pulley are
welded to the axle BE that is supported by bearings at C
and D. If a 720–N vertical load is applied at A when the
lever is horizontal, determine (a) the tension in the cord,
(b) the reactions at C and D. Assume that the bearing
at D does not exert any axial thrust.
SOLUTION
Dimensions in mm
We have six unknowns and six equations of equilibrium. —OK
0: ( 120 ) ( ) (120 160 ) (80 200 ) ( 720 ) 0
C xy
DD TΣ = − × + + − × + − ×− =M k i j j ki k i j
..
Equating to zero the coefficients of the unit vectors:
:k
3
120 144 10 0T− +×=
(a)
1200 NT=
:i
3
120 57.6 10 0 480 N
yy
DD+ ×= =−
: 120 160(1200 N) 0
x
D−− =j
1600 N
x
D= −
0:
x
FΣ=
0
xx
CDT+ +=
1600 1200 400 N
x
C=−=
0:
y
FΣ=
720 0
yy
CD+− =
480 720 1200 N
y
C=+=
0:
z
FΣ=
0
z
C=
(b)
(400 N) (1200 N) ; (1600 N) (480 N)=+ =−−C i jD i j
PROBLEM 4.56
Solve Problem 4.55, assuming that the axle has been
rotated clockwise in its bearings by 30° and that the
720-N load remains vertical.
PROBLEM 4.55 A 200–mm lever and a 240–mm–
diameter pulley are welded to the axle BE that is
supported by bearings at C and D. If a 720–N vertical
load is applied at A when the lever is horizontal,
determine (a) the tension in the cord, (b) the reactions at
C and D. Assume that the bearing at D does not exert
any axial thrust.
SOLUTION
Dimensions in mm
We have six unknowns and six equations of equilibrium.
(a)
0: ( 120 ) ( ) (120 160 ) (80 173.21 ) ( 720 ) 0
C xy
DD TΣ = − × + + − × + − ×− =M k i j j ki k i j
33
120 120 120 160 57.6 10 124.71 10 0
xy
D DTT− + − − +×+ × =j ikj i k
Equating to zero the coefficients of the unit vectors,
3
: 120 124.71 10 0 1039.2 NTT− + ×= =k
1039 NT=
3
: 120 57.6 10 0 480 N
yy
DD+ ×= =−i
: 120 160(1039.2)
x
D−−
j
1385.6 N
x
D= −
0:
x
FΣ=
0
xx
CDT+ +=
1385.6 1039.2 346.4
x
C=−=
0:
y
FΣ=
720 0
yy
CD+− =
480 720 1200 N
y
C=+=
0:
z
FΣ=
0
z
C=
(b)
(346 N) (1200 N) (1386 N) (480 N)=+ =−−C i jD i j
PROBLEM 4.57
The rectangular plate shown weighs 80 lb and is supported by three
vertical wires. Determine the tension in each wire.
SOLUTION
Free–Body Diagram:
///
0: ( 80 lb) 0
B ABA CBC GB
T TrΣ = × + × + ×− =M r jr j j
(60 in.) [(60 in.) (15 in.) ] [(30 in.) (30 in.) ] ( 80 lb) 0
AC
TT× + + × + + ×− =kj i k j i k j
60 60 15 2400 2400 0
AC C
TT T−+ −− + =
ikik i
Equating to zero the coefficients of the unit vectors:
:i
60 15(40) 2400 0
A
T− +=
30.0 lb
A
T=
:k
60 2400 0
C
T−=
40.0 lb
C
T=
0:
y
FΣ=
80 lb 0
ABC
TTT++− =
30 lb 40 lb 80 lb 0
B
T++ − =
10.00 lb
B
T=
PROBLEM 4.58
The rectangular plate shown weighs 80 lb and is supported by three
vertical wires. Determine the weight and location of the lightest block
that should be placed on the plate if the tensions in the three wires are
to be equal.
SOLUTION
Free–Body Diagram:
Let
b
W−j
be the weight of the block and x and z the block’s coordinates.
SOLUTION Continued
Solving (4) and (5) for
/
b
WW
and recalling of
0 60 in.,x≤≤
0 90 in.,z≤≤
(4):
55
0.125
40 40 0
b
W
Wz
=≥=
−−
(5):
10 10 0.5
20 20 0
b
W
Wx
=≥=
−−
Thus,
min
( ) 0.5 0.5(80) 40 lb
b
WW
= = =
min
( ) 40.0 lb
b
W=
Making
0.5
b
WW=
in (4) and (5):
5 ( 40)(0.5 ) 0Wz W+− =
30.0 in.z=
10 ( 20)(0.5 ) 0Wx W− −− =
0 in.x=
PROBLEM 4.59
An opening in a floor is covered by a
1 1.2-m×
sheet of
plywood of mass 18 kg. The sheet is hinged at A and B and
is maintained in a position slightly above the floor by a
small block C. Determine the vertical component of the
reaction (a) at A, (b) at B, (c) at C.
SOLUTION
/
/
/
0.6
0.8 1.05
0.3 0.6
BA
CA
GA
=
= +
= +
ri
r ik
r ik
(18 kg)9.81
176.58 N
W mg
W
= =
=
///
0: ( ) 0
A BA CA GA
M BC WΣ = × + × + ×− =
r jr jr j
(0.6 ) (0.8 1.05 ) (0.3 0.6 ) ( ) 0BC W× + + × + + ×− =ij i kj i k j
0.6 0.8 1.05 0.3 0.6 0B C CW W+ − − +=kk iki
Equate coefficients of unit vectors to zero:
0.6
: 1.05 0.6 0 176.58 N 100.90 N
1.05
CW C
+== =
i
: 0.6 0.8 0.3 0BCW
+− =k
0.6 0.8(100.90 N) 0.3(176.58 N) 0 46.24 NBB+−==−
0: 0
y
F ABCWΣ = ++− =
46.24 N 100.90 N 176.58 N 0 121.92 NAA−+ + ==
( ) 121.9 N ( ) 46.2 N ( ) 100.9 NaA bB cC= =−=
consent of McGraw–Hill Education.
PROBLEM 4.60
Solve Problem 4.59, assuming that the small block C is
moved and placed under edge DE at a point 0.15 m from
corner E.
PROBLEM 4.59 An opening in a floor is covered by a
1 1.2-m×
sheet of plywood of mass 18 kg. The sheet is
hinged at A and B and is maintained in a position slightly
above the floor by a small block C. Determine the vertical
component of the reaction (a) at A, (b) at B, (c) at C.
SOLUTION
/
/
/
0.6
0.65 1.2
0.3 0.6
BA
CA
GA
=
= +
= +
ri
r ik
r ik
2
(18 kg) 9.81m/s
176.58 N
W mg
W
= =
=
///
0: ( ) 0
A BA CA GA
M BC WΣ = × + × + ×− =
r jr jr j
0.6 (0.65 1.2 ) (0.3 0.6 ) ( ) 0BC W× + + × + + ×− =ij i k j i k j
0.6 0.65 1.2 0.3 0.6 0B C CW W+ −− + =k kik i
Equate coefficients of unit vectors to zero:
0.6
: 1.2 0.6 0 176.58 N 88.29 N
1.2
CW C
−+ = = =
i
: 0.6 0.65 0.3 0B CW+ −=k
0.6 0.65(88.29 N) 0.3(176.58 N) 0 7.36 NBB+−==−
0: 0
y
F ABCWΣ = ++− =
7.36 N 88.29 N 176.58 N 0 95.648 NAA−+ − = =
( ) 95.6 N ( ) 7.36 N ( ) 88.3 NaA bB cC= =−=
consent of McGraw–Hill Education.
PROBLEM 4.61
A 48-in. boom is held by a ball-and–socket joint at C and by
two cables BF and DAE; cable DAE passes around a
frictionless pulley at A. For the loading shown, determine
the tension in each cable and the reaction at C.
SOLUTION
Free–Body Diagram:
Five unknowns and six equations of equilibrium, but equilibrium is
maintained
( 0).
AC
MΣ=