PROBLEM 4.103
Two transmission belts pass over a double–sheaved pulley
that is attached to an axle supported by bearings at A and
D. The radius of the inner sheave is 125 mm and the radius
of the outer sheave is 250 mm. Knowing that when the
system is at rest, the tension is 90 N in both portions of
PROBLEM 4.104
A thin ring of mass 2 kg and radius r = 140 mm is held against a frictionless
wall by a 125–mm string AB. Determine (a) the distance d, (b) the tension in
the string, (c) the reaction at C.
SOLUTION
Free–Body Diagram:
(Three-force body)
PROBLEM 4.105
The table shown weighs 30 lb and has a diameter of 4 ft. It is supported
by three legs equally spaced around the edge. A vertical load P of
magnitude 100 lb is applied to the top of the table at D. Determine the
maximum value of a if the table is not to tip over. Show, on a sketch,
the area of the table over which P can act without tipping the table.
SOLUTION
2 ft sin30 1ftr br= = °=
in shaded area for no tipping.
PROBLEM 4.106
A vertical load P is applied at end B of rod BC. The constant of the
spring is k, and the spring is unstretched when θ = 60°. (a) Neglecting
the weight of the rod, express the angle θ corresponding to the
equilibrium position terms of P, k, and l. (b) Determine the value of θ
corresponding to equilibrium if P =
1
4
kl.
SOLUTION
Free–Body Diagram:
(a) Triangle ABC is isosceles. We have
2( ) 2 sin ; cos
22
AB AD l CD l
θθ
= = =
Elongation of spring:
()() 60
2 sin 2 sin30
2
x AB AB
ll
θθ
θ
=−=°
= −°
1
2 sin 22
T k x kl
θ
= = −
0: cos ( sin ) 0
2
C
M T l Pl
θ
θ
Σ= − =
SOLUTION Continued
1
2
1
2 sin cos 2sin cos 0
22 2 2 2
cos 0 or 2( )sin 0
22
180 (trivial) sin 2
kl l Pl
kl P kl
kl
kl P
θ θ θθ
θθ
θ
θ
−− =
= − −=
=°=
−
1
1
2sin /( )
2kl kl P
θ
−
= −
(b) For
1,
4
P kl=
1
2
3
4
2
sin 23
kl
kl
θ
= =
41.8
2
θ
= °
83.6
θ
= °
PROBLEM 4.107
A force P is applied to a bent rod ABC, which may be supported in four different ways as shown. In each case,
if possible, determine the reactions at the supports.
SOLUTION
(a)
0: ( sin 45 )2 (cos 45 ) 0
Aa
M PC a aΣ = −+ ° + °=
32
CP=
2
3
CP=
0.471P=C
45°
21
0: 32
xx
FA P
Σ= −
3
xP
A=
21 2
0: 33
2
yy y
P
F AP P A
Σ = −+ =
0.745P=A
63.4°
(b)
0: ( cos30 )2 ( sin30 ) 0
C
M Pa A a A a
Σ = + − ° + °=
(1.732 0.5) 0.812A PA P−= =
0.812P=A
60.0°
0: (0.812 )sin30 0
xx
F PCΣ = °+ =
0.406
x
CP= −
0: (0.812 )cos30 0
yy
F P PCΣ = °− + =
0.297
y
CP= −
0.503P=C
36.2°
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
SOLUTION Continued
(c)
0: ( cos30 )2 ( sin30 ) 0
C
M Pa A a A a
Σ = + − ° + °=
(1.732 0.5) 0.448A PA P+= =
0.448AP=
60.0°
0: (0.448 )sin30 0 0.224
x xx
F P C CPΣ = − °+ = =
0: (0.448 )cos30 0 0.612
y yy
F P PC C PΣ = °− + = =
0.652P
=
C
69.9°
(d) Force T exerted by wire and reactions A and C all intersect at Point D.
0: 0
Da
MPΣ= =
Equilibrium is not maintained.
consent of McGraw–Hill Education.
PROBLEM 4.108
The rigid L–shaped member ABF is supported by
a ball–and–socket joint at A and by three cables.
For the loading shown, determine the tension in
each cable and the reaction at A.
SOLUTION
Free–Body Diagram:
/
/
/
/
/
12
12 8
12 16
12 24
12 32
BA
FA
DA
EA
FA
=
= −
= −
= −
= −
ri
r jk
r ik
r ik
r ik
12 9
15 in.
0.8 0.6
BG
BG
BG
=−+
=
=−+
ik
λ ik
12 16 ; 20 in.; 0.6 0.8
12 16 ; 20 in.; 0.6 0.8
DH
FJ
DH DH
FJ FJ
λ
λ
=−+ = =− +
=−+ = =− +
ij i j
ij i j
//
//
0:
( 24 ) ( 24 ) 0
12 0 0 12 0 16 12 0 32
0.8 0 0.6 0.6 0.8 0 0.6 0.8 0
12 0 8 12 0 24 0
0 24 0 0 24 0
A B A BG BG DH DH DH F A FJ FJ
FA EA
BG DH FJ
TTT
λλλ
Σ= × +× +×
+ ×− + ×− =
+ −+ −
−− −
+ −+ − =
−−
M rT rT rT
r jr j
ijk i jk i jk
ijk ijk
Coefficient of i:
12.8 25.6 192 576 0
DH FJ
TT+ + −−=
(1)
Coefficient of k:
9.6 9.6 288 288 0
DH FJ
TT+ + −−=
(2)
3
4
Eq. (1) − Eq. (2):
9.6 0
FJ
T=
0
FJ
T=
SOLUTION Continued
From Eq. (1):
12.8 268 0
DH
T−=
60 lb
DH
T=
Coefficient of j:
7.2 (16 0.6)(60.0 lb) 0
BG
T
− +× =
80.0 lb
BG
T=
0: 24 24 0
BG BG DH DH FJ
TT TΣ= + + + − − =F A jjλλ
Coefficient of i:
(80)( 0.8) (60.0)( 0.6) 0 100.0 lb
xx
AA+ −+ −= =
Coefficient of j:
(60.0)(0.8) 24 24 0
y
A+ −−=
0
y
A=
Coefficient of k:
(80.0)( 0.6) 0
z
A+ +=
48.0 lb
z
A= −
(100.0 lb) (48.0 lb)= −A ik
Note: The value
0
y
A=
can be confirmed by considering
0.
BF
MΣ=
consent of McGraw–Hill Education.
PROBLEM 4.109
A 1.2-m plank with a mass of 3 kg rests on two joists. Knowing that the
coefficient of static friction between the plank and the joists is 0.30,
determine the magnitude of the horizontal force required to move the
plank when (a) a = 750 mm, (b) a = 900 mm.
SOLUTION
Free body member AB:
In the vertical plane,
0: 0 hence (1)
22
AC C
LL
M Na W N W a
Σ= − = =
2
0: 0 hence (2)
2
y AC A
aL
F NNW NW a
−
Σ= + −= =
In the horizontal plane,
0: 0 hence F (3)
AC C
L
M F a PL P a
Σ= −= =
0: 0 hence (4)
Z AC A
La
F F PF F Pa
−
Σ = +− = =
(a) Substituting given data:
()
()
2
0.75 m, 1.2 m, 0.30, 3 kg 9.81 m/s 29.430 N
s
aL W
m
= = = = =
0.8 , 0.2 , 1.6 , 0.6
C A CA
N WN WF PF P= = = =
To slip at A,
( )
or 0.6 0.30 0.2 0.1
A sA
F N P WP W
m
= = =
To slip at C,
( )
or 1.6 0.30 0.8 0.15
C sC
FN P WP W
m
= = =
The plank will slip at A
2.94 NP=
consent of McGraw–Hill Education.
PROBLEM 4.109 (Continued)
(b) Substituting given data:
( )
( )
2
0.9 m, 1.2 m, 0.30, 3 kg 9.81 m/s 29.430 N
s
aL W
m
= = = = =
2 1 41
, , ,
3 3 33
C ACA
N WN WF PF P= = = =
To slip at
A,
11
or 0.30 0.3
33
A sA
F N P WPW
m
= = =
To slip at
C,
42
or 0.30 0.15
33
C sC
FN P WP W
m
= = =
The plank will slip at C for
( )
0.15 29.43 NP=
4.41 NP=
consent of McGraw–Hill Education.
PROBLEM 4.110
Two 10–lb blocks A and B are connected by a slender rod of negligible weight. The
an angle
30 .
θ
= °
with the vertical. (a) Show that the system is in equilibrium
when
0.P=
(b) Determine the largest value of P for which equilibrium is
maintained.
SOLUTION
FBD block
B:
(
a) Since
2.69 lbP=
to initiate motion, equilibrium exists with
0P=
(
b) For
max ,P
motion impends at both surfaces:
Block B:
0: 10lb cos 30 0Σ = − − °=
y B AB
FN F
3
10 lb 2
B AB
NF= +
(1)
Impending motion:
0.3
B sB B
FN N
µ
= =
0: sin 30 0
Σ = − °=
x B AB
F FF
2 0.6
AB B B
FF N= =
(2)
Solving Eqs. (1) and (2):
3
10 lb (0.6 ) 20.8166 lb
2
BB
NN=+=
FBD block
A:
Then
0.6 12.4900 lb
AB B
FN= =
Block A:
0: sin 30 0
Σ = °− =
x AB A
FF N
11
(12.4900 lb) 6.2450 lb
22
A AB
NF= = =
Impending motion:
0.3(6.2450 lb) 1.8735 lb
A sA
FN
µ
= = =
0: cos 30 10 lb 0Σ = + °− − =
y A AB
F FF P
310 lb
23
1.8735 lb (12.4900 lb) 10 lb
2
2.69 lb
=+−
=+−
=
A AB
PF F
2.69 lb
P=
consent of McGraw–Hill Education.
PROBLEM 4.104
A thin ring of mass 2 kg and radius r = 140 mm is held against a frictionless
wall by a 125–mm string AB. Determine (a) the distance d, (b) the tension in
the string, (c) the reaction at C.
SOLUTION
Free–Body Diagram:
(Three-force body)
PROBLEM 4.105
The table shown weighs 30 lb and has a diameter of 4 ft. It is supported
by three legs equally spaced around the edge. A vertical load P of
magnitude 100 lb is applied to the top of the table at D. Determine the
maximum value of a if the table is not to tip over. Show, on a sketch,
the area of the table over which P can act without tipping the table.
SOLUTION
2 ft sin30 1ftr br= = °=
in shaded area for no tipping.
PROBLEM 4.106
A vertical load P is applied at end B of rod BC. The constant of the
spring is k, and the spring is unstretched when θ = 60°. (a) Neglecting
the weight of the rod, express the angle θ corresponding to the
equilibrium position terms of P, k, and l. (b) Determine the value of θ
corresponding to equilibrium if P =
1
4
kl.
SOLUTION
Free–Body Diagram:
(a) Triangle ABC is isosceles. We have
2( ) 2 sin ; cos
22
AB AD l CD l
θθ
= = =
Elongation of spring:
()() 60
2 sin 2 sin30
2
x AB AB
ll
θθ
θ
=−=°
= −°
1
2 sin 22
T k x kl
θ
= = −
0: cos ( sin ) 0
2
C
M T l Pl
θ
θ
Σ= − =
SOLUTION Continued
1
2
1
2 sin cos 2sin cos 0
22 2 2 2
cos 0 or 2( )sin 0
22
180 (trivial) sin 2
kl l Pl
kl P kl
kl
kl P
θ θ θθ
θθ
θ
θ
−− =
= − −=
=°=
−
1
1
2sin /( )
2kl kl P
θ
−
= −
(b) For
1,
4
P kl=
1
2
3
4
2
sin 23
kl
kl
θ
= =
41.8
2
θ
= °
83.6
θ
= °
PROBLEM 4.107
A force P is applied to a bent rod ABC, which may be supported in four different ways as shown. In each case,
if possible, determine the reactions at the supports.
SOLUTION
(a)
0: ( sin 45 )2 (cos 45 ) 0
Aa
M PC a aΣ = −+ ° + °=
32
CP=
2
3
CP=
0.471P=C
45°
21
0: 32
xx
FA P
Σ= −
3
xP
A=
21 2
0: 33
2
yy y
P
F AP P A
Σ = −+ =
0.745P=A
63.4°
(b)
0: ( cos30 )2 ( sin30 ) 0
C
M Pa A a A a
Σ = + − ° + °=
(1.732 0.5) 0.812A PA P−= =
0.812P=A
60.0°
0: (0.812 )sin30 0
xx
F PCΣ = °+ =
0.406
x
CP= −
0: (0.812 )cos30 0
yy
F P PCΣ = °− + =
0.297
y
CP= −
0.503P=C
36.2°
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
SOLUTION Continued
(c)
0: ( cos30 )2 ( sin30 ) 0
C
M Pa A a A a
Σ = + − ° + °=
(1.732 0.5) 0.448A PA P+= =
0.448AP=
60.0°
0: (0.448 )sin30 0 0.224
x xx
F P C CPΣ = − °+ = =
0: (0.448 )cos30 0 0.612
y yy
F P PC C PΣ = °− + = =
0.652P
=
C
69.9°
(d) Force T exerted by wire and reactions A and C all intersect at Point D.
0: 0
Da
MPΣ= =
Equilibrium is not maintained.
consent of McGraw–Hill Education.
PROBLEM 4.108
The rigid L–shaped member ABF is supported by
a ball–and–socket joint at A and by three cables.
For the loading shown, determine the tension in
each cable and the reaction at A.
SOLUTION
Free–Body Diagram:
/
/
/
/
/
12
12 8
12 16
12 24
12 32
BA
FA
DA
EA
FA
=
= −
= −
= −
= −
ri
r jk
r ik
r ik
r ik
12 9
15 in.
0.8 0.6
BG
BG
BG
=−+
=
=−+
ik
λ ik
12 16 ; 20 in.; 0.6 0.8
12 16 ; 20 in.; 0.6 0.8
DH
FJ
DH DH
FJ FJ
λ
λ
=−+ = =− +
=−+ = =− +
ij i j
ij i j
//
//
0:
( 24 ) ( 24 ) 0
12 0 0 12 0 16 12 0 32
0.8 0 0.6 0.6 0.8 0 0.6 0.8 0
12 0 8 12 0 24 0
0 24 0 0 24 0
A B A BG BG DH DH DH F A FJ FJ
FA EA
BG DH FJ
TTT
λλλ
Σ= × +× +×
+ ×− + ×− =
+ −+ −
−− −
+ −+ − =
−−
M rT rT rT
r jr j
ijk i jk i jk
ijk ijk
Coefficient of i:
12.8 25.6 192 576 0
DH FJ
TT+ + −−=
(1)
Coefficient of k:
9.6 9.6 288 288 0
DH FJ
TT+ + −−=
(2)
3
4
Eq. (1) − Eq. (2):
9.6 0
FJ
T=
0
FJ
T=
SOLUTION Continued
From Eq. (1):
12.8 268 0
DH
T−=
60 lb
DH
T=
Coefficient of j:
7.2 (16 0.6)(60.0 lb) 0
BG
T
− +× =
80.0 lb
BG
T=
0: 24 24 0
BG BG DH DH FJ
TT TΣ= + + + − − =F A jjλλ
Coefficient of i:
(80)( 0.8) (60.0)( 0.6) 0 100.0 lb
xx
AA+ −+ −= =
Coefficient of j:
(60.0)(0.8) 24 24 0
y
A+ −−=
0
y
A=
Coefficient of k:
(80.0)( 0.6) 0
z
A+ +=
48.0 lb
z
A= −
(100.0 lb) (48.0 lb)= −A ik
Note: The value
0
y
A=
can be confirmed by considering
0.
BF
MΣ=
consent of McGraw–Hill Education.
PROBLEM 4.109
A 1.2-m plank with a mass of 3 kg rests on two joists. Knowing that the
coefficient of static friction between the plank and the joists is 0.30,
determine the magnitude of the horizontal force required to move the
plank when (a) a = 750 mm, (b) a = 900 mm.
SOLUTION
Free body member AB:
In the vertical plane,
0: 0 hence (1)
22
AC C
LL
M Na W N W a
Σ= − = =
2
0: 0 hence (2)
2
y AC A
aL
F NNW NW a
−
Σ= + −= =
In the horizontal plane,
0: 0 hence F (3)
AC C
L
M F a PL P a
Σ= −= =
0: 0 hence (4)
Z AC A
La
F F PF F Pa
−
Σ = +− = =
(a) Substituting given data:
()
()
2
0.75 m, 1.2 m, 0.30, 3 kg 9.81 m/s 29.430 N
s
aL W
m
= = = = =
0.8 , 0.2 , 1.6 , 0.6
C A CA
N WN WF PF P= = = =
To slip at A,
( )
or 0.6 0.30 0.2 0.1
A sA
F N P WP W
m
= = =
To slip at C,
( )
or 1.6 0.30 0.8 0.15
C sC
FN P WP W
m
= = =
The plank will slip at A
2.94 NP=
consent of McGraw–Hill Education.
PROBLEM 4.109 (Continued)
(b) Substituting given data:
( )
( )
2
0.9 m, 1.2 m, 0.30, 3 kg 9.81 m/s 29.430 N
s
aL W
m
= = = = =
2 1 41
, , ,
3 3 33
C ACA
N WN WF PF P= = = =
To slip at
A,
11
or 0.30 0.3
33
A sA
F N P WPW
m
= = =
To slip at
C,
42
or 0.30 0.15
33
C sC
FN P WP W
m
= = =
The plank will slip at C for
( )
0.15 29.43 NP=
4.41 NP=
consent of McGraw–Hill Education.
PROBLEM 4.110
Two 10–lb blocks A and B are connected by a slender rod of negligible weight. The
an angle
30 .
θ
= °
with the vertical. (a) Show that the system is in equilibrium
when
0.P=
(b) Determine the largest value of P for which equilibrium is
maintained.
SOLUTION
FBD block
B:
(
a) Since
2.69 lbP=
to initiate motion, equilibrium exists with
0P=
(
b) For
max ,P
motion impends at both surfaces:
Block B:
0: 10lb cos 30 0Σ = − − °=
y B AB
FN F
3
10 lb 2
B AB
NF= +
(1)
Impending motion:
0.3
B sB B
FN N
µ
= =
0: sin 30 0
Σ = − °=
x B AB
F FF
2 0.6
AB B B
FF N= =
(2)
Solving Eqs. (1) and (2):
3
10 lb (0.6 ) 20.8166 lb
2
BB
NN=+=
FBD block
A:
Then
0.6 12.4900 lb
AB B
FN= =
Block A:
0: sin 30 0
Σ = °− =
x AB A
FF N
11
(12.4900 lb) 6.2450 lb
22
A AB
NF= = =
Impending motion:
0.3(6.2450 lb) 1.8735 lb
A sA
FN
µ
= = =
0: cos 30 10 lb 0Σ = + °− − =
y A AB
F FF P
310 lb
23
1.8735 lb (12.4900 lb) 10 lb
2
2.69 lb
=+−
=+−
=
A AB
PF F
2.69 lb
P=
consent of McGraw–Hill Education.