PROBLEM 4.12
A lever AB is hinged at C and attached to a control cable at A. If
the lever is subjected to a 500–N horizontal force at B,
determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
Triangle ACD is isosceles with
90 30 120C= °+ °= °�
1(180 120 ) 30 .
2
AD
= = °− ° = °
��
(a) We resolve
AD
F
into components along AB and perpendicular to AB.
0: ( sin30 )(250 mm) (500 N)(100 mm) 0
C AD
MFΣ= ° − =
400 N
AD
F=
(b)
0: (400 N)cos60 500 N 0
xx
FCΣ = − °+ − =
300 N
x
C= +
0: (400 N)sin60° 0
yy
FCΣ= − + =
346.4 N
y
C= +
458 N=C
49.1°
consent of McGraw–Hill Education.
PROBLEM 4.13
Determine the reactions at A and B when (a)
α
= 0, (b)
α
= 90°,
(c)
α
= 30°.
SOLUTION
(a)
0
α
=
0: (20 in.) 75 lb(10 in.) 0
37.5 lb
A
MB
B
Σ= − =
=
0: 0
xx
FAΣ= =
+
0: 75 lb 37.5 lb 0
37.5 lb
yy
y
FA
A
Σ= − + =
=
37.5 lb= =AB
(b)
90
α
= °
0: (12 in.) 75 lb(10 in.) 0
62.5 lb
A
MB
B
Σ= − =
=
0: 0
62.5 lb
xx
x
F AB
A
Σ= −=
=
0: 75 lb 0
75 lb
yy
y
FA
A
Σ= − =
=
22
22
(62.5 lb) (75 lb)
97.6 lb
xy
A AA= +
= +
=
75
tan 62.5
50.2
θ
θ
=
= °
97.6 lb=A
50.2 ; 62.5 lb°=B
consent of McGraw–Hill Education.
SOLUTION Continued
(c)
30
α
= °
0: ( cos 30°)(20 in.) ( sin30 )(12 in.)
(75 lb)(10 in.) 0
A
MB BΣ= + °
−=
32.161 lbB=
0: (32.161)sin30 0
16.0805 lb
xx
x
FA
A
Σ = − °=
=
0: (32.161)cos30 75 0
47.148 lb
yy
y
FA
A
Σ = + °− =
=
22
22
(16.0805) (47.148)
49.8 lb
xy
A AA= +
= +
=
47.148
tan 16.0805
71.2
θ
θ
=
= °
49.8 lb=A
71.2 ; 32.2 lb°=B
60.0°
consent of McGraw–Hill Education.
PROBLEM 4.14
The bracket BCD is hinged at C and attached to a control
cable at B. For the loading shown, determine (a) the
tension in the cable, (b) the reaction at C.
SOLUTION
At B:
0.18 m
0.24 m
y
x
T
T=
3
4
yx
TT
=
(1)
(a)
0: (0.18 m) (240 N)(0.4 m) (240 N)(0.8 m) 0
Cx
MT
Σ= − − =
1600 N
x
T= +
From Eq. (1):
3(1600 N) 1200 N
4
y
T= =
22 2 2
1600 1200 2000 N
xy
T TT= += + =
2.00 kNT=
(b)
0: 0
x xx
F CT
Σ= −=
1600 N 0 1600 N
xx
CC−==+
1600 N
x=C
0: 240 N 240 N 0
y yy
F CTΣ= −− − =
1200 N 480 N 0
y
C− −=
1680 N
y
C= +
1680 N
y=C
46.4
2320 NC
α
= °
=
2.32 kN=C
46.4°
consent of McGraw–Hill Education.
PROBLEM 4.15
Solve Problem 4.14, assuming that
0.32 m.a=
PROBLEM 4.14 The bracket BCD is hinged at C and
attached to a control cable at B. For the loading shown,
determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
(a) At B:
0.32 m
0.24 m
4
3
y
x
yx
T
T
TT
=
=
0: (0.32 m) (240 N)(0.4 m) (240 N)(0.8 m) 0
Cx
MTΣ= − − =
900 N
x
T=
From Eq. (1):
4(900 N) 1200 N
3
y
T= =
22 2 2
900 1200 1500 N
xy
T TT= += + =
1.500 kNT=
(b)
0: 0
x xx
F CTΣ= −=
900 N 0 900 N
xx
CC−==+
900 N
x=C
0: 240 N 240 N 0
y yy
F CTΣ= −− − =
1200 N 480 N 0
y
C− −=
1680 N
y
C= +
1680 N
y=C
61.8
1906 NC
α
= °
=
1.906 kN=C
61.8°
consent of McGraw–Hill Education.
PROBLEM 4.16
Determine the reactions at A and B when (a)
0,h=
(b)
200 mm.h=
SOLUTION
Free–Body Diagram:
0: ( cos60 )(0.5 m) ( sin60 ) (150 N)(0.25 m) 0
A
M B BhΣ = ° − °− =
37.5
0.25 0.866
Bh
=−
(1)
(a) When
0,h=
From Eq. (1):
37.5 150 N
0.25
B= =
150.0 N=B
30.0°
0: sin60 0
yx
F ABΣ = − °=
(150)sin60 129.9 N
x
A= °=
129.9 N
x
=A
0: 150 cos60 0
yy
FA BΣ = − + °=
150 (150)cos60 75 N
y
A= − °=
75 N
y
=A
30
150.0 NA
α
= °
=
150.0 N=A
30.0°
(b) When h = 200 mm = 0.2 m,
From Eq. (1):
37.5 488.3 N
0.25 0.866(0.2)
B= =
−
488 N=B
30.0°
0: sin 60 0
xx
F ABΣ = − °=
(488.3)sin60 422.88 N
x
A= °=
422.88 N
x
=A
0: 150 cos60 0
yy
FA BΣ = − + °=
150 (488.3)cos60 94.15 N
y
A= − °=−
94.15 N
y=A
12.55
433.2 NA
α
= °
=
433 N=A
12.55°
consent of McGraw–Hill Education.
PROBLEM 4.17
A light bar AD is suspended from a cable BE and supports a
50-lb block at C. The ends A and D of the bar are in contact
with frictionless vertical walls. Determine the tension in cable
BE and the reactions at A and D.
SOLUTION
Free–Body Diagram:
0:
x
F ADΣ= =
0:
y
FΣ=
50.0 lb
BE
T=
We note that the forces shown form two couples.
0: (8 in.) (50 lb)(3 in.) 0MAΣ= − =
18.75 lbA=
18.75 lb=A
18.75 lb=D
consent of McGraw–Hill Education.
PROBLEM 4.18
Bar AD is attached at A and C to collars that can move freely
on the rods shown. If the cord BE is vertical (
α
= 0), determine
the tension in the cord and the reactions at A and C.
SOLUTION
Free–Body Diagram:
0: cos30 (80 N)cos30 0
y
FTΣ = − °+ °=
80 NT=
80.0 NT=
0: ( sin30 )(0.4 m) (80 N)(0.2 m) (80 N)(0.2 m) 0
C
MAΣ= ° − − =
160 NA= +
160.0 N=A
30.0°
0: (80 N)(0.2 m) (80 N)(0.6 m) ( sin30 )(0.4 m) 0
A
MCΣ= − + ° =
160 NC= +
160.0 N=C
30.0°
consent of McGraw–Hill Education.
PROBLEM 4.19
Solve Problem 4.18 if the cord BE is parallel to the rods (
α
= 30°).
PROBLEM 4.18 Bar AD is attached at A and C to collars that
can move freely on the rods shown. If the cord BE is vertical
(
α
= 0), determine the tension in the cord and the reactions at A
and C.
SOLUTION
Free–Body Diagram:
0: (80 N)cos30 0
y
FTΣ = − + °=
69.282 NT=
69.3 NT=
0: (69.282 N)cos30 (0.2 m)
(80 N)(0.2 m) ( sin30 )(0.4 m) 0
C
M
A
Σ= − °
− +°=
140.000 NA= +
140.0 N=A
30.0°
0: (69.282 N)cos30 (0.2 m)
(80 N)(0.6 m) ( sin30 )(0.4 m) 0
A
M
C
Σ= + °
− + °=
180.000 NC= +
180.0 N=C
30.0°
consent of McGraw–Hill Education.
PROBLEM 4.12
A lever AB is hinged at C and attached to a control cable at A. If
the lever is subjected to a 500–N horizontal force at B,
determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
Triangle ACD is isosceles with
90 30 120C= °+ °= °�
1(180 120 ) 30 .
2
AD
= = °− ° = °
��
(a) We resolve
AD
F
into components along AB and perpendicular to AB.
0: ( sin30 )(250 mm) (500 N)(100 mm) 0
C AD
MFΣ= ° − =
400 N
AD
F=
(b)
0: (400 N)cos60 500 N 0
xx
FCΣ = − °+ − =
300 N
x
C= +
0: (400 N)sin60° 0
yy
FCΣ= − + =
346.4 N
y
C= +
458 N=C
49.1°
consent of McGraw–Hill Education.
PROBLEM 4.13
Determine the reactions at A and B when (a)
α
= 0, (b)
α
= 90°,
(c)
α
= 30°.
SOLUTION
(a)
0
α
=
0: (20 in.) 75 lb(10 in.) 0
37.5 lb
A
MB
B
Σ= − =
=
0: 0
xx
FAΣ= =
+
0: 75 lb 37.5 lb 0
37.5 lb
yy
y
FA
A
Σ= − + =
=
37.5 lb= =AB
(b)
90
α
= °
0: (12 in.) 75 lb(10 in.) 0
62.5 lb
A
MB
B
Σ= − =
=
0: 0
62.5 lb
xx
x
F AB
A
Σ= −=
=
0: 75 lb 0
75 lb
yy
y
FA
A
Σ= − =
=
22
22
(62.5 lb) (75 lb)
97.6 lb
xy
A AA= +
= +
=
75
tan 62.5
50.2
θ
θ
=
= °
97.6 lb=A
50.2 ; 62.5 lb°=B
consent of McGraw–Hill Education.
SOLUTION Continued
(c)
30
α
= °
0: ( cos 30°)(20 in.) ( sin30 )(12 in.)
(75 lb)(10 in.) 0
A
MB BΣ= + °
−=
32.161 lbB=
0: (32.161)sin30 0
16.0805 lb
xx
x
FA
A
Σ = − °=
=
0: (32.161)cos30 75 0
47.148 lb
yy
y
FA
A
Σ = + °− =
=
22
22
(16.0805) (47.148)
49.8 lb
xy
A AA= +
= +
=
47.148
tan 16.0805
71.2
θ
θ
=
= °
49.8 lb=A
71.2 ; 32.2 lb°=B
60.0°
consent of McGraw–Hill Education.
PROBLEM 4.14
The bracket BCD is hinged at C and attached to a control
cable at B. For the loading shown, determine (a) the
tension in the cable, (b) the reaction at C.
SOLUTION
At B:
0.18 m
0.24 m
y
x
T
T=
3
4
yx
TT
=
(1)
(a)
0: (0.18 m) (240 N)(0.4 m) (240 N)(0.8 m) 0
Cx
MT
Σ= − − =
1600 N
x
T= +
From Eq. (1):
3(1600 N) 1200 N
4
y
T= =
22 2 2
1600 1200 2000 N
xy
T TT= += + =
2.00 kNT=
(b)
0: 0
x xx
F CT
Σ= −=
1600 N 0 1600 N
xx
CC−==+
1600 N
x=C
0: 240 N 240 N 0
y yy
F CTΣ= −− − =
1200 N 480 N 0
y
C− −=
1680 N
y
C= +
1680 N
y=C
46.4
2320 NC
α
= °
=
2.32 kN=C
46.4°
consent of McGraw–Hill Education.
PROBLEM 4.15
Solve Problem 4.14, assuming that
0.32 m.a=
PROBLEM 4.14 The bracket BCD is hinged at C and
attached to a control cable at B. For the loading shown,
determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
(a) At B:
0.32 m
0.24 m
4
3
y
x
yx
T
T
TT
=
=
0: (0.32 m) (240 N)(0.4 m) (240 N)(0.8 m) 0
Cx
MTΣ= − − =
900 N
x
T=
From Eq. (1):
4(900 N) 1200 N
3
y
T= =
22 2 2
900 1200 1500 N
xy
T TT= += + =
1.500 kNT=
(b)
0: 0
x xx
F CTΣ= −=
900 N 0 900 N
xx
CC−==+
900 N
x=C
0: 240 N 240 N 0
y yy
F CTΣ= −− − =
1200 N 480 N 0
y
C− −=
1680 N
y
C= +
1680 N
y=C
61.8
1906 NC
α
= °
=
1.906 kN=C
61.8°
consent of McGraw–Hill Education.
PROBLEM 4.16
Determine the reactions at A and B when (a)
0,h=
(b)
200 mm.h=
SOLUTION
Free–Body Diagram:
0: ( cos60 )(0.5 m) ( sin60 ) (150 N)(0.25 m) 0
A
M B BhΣ = ° − °− =
37.5
0.25 0.866
Bh
=−
(1)
(a) When
0,h=
From Eq. (1):
37.5 150 N
0.25
B= =
150.0 N=B
30.0°
0: sin60 0
yx
F ABΣ = − °=
(150)sin60 129.9 N
x
A= °=
129.9 N
x
=A
0: 150 cos60 0
yy
FA BΣ = − + °=
150 (150)cos60 75 N
y
A= − °=
75 N
y
=A
30
150.0 NA
α
= °
=
150.0 N=A
30.0°
(b) When h = 200 mm = 0.2 m,
From Eq. (1):
37.5 488.3 N
0.25 0.866(0.2)
B= =
−
488 N=B
30.0°
0: sin 60 0
xx
F ABΣ = − °=
(488.3)sin60 422.88 N
x
A= °=
422.88 N
x
=A
0: 150 cos60 0
yy
FA BΣ = − + °=
150 (488.3)cos60 94.15 N
y
A= − °=−
94.15 N
y=A
12.55
433.2 NA
α
= °
=
433 N=A
12.55°
consent of McGraw–Hill Education.
PROBLEM 4.17
A light bar AD is suspended from a cable BE and supports a
50-lb block at C. The ends A and D of the bar are in contact
with frictionless vertical walls. Determine the tension in cable
BE and the reactions at A and D.
SOLUTION
Free–Body Diagram:
0:
x
F ADΣ= =
0:
y
FΣ=
50.0 lb
BE
T=
We note that the forces shown form two couples.
0: (8 in.) (50 lb)(3 in.) 0MAΣ= − =
18.75 lbA=
18.75 lb=A
18.75 lb=D
consent of McGraw–Hill Education.
PROBLEM 4.18
Bar AD is attached at A and C to collars that can move freely
on the rods shown. If the cord BE is vertical (
α
= 0), determine
the tension in the cord and the reactions at A and C.
SOLUTION
Free–Body Diagram:
0: cos30 (80 N)cos30 0
y
FTΣ = − °+ °=
80 NT=
80.0 NT=
0: ( sin30 )(0.4 m) (80 N)(0.2 m) (80 N)(0.2 m) 0
C
MAΣ= ° − − =
160 NA= +
160.0 N=A
30.0°
0: (80 N)(0.2 m) (80 N)(0.6 m) ( sin30 )(0.4 m) 0
A
MCΣ= − + ° =
160 NC= +
160.0 N=C
30.0°
consent of McGraw–Hill Education.
PROBLEM 4.19
Solve Problem 4.18 if the cord BE is parallel to the rods (
α
= 30°).
PROBLEM 4.18 Bar AD is attached at A and C to collars that
can move freely on the rods shown. If the cord BE is vertical
(
α
= 0), determine the tension in the cord and the reactions at A
and C.
SOLUTION
Free–Body Diagram:
0: (80 N)cos30 0
y
FTΣ = − + °=
69.282 NT=
69.3 NT=
0: (69.282 N)cos30 (0.2 m)
(80 N)(0.2 m) ( sin30 )(0.4 m) 0
C
M
A
Σ= − °
− +°=
140.000 NA= +
140.0 N=A
30.0°
0: (69.282 N)cos30 (0.2 m)
(80 N)(0.6 m) ( sin30 )(0.4 m) 0
A
M
C
Σ= + °
− + °=
180.000 NC= +
180.0 N=C
30.0°
consent of McGraw–Hill Education.