PROBLEM 4.95
Two slender rods of negligible weight are pin-connected at C and
attached to blocks A and B, each of weight W. Knowing that
80
θ
= °
and that the coefficient of static friction between the
blocks and the horizontal surface is 0.30, determine the largest
value of P for which equilibrium is maintained.
SOLUTION
FBD pin C:
sin 20 0.34202
cos 20 0.93969
AC
BC
FP P
FP P
= °=
= °=
0: sin 30 0
y A AC
F N WFΣ = − − °=
or
0.34202 sin 30 0.171010
A
NW P W P= + °= +
FBD block A:
0: cos 30 0
x A AC
F FFΣ = − °=
or
0.34202 cos 30 0.29620
A
FP P= °=
For impending motion at A:
A sA
FN
µ
=
Then
0.29620
: 0.171010 0.3
A
As
F
NW P P
µ
=+=
or
1.22500
PW=
0: cos 30 0
y B BC
F N WFΣ = − − °=
0.93969 cos 30 0.81380
B
NWPWP= + °= +
0: sin 30 0
x BC B
FF FΣ = °− =
consent of McGraw–Hill Education.
SOLUTION Continued
FBD block B:
For impending motion at B:
B sB
FN
µ
=
Then
0.46985
: 0.81380 0.3
B
Bs
FP
NW P
µ
=+=
or
1.32914PW=
Thus, maximum P for equilibrium
max
1.225PW=
consent of McGraw–Hill Education.
PROBLEM 4.96
Two slender rods of negligible weight are pin–connected at C and
attached to blocks A and B, each of weight W. Knowing that
P = 1.260W and that the coefficient of static friction between the
blocks and the horizontal surface is 0.30, determine the range of
values of θ, between 0 and 180°, for which equilibrium is
maintained.
SOLUTION
AC and BC are two–force members
Free body: Joint C
Force triangle:
From Force triangle:
sin( 60 ) 1.26 sin( 60 )
AB
FP W
θθ
= −°= −°
(1)
cos( 60 ) 1.26 cos( 60 )
BC
FP W
θθ
= −°= −°
(2)
We shall, in turn, seek
θ
corresponding to impending motion of each block
For motion of A impending to left
from solution of Prob. 8.46:
0.419
AC
FW=
EQ (1):
0.419 1.26 sin( 60 )
sin( 60 ) 0.33254
60 19.423
AC
F WW
θ
θ
θ
= = −°
− °=
− °= °
79.42
θ
= °
For motion of B impending to right.
from solution of Prob. 8.46:
1.249
BC
FW=
Eq. (2):
1.249 1.26 cos( 60 )
cos( 60 ) 0.99127
BC
F WW
θ
θ
= = −°
− °=
60 7.58
θ
− °=± °
60 7.58
θ
− °=+ °
67.6
θ
= °
60 7.58
θ
− °=− °
52.4
θ
= °
consent of McGraw–Hill Education.
SOLUTION Continued
For motion of A impending to right
180 60 16.7 103.3
γ
= °− °− °= °
Law of sines:
sin16.7 sin103.3
0.29528
AB
AB
FW
FW
−=
°°
= −
Note: Direction of
AB
F+
is kept same as in free body of Joint C.
Eq. (1):
0.29528 1.26 sin( 60 )
sin( 60 ) 0.23435
AB
F WW
θ
θ
=− = −°
− °=−
( 60 ) 13.553
θ
− °=− °
46.4
θ
= °
Summary:
A moves
to right
No
motion
B moves
to right
No
motion
A moves
to left
θ
46.4°
52.4°
67.6°
79.4°
No motion for:
46.4 52.4 and 67.6° 79.4°
θθ
°≤≤ ° ≤≤
consent of McGraw–Hill Education.
PROBLEM 4.97
The cylinder shown is of weight W and radius r, and the coefficient of static
friction
s
µ
is the same at A and B. Determine the magnitude of the largest couple
M that can be applied to the cylinder if it is not to rotate.
SOLUTION
FBD cylinder:
For maximum M, motion impends at both A and B
A sA
B sB
FN
FN
µ
µ
=
=
0: 0
x AB
A B sB
F NF
NF N
µ
Σ= − =
= =
2
AsAsB
FN N
µµ
= =
2
0: 0
y BA
B sB
F N FW
N NW
µ
Σ= + −=
+=
or
2
1
Bs
W
N
µ
=+
and
2
2
2
1
1
s
Bs
s
A
W
F
W
F
µ
µ
µ
µ
=+
=+
0: ( ) 0
C AB
M M rF FΣ= − + =
( )
22
1
ss s
W
Mr
µµ µ
= + +
max 2
1
1
s
ss
M Wr
µ
µµ
+
=+
consent of McGraw–Hill Education.
PROBLEM 4.98
The cylinder shown is of weight W and radius r. Express in terms W and r the
magnitude of the largest couple M that can be applied to the cylinder if it is not to
rotate, assuming the coefficient of static friction to be (a) zero at A and 0.30 at B,
(b) 0.25 at A and 0.30 at B.
SOLUTION
FBD cylinder:
For maximum M, motion impends at both A and B
A AA
B BB
FN
FN
µ
µ
=
=
0: 0
x AB
A B BB
F NF
NF N
µ
Σ= − =
= =
A AA ABB
FN N
µ µµ
= =
0: 0
(1 )
y BA
B AB
F N FW
NW
µµ
Σ= + −=
+=
or
1
1
BAB
NW
µµ
=+
and
1
1
B
B BB AB
AB
A AB B AB
FN W
FN W
µ
µµµ
µµ
µµ µµ
= = +
= = +
0: ( ) 0
C AB
M M rF FΣ= − + =
1
1
A
BAB
M Wr
µ
µµµ
+
=+
(a) For
0 and 0.30:
AB
µµ
= =
0.300M Wr=
(b) For
0.25 and 0.30:
AB
µµ
= =
0.349M Wr=
consent of McGraw–Hill Education.
PROBLEM 4.99
A T–shaped bracket supports the four loads shown. Determine the
reactions at A and B (a) if
10 in.,a=
(b) if
7 in.a=
SOLUTION
Free–Body Diagram:
0: 0
xx
FBΣ= =
0: (40 lb)(6 in.) (30 lb) (10 lb)( 8 in.) (12 in.) 0
B
M aa AΣ= − − + + =
(40 160)
12
a
A−
=
(1)
0: (40 lb)(6 in.) (50 lb)(12 in.) (30 lb)( 12 in.)
(10 lb)( 20 in.) (12 in.) 0
A
y
Ma
aB
Σ= − − − +
− ++ =
(1400 40 )
12
ya
B+
=
Since
(1400 40 )
0, 12
x
a
BB +
= =
(2)
(a)
For 10 in.,a=
Eq. (1):
(40 10 160) 20.0 lb
12
A×−
= = +
20.0 lb=A
Eq. (2):
(1400 40 10) 150.0 lb
12
B+×
= = +
150.0 lb=B
(b)
For 7 in.,a=
Eq. (1):
(40 7 160) 10.00 lb
12
A×−
= = +
10.00 lb=A
Eq. (2):
(1400 40 7) 140.0 lb
12
B+×
= = +
140.0 lb=B
consent of McGraw–Hill Education.
PROBLEM 4.100
Neglecting friction and the radius of the pulley,
determine (a) the tension in cable ADB, (b) the reaction
at C.
SOLUTION
Free–Body Diagram:
Dimensions in mm
Geometry:
Distance:
22
(0.36) (0.150) 0.39 mAD =+=
Distance:
22
(0.2) (0.15) 0.25 mBD =+=
Equilibrium for beam:
(a)
0.15 0.15
0: (120 N)(0.28 m) (0.36 m) (0.2 m) 0
0.39 0.25
C
M TT
Σ= − − =
130.000 NT=
or
130.0 NT=
(b)
0.36 0.2
0: (130.000 N) (130.000 N) 0
0.39 0.25
xx
FC
Σ= + + =
224.00 N
x
C= −
0.15 0.15
0: (130.00 N) (130.00 N) 120 N 0
0.39 0.25
yy
FC
Σ= + + − =
8.0000 N
y
C= −
Thus,
22 2 2
( 224) ( 8) 224.14 N
xy
C CC= + = − +− =
and
11
8
tan tan 2.0454
224
y
x
C
C
θ
−−
= = = °
224 N=
C
2.05°
consent of McGraw–Hill Education.
PROBLEM 4.101
Member ABC is supported by a pin and bracket at B and by an
inextensible cord attached at A and C and passing over a frictionless
pulley at D. The tension may be assumed to be the same in portions
AD and CD of the cord. For the loading shown and neglecting the size
of the pulley, determine the tension in the cord and the reaction at B.
SOLUTION
Free–Body Diagram:
Force Triangle
PROBLEM 4.102
A movable bracket is held at rest by a cable attached at C and by
frictionless rollers at A and B. For the loading shown, determine
(a) the tension in the cable, (b) the reactions at A and B.
SOLUTION
Free–Body Diagram:
(a)
0: 600 N 0
y
FTΣ= − =
600 NT=
(b)
0: 0
x
F BA BAΣ = −= ∴ =
0: (600 N)(600 mm) (90 mm) 0MAΣ= − =
4000 NA=
4000 NB∴=
4.00 kN=A
;
4.00 kN=B
consent of McGraw–Hill Education.
SOLUTION Continued
FBD block B:
For impending motion at B:
B sB
FN
µ
=
Then
0.46985
: 0.81380 0.3
B
Bs
FP
NW P
µ
=+=
or
1.32914PW=
Thus, maximum P for equilibrium
max
1.225PW=
consent of McGraw–Hill Education.
PROBLEM 4.96
Two slender rods of negligible weight are pin–connected at C and
attached to blocks A and B, each of weight W. Knowing that
P = 1.260W and that the coefficient of static friction between the
blocks and the horizontal surface is 0.30, determine the range of
values of θ, between 0 and 180°, for which equilibrium is
maintained.
SOLUTION
AC and BC are two–force members
Free body: Joint C
Force triangle:
From Force triangle:
sin( 60 ) 1.26 sin( 60 )
AB
FP W
θθ
= −°= −°
(1)
cos( 60 ) 1.26 cos( 60 )
BC
FP W
θθ
= −°= −°
(2)
We shall, in turn, seek
θ
corresponding to impending motion of each block
For motion of A impending to left
from solution of Prob. 8.46:
0.419
AC
FW=
EQ (1):
0.419 1.26 sin( 60 )
sin( 60 ) 0.33254
60 19.423
AC
F WW
θ
θ
θ
= = −°
− °=
− °= °
79.42
θ
= °
For motion of B impending to right.
from solution of Prob. 8.46:
1.249
BC
FW=
Eq. (2):
1.249 1.26 cos( 60 )
cos( 60 ) 0.99127
BC
F WW
θ
θ
= = −°
− °=
60 7.58
θ
− °=± °
60 7.58
θ
− °=+ °
67.6
θ
= °
60 7.58
θ
− °=− °
52.4
θ
= °
consent of McGraw–Hill Education.
SOLUTION Continued
For motion of A impending to right
180 60 16.7 103.3
γ
= °− °− °= °
Law of sines:
sin16.7 sin103.3
0.29528
AB
AB
FW
FW
−=
°°
= −
Note: Direction of
AB
F+
is kept same as in free body of Joint C.
Eq. (1):
0.29528 1.26 sin( 60 )
sin( 60 ) 0.23435
AB
F WW
θ
θ
=− = −°
− °=−
( 60 ) 13.553
θ
− °=− °
46.4
θ
= °
Summary:
A moves
to right
No
motion
B moves
to right
No
motion
A moves
to left
θ
46.4°
52.4°
67.6°
79.4°
No motion for:
46.4 52.4 and 67.6° 79.4°
θθ
°≤≤ ° ≤≤
consent of McGraw–Hill Education.
PROBLEM 4.97
The cylinder shown is of weight W and radius r, and the coefficient of static
friction
s
µ
is the same at A and B. Determine the magnitude of the largest couple
M that can be applied to the cylinder if it is not to rotate.
SOLUTION
FBD cylinder:
For maximum M, motion impends at both A and B
A sA
B sB
FN
FN
µ
µ
=
=
0: 0
x AB
A B sB
F NF
NF N
µ
Σ= − =
= =
2
AsAsB
FN N
µµ
= =
2
0: 0
y BA
B sB
F N FW
N NW
µ
Σ= + −=
+=
or
2
1
Bs
W
N
µ
=+
and
2
2
2
1
1
s
Bs
s
A
W
F
W
F
µ
µ
µ
µ
=+
=+
0: ( ) 0
C AB
M M rF FΣ= − + =
( )
22
1
ss s
W
Mr
µµ µ
= + +
max 2
1
1
s
ss
M Wr
µ
µµ
+
=+
consent of McGraw–Hill Education.
PROBLEM 4.98
The cylinder shown is of weight W and radius r. Express in terms W and r the
magnitude of the largest couple M that can be applied to the cylinder if it is not to
rotate, assuming the coefficient of static friction to be (a) zero at A and 0.30 at B,
(b) 0.25 at A and 0.30 at B.
SOLUTION
FBD cylinder:
For maximum M, motion impends at both A and B
A AA
B BB
FN
FN
µ
µ
=
=
0: 0
x AB
A B BB
F NF
NF N
µ
Σ= − =
= =
A AA ABB
FN N
µ µµ
= =
0: 0
(1 )
y BA
B AB
F N FW
NW
µµ
Σ= + −=
+=
or
1
1
BAB
NW
µµ
=+
and
1
1
B
B BB AB
AB
A AB B AB
FN W
FN W
µ
µµµ
µµ
µµ µµ
= = +
= = +
0: ( ) 0
C AB
M M rF FΣ= − + =
1
1
A
BAB
M Wr
µ
µµµ
+
=+
(a) For
0 and 0.30:
AB
µµ
= =
0.300M Wr=
(b) For
0.25 and 0.30:
AB
µµ
= =
0.349M Wr=
consent of McGraw–Hill Education.
PROBLEM 4.99
A T–shaped bracket supports the four loads shown. Determine the
reactions at A and B (a) if
10 in.,a=
(b) if
7 in.a=
SOLUTION
Free–Body Diagram:
0: 0
xx
FBΣ= =
0: (40 lb)(6 in.) (30 lb) (10 lb)( 8 in.) (12 in.) 0
B
M aa AΣ= − − + + =
(40 160)
12
a
A−
=
(1)
0: (40 lb)(6 in.) (50 lb)(12 in.) (30 lb)( 12 in.)
(10 lb)( 20 in.) (12 in.) 0
A
y
Ma
aB
Σ= − − − +
− ++ =
(1400 40 )
12
ya
B+
=
Since
(1400 40 )
0, 12
x
a
BB +
= =
(2)
(a)
For 10 in.,a=
Eq. (1):
(40 10 160) 20.0 lb
12
A×−
= = +
20.0 lb=A
Eq. (2):
(1400 40 10) 150.0 lb
12
B+×
= = +
150.0 lb=B
(b)
For 7 in.,a=
Eq. (1):
(40 7 160) 10.00 lb
12
A×−
= = +
10.00 lb=A
Eq. (2):
(1400 40 7) 140.0 lb
12
B+×
= = +
140.0 lb=B
consent of McGraw–Hill Education.
PROBLEM 4.100
Neglecting friction and the radius of the pulley,
determine (a) the tension in cable ADB, (b) the reaction
at C.
SOLUTION
Free–Body Diagram:
Dimensions in mm
Geometry:
Distance:
22
(0.36) (0.150) 0.39 mAD =+=
Distance:
22
(0.2) (0.15) 0.25 mBD =+=
Equilibrium for beam:
(a)
0.15 0.15
0: (120 N)(0.28 m) (0.36 m) (0.2 m) 0
0.39 0.25
C
M TT
Σ= − − =
130.000 NT=
or
130.0 NT=
(b)
0.36 0.2
0: (130.000 N) (130.000 N) 0
0.39 0.25
xx
FC
Σ= + + =
224.00 N
x
C= −
0.15 0.15
0: (130.00 N) (130.00 N) 120 N 0
0.39 0.25
yy
FC
Σ= + + − =
8.0000 N
y
C= −
Thus,
22 2 2
( 224) ( 8) 224.14 N
xy
C CC= + = − +− =
and
11
8
tan tan 2.0454
224
y
x
C
C
θ
−−
= = = °
224 N=
C
2.05°
consent of McGraw–Hill Education.
PROBLEM 4.101
Member ABC is supported by a pin and bracket at B and by an
inextensible cord attached at A and C and passing over a frictionless
pulley at D. The tension may be assumed to be the same in portions
AD and CD of the cord. For the loading shown and neglecting the size
of the pulley, determine the tension in the cord and the reaction at B.
SOLUTION
Free–Body Diagram:
Force Triangle
PROBLEM 4.102
A movable bracket is held at rest by a cable attached at C and by
frictionless rollers at A and B. For the loading shown, determine
(a) the tension in the cable, (b) the reactions at A and B.
SOLUTION
Free–Body Diagram:
(a)
0: 600 N 0
y
FTΣ= − =
600 NT=
(b)
0: 0
x
F BA BAΣ = −= ∴ =
0: (600 N)(600 mm) (90 mm) 0MAΣ= − =
4000 NA=
4000 NB∴=
4.00 kN=A
;
4.00 kN=B
consent of McGraw–Hill Education.