PROBLEM 4.45
Solve Problem 4.44, assuming that the 170–N force applied at B is horizontal
and directed to the left.
PROBLEM 4.44 Rod AB is supported by a pin and bracket at A and rests
against a frictionless peg at C. Determine the reactions at A and C when a
170-N vertical force is applied at B.
PROBLEM 4.46
Determine the reactions at A and B when
50°.
β
=
SOLUTION
Free–Body Diagram: (Three-force body)
Reaction A must pass through Point D where the 100–N force
and B intersect.
In right ∆ BCD:
90 75 15
250tan 75 933.01mmBD
a
= °− °= °
= °=
In right ∆ ABD:
150 mm
tan 933.01mm
9.1333
AB
BD
γ
γ
= =
= °
Force Triangle
Law of sines:
100 N
sin9.1333° sin15 sin155.867
163.1 N; 257.6 N
AB
AB
= =
°°
= =
163.1 N=A
74.1°
258 N=B
65.0°
Dimensions in mm
consent of McGraw–Hill Education.
PROBLEM 4.47
Determine the reactions at A and B when
β
= 80°.
SOLUTION
Free–Body Diagram:
(Three-force body)
Reaction A must pass through D where the 100-N force and B intersect.
In right triangle BCD:
90 75 15
α
= °− °= °
tan75 250 tan75
933.01 mm
BD BC
BD
= °= °
=
In right triangle ABD:
150 mm
tan 933.01 mm
AB
BD
γ
= =
9.1333
γ
= °
consent of McGraw–Hill Education.
SOLUTION Continued
Force Triangle
Law of sines:
100 N
sin 9.1333° sin15 sin155.867
AB
= =
°°
PROBLEM 4.48
A slender rod of length L and weight W is attached to a
collar at A and is fitted with a small wheel at B. Knowing
that the wheel rolls freely along a cylindrical surface of
radius R, and neglecting friction, derive an equation in
θ
, L,
and R that must be satisfied when the rod is in equilibrium.
SOLUTION
Free–Body Diagram (Three-force body)
cosAC AE L
θ
= =
In ∆ ABC:
22 2
2 22
222
222
22
()() ()
(2 cos ) ( sin )
4cos sin
4cos 1 cos
3cos 1
CE BE BC
L LR
R
L
R
L
R
L
θθ
θθ
θθ
θ
+=
+=
= +
= +−
= +
2
21
cos 1
3R
L
θ
=−
PROBLEM 4.49
Knowing that for the rod of Problem 4.48, L = 15 in., R = 20
in., and W = 10 lb, determine (a) the angle
θ
corresponding
to equilibrium, (b) the reactions at A and B.
SOLUTION
equation:
2
2
1
cos 1
3R
L
θ
=−
For
15 in., 20 in., and 10 lb,LR W= = =
(a)
2
21 20 in.
cos 1 ; 59.39
3 15 in.
θθ
= −=°
59.4
θ
= °
In ∆ ABC:
sin 1
tan tan
2 cos 2
1
tan tan59.39 0.8452
2
40.2
BE L
CE L
θ
aθ
θ
a
a
= = =
= °=
= °
Force Triangle
tan (10 lb)tan 40.2 8.45 lb
(10 lb) 13.09 lb
cos cos 40.2
AW
W
B
a
a
= = °=
= = =
°
(b)
8.45 lb=A
13.09 lb=B
49.8°
consent of McGraw–Hill Education.
PROBLEM 4.50
A uniform rod AB of length 2R rests inside a hemispherical bowl of
radius R as shown. Neglecting friction, determine the angle
θ
corresponding to equilibrium.
SOLUTION
Note that the angle
α
of triangle DOA is the central angle corresponding to the inscribed angle
θ
of
triangle DCA.
2
αθ
=
The horizontal projections of
, ( ),
AE
AE x
and
, ( ),
AG
AG x
are equal.
AE AG A
xxx= =
or
( )cos2 ( )cosAE AG
θθ
=
and
(2 )cos2 cosRR
θθ
=
Now
2
cos2 2cos 1
θθ
= −
then
2
4cos 2 cos
θθ
−=
or
2
4cos cos 2 0
θθ
− −=
Applying the quadratic equation,
cos 0.84307 and cos 0.59307
θθ
= = −
32.534 and 126.375 (Discard)
θθ
=°=°
or
32.5
θ
= °
consent of McGraw–Hill Education.
PROBLEM 4.51
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
belt B and 150 N in both portions of belt C, determine the
reactions at A and D. Assume that the bearing at D does
not exert any axial thrust.
SOLUTION
We replace
B
T
and
B
′
T
by their resultant
( 180 N)−j
and
C
T
and
C
′
T
by their resultant
( 300 N) .−k
Dimensions in mm
equilibrium is maintained
( 0).
x
M
Σ=
0: (150 ) ( 180 ) (250 ) ( 300 ) (450 ) ( ) 0
A yz
DDΣ = ×− + ×− + × + =
M i j i k i jk
33
27 10 75 10 450 450 0
yz
DD−× + × + − =k jkj
Equating coefficients of j and k to zero,
:j
3
75 10 450 0
z
D×− =
166.7 N
z
D=
3
: 27 10 450 0
y
D−× + =k
60.0 N
y
D=
0:
x
FΣ=
0
x
A=
0: 180 N 0
y yy
F ADΣ= + − =
180 60 120.0 N
y
A= −=
0: 300 N 0
z zz
F ADΣ= + − =
300 166.7 133.3 N
z
A=−=
(120.0 N) (133.3 N) ; (60.0 N) (166.7 N)=+=+A j kD j k
consent of McGraw–Hill Education.
PROBLEM 4.52
Solve Problem 4.51, assuming that the pulley rotates at a
constant rate and that TB = 104 N, T′B = 84 N, TC = 175
N.
PROBLEM 4.51 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 belt B and 150 N in both portions of
belt C, determine the reactions at A 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: (150 250 ) ( 104 ) (150 250 ) ( 84 )
(250 125 ) ( 175 ) (250 125 ) ( )
450 ( ) 0
A
C
yz
T
DD
Σ = + ×− + − ×−
+ + ×− + − ×−
+×+ =
M ik j ik j
j k ij
i
i jk
150(104 84) 250(104 84) 250(175 ) 125(175 )
450 450 0
CC
yz
TT
DD
′′
− + + −+ + − −
+ −=
ki j
kj
: 250(104 84) 125(175 ) 0 175 40 135;
C CC
T TT
′ ′′
−− −= == =i
: 250(175 135) 450 0
z
D+− =j
172.2 N
z
D=
: 150(104 84) 450 0
y
D− ++ =k
62.7 N
y
D=
consent of McGraw–Hill Education.
SOLUTION Continued
0:
x
FΣ=
0
x
A=
0:
y
FΣ=
104 84 62.7 0
y
A− −+ =
125.3 N
y
A=
0:
z
FΣ=
175 135 172.2 0
z
A−−+ =
137.8 N
z
A=
(125.3 N) (137.8 N) ; (62.7 N) (172.2 N)=+=+A j kD j k
consent of McGraw–Hill Education.
PROBLEM 4.46
Determine the reactions at A and B when
50°.
β
=
SOLUTION
Free–Body Diagram: (Three-force body)
Reaction A must pass through Point D where the 100–N force
and B intersect.
In right ∆ BCD:
90 75 15
250tan 75 933.01mmBD
a
= °− °= °
= °=
In right ∆ ABD:
150 mm
tan 933.01mm
9.1333
AB
BD
γ
γ
= =
= °
Force Triangle
Law of sines:
100 N
sin9.1333° sin15 sin155.867
163.1 N; 257.6 N
AB
AB
= =
°°
= =
163.1 N=A
74.1°
258 N=B
65.0°
Dimensions in mm
consent of McGraw–Hill Education.
PROBLEM 4.47
Determine the reactions at A and B when
β
= 80°.
SOLUTION
Free–Body Diagram:
(Three-force body)
Reaction A must pass through D where the 100-N force and B intersect.
In right triangle BCD:
90 75 15
α
= °− °= °
tan75 250 tan75
933.01 mm
BD BC
BD
= °= °
=
In right triangle ABD:
150 mm
tan 933.01 mm
AB
BD
γ
= =
9.1333
γ
= °
consent of McGraw–Hill Education.
SOLUTION Continued
Force Triangle
Law of sines:
100 N
sin 9.1333° sin15 sin155.867
AB
= =
°°
PROBLEM 4.48
A slender rod of length L and weight W is attached to a
collar at A and is fitted with a small wheel at B. Knowing
that the wheel rolls freely along a cylindrical surface of
radius R, and neglecting friction, derive an equation in
θ
, L,
and R that must be satisfied when the rod is in equilibrium.
SOLUTION
Free–Body Diagram (Three-force body)
cosAC AE L
θ
= =
In ∆ ABC:
22 2
2 22
222
222
22
()() ()
(2 cos ) ( sin )
4cos sin
4cos 1 cos
3cos 1
CE BE BC
L LR
R
L
R
L
R
L
θθ
θθ
θθ
θ
+=
+=
= +
= +−
= +
2
21
cos 1
3R
L
θ
=−
PROBLEM 4.49
Knowing that for the rod of Problem 4.48, L = 15 in., R = 20
in., and W = 10 lb, determine (a) the angle
θ
corresponding
to equilibrium, (b) the reactions at A and B.
SOLUTION
equation:
2
2
1
cos 1
3R
L
θ
=−
For
15 in., 20 in., and 10 lb,LR W= = =
(a)
2
21 20 in.
cos 1 ; 59.39
3 15 in.
θθ
= −=°
59.4
θ
= °
In ∆ ABC:
sin 1
tan tan
2 cos 2
1
tan tan59.39 0.8452
2
40.2
BE L
CE L
θ
aθ
θ
a
a
= = =
= °=
= °
Force Triangle
tan (10 lb)tan 40.2 8.45 lb
(10 lb) 13.09 lb
cos cos 40.2
AW
W
B
a
a
= = °=
= = =
°
(b)
8.45 lb=A
13.09 lb=B
49.8°
consent of McGraw–Hill Education.
PROBLEM 4.50
A uniform rod AB of length 2R rests inside a hemispherical bowl of
radius R as shown. Neglecting friction, determine the angle
θ
corresponding to equilibrium.
SOLUTION
Note that the angle
α
of triangle DOA is the central angle corresponding to the inscribed angle
θ
of
triangle DCA.
2
αθ
=
The horizontal projections of
, ( ),
AE
AE x
and
, ( ),
AG
AG x
are equal.
AE AG A
xxx= =
or
( )cos2 ( )cosAE AG
θθ
=
and
(2 )cos2 cosRR
θθ
=
Now
2
cos2 2cos 1
θθ
= −
then
2
4cos 2 cos
θθ
−=
or
2
4cos cos 2 0
θθ
− −=
Applying the quadratic equation,
cos 0.84307 and cos 0.59307
θθ
= = −
32.534 and 126.375 (Discard)
θθ
=°=°
or
32.5
θ
= °
consent of McGraw–Hill Education.
PROBLEM 4.51
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
belt B and 150 N in both portions of belt C, determine the
reactions at A and D. Assume that the bearing at D does
not exert any axial thrust.
SOLUTION
We replace
B
T
and
B
′
T
by their resultant
( 180 N)−j
and
C
T
and
C
′
T
by their resultant
( 300 N) .−k
Dimensions in mm
equilibrium is maintained
( 0).
x
M
Σ=
0: (150 ) ( 180 ) (250 ) ( 300 ) (450 ) ( ) 0
A yz
DDΣ = ×− + ×− + × + =
M i j i k i jk
33
27 10 75 10 450 450 0
yz
DD−× + × + − =k jkj
Equating coefficients of j and k to zero,
:j
3
75 10 450 0
z
D×− =
166.7 N
z
D=
3
: 27 10 450 0
y
D−× + =k
60.0 N
y
D=
0:
x
FΣ=
0
x
A=
0: 180 N 0
y yy
F ADΣ= + − =
180 60 120.0 N
y
A= −=
0: 300 N 0
z zz
F ADΣ= + − =
300 166.7 133.3 N
z
A=−=
(120.0 N) (133.3 N) ; (60.0 N) (166.7 N)=+=+A j kD j k
consent of McGraw–Hill Education.
PROBLEM 4.52
Solve Problem 4.51, assuming that the pulley rotates at a
constant rate and that TB = 104 N, T′B = 84 N, TC = 175
N.
PROBLEM 4.51 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 belt B and 150 N in both portions of
belt C, determine the reactions at A 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: (150 250 ) ( 104 ) (150 250 ) ( 84 )
(250 125 ) ( 175 ) (250 125 ) ( )
450 ( ) 0
A
C
yz
T
DD
Σ = + ×− + − ×−
+ + ×− + − ×−
+×+ =
M ik j ik j
j k ij
i
i jk
150(104 84) 250(104 84) 250(175 ) 125(175 )
450 450 0
CC
yz
TT
DD
′′
− + + −+ + − −
+ −=
ki j
kj
: 250(104 84) 125(175 ) 0 175 40 135;
C CC
T TT
′ ′′
−− −= == =i
: 250(175 135) 450 0
z
D+− =j
172.2 N
z
D=
: 150(104 84) 450 0
y
D− ++ =k
62.7 N
y
D=
consent of McGraw–Hill Education.
SOLUTION Continued
0:
x
FΣ=
0
x
A=
0:
y
FΣ=
104 84 62.7 0
y
A− −+ =
125.3 N
y
A=
0:
z
FΣ=
175 135 172.2 0
z
A−−+ =
137.8 N
z
A=
(125.3 N) (137.8 N) ; (62.7 N) (172.2 N)=+=+A j kD j k
consent of McGraw–Hill Education.