PROBLEM 4.27
Determine the reactions at B and C when a = 30 mm.
PROBLEM 4.28
The spanner shown is used to rotate a shaft. A pin fits in a
hole at A, while a flat, frictionless surface rests against the
shaft at B. If a 60–lb force P is exerted on the spanner at D,
find the reactions at A and B.
SOLUTION
Free–Body Diagram:
(Three-force body)
The line of action of A must pass through D, where B and P intersect.
3sin50
tan 3cos50 15
0.135756
7.7310
a
a
°
=°+
=
= °
60 lb
sin 7.7310°
446.02 lb
60 lb
tan 7.7310°
441.97 lb
A
B
=
=
=
=
Force triangle
446 lb=A
7.73°
442 lb=B
consent of McGraw–Hill Education.
PROBLEM 4.29
A 12–ft wooden beam weighing 80 lb is supported by a pin and
bracket at A and by cable BC. Find the reaction at A and the tension
in the cable.
SOLUTION
Since CB is a two–force member, the force it exerts on member AB is directed along CB.
Free–Body Diagram of AB: (Three–Force member)
Triangle CFD:
( )
( )
26 ft 1.50 ft
8
BG
DG AC
AB
DG
=
= =
Triangle EAD:
1.50
tan 6
14.0362
DG
AE
a
a
= =
=
A
Triangle EGB: Force triangle
80 lb
sin53.13 sin75.964 sin 50.906°
AT
= =
°°
82.5 lb=A
14.04°
100.0 lbT=
consent of McGraw–Hill Education.
PROBLEM 4.30
A T–shaped bracket supports a 300–N load as shown. Determine
the reactions at A and C when α = 45°.
SOLUTION
Free–Body Diagram:
(Three-force body)
The line of action of C must pass through E, where A and the 300-N force intersect.
SOLUTION Continued
Force Triangle
Law of sines:
300 N
sin32.905 sin135 sin12.0948
AC
= =
°° °
778 N=A
;
1012 N=C
77.9°
consent of McGraw–Hill Education.
PROBLEM 4.31
One end of rod AB rests in the corner A and the other end is attached to
cord BD. If the rod supports a 150–N load at its midpoint C, find the
reaction at A and the tension in the cord.
SOLUTION
Free–Body Diagram: (Three-force body) Dimensions in mm
The line of action of reaction at A must pass through E, where T and the 150–N load intersect.
Force triangle
460
tan 240
62.447
100
tan 240
22.620
EF
AF
EH
DH
a
a
β
β
= =
= °
= =
= °
150 N
sin67.380 sin 27.553 sin 85.067°
AT
= =
°°
139.0 N=A
62.4°
69.6 NT=
PROBLEM 4.32
Using the method of Section 4.2B, solve Problem 4.12.
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
Reaction at C must pass through E, where FAD and 500-N force intersect.
Since
250 mm,AC CD= =
triangle
ACD
is isosceles.
We have
90 30 120C= °+ °= °�
and
1(180 120 ) 30
2
AD= = °− ° = °��
Dimensions in mm
On the other hand, from triangle BCF:
( )sin30 200 sin 30 100 mm
250 100 150 mm
CF BC
FD CD CF
= °= °=
=−= −=
From triangle EFD, and since
30 :D= °�
( )tan30 150tan30 86.60 mmEF FD= °= °=
From triangle EFC:
100 mm
tan 86.60 mm
49.11
CF
EF
a
a
= =
= °
Force triangle
Law of sines
500 N
sin 49.11 sin 60 sin 70.89°
400 N, 458 N
AD
AD
FC
FC
= =
°°
= =
(a)
400 N
AD
F=
(b)
458 N=C
49.1°
Free–Body Diagram:
(Three-Force body)
consent of McGraw–Hill Education.
PROBLEM 4.33
Using the method of Section 4.2B, solve Problem 4.16.
PROBLEM 4.16 Determine the reactions at A and B when
(a)
0,h=
(b)
200 mm.h=
SOLUTION
(a) h = 0
Free–Body Diagram:
PROBLEM 4.34
A 40–lb roller, of diameter 8 in., which is to be used on a tile floor, is
resting directly on the subflooring as shown. Knowing that the thickness
of each tile is 0.3 in., determine the force P required to move the roller
onto the tiles if the roller is (a) pushed to the left, (b) pulled to the right.
SOLUTION
Force Triangle
tile,
13.7 in.
cos 4 in.
22.332
α
α
−
=
= °
(a) Roller pushed to left (three-force body):
Forces must pass through O.
Law of sines:
40 lb ; 24.87 lb
sin37.668 sin 22.332
PP= =
°°
24.9 lb=P
30.0°
PROBLEM 4.35
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
Reaction at B must pass through D.
7 in.
tan 12 in.
30.256
7 in.
tan 24 in.
16.26
a
a
β
β
=
= °
=
= °
Force Triangle
Law of sines:
72 lb
sin59.744 sin13.996 sin106.26
(sin13.996 ) ( 72 lb)(sin 59.744°)
(0.24185) ( 72)(0.86378)
TT B
TT
TT
−
= =
°°
°= −
= −
100.00 lbT=
100.0 lbT=
sin 106.26°
(100 lb) sin59.744
111.14 lb
B=°
=
111.1 lb
=B
30.3°
Free–Body Diagram:
PROBLEM 4.28
The spanner shown is used to rotate a shaft. A pin fits in a
hole at A, while a flat, frictionless surface rests against the
shaft at B. If a 60–lb force P is exerted on the spanner at D,
find the reactions at A and B.
SOLUTION
Free–Body Diagram:
(Three-force body)
The line of action of A must pass through D, where B and P intersect.
3sin50
tan 3cos50 15
0.135756
7.7310
a
a
°
=°+
=
= °
60 lb
sin 7.7310°
446.02 lb
60 lb
tan 7.7310°
441.97 lb
A
B
=
=
=
=
Force triangle
446 lb=A
7.73°
442 lb=B
consent of McGraw–Hill Education.
PROBLEM 4.29
A 12–ft wooden beam weighing 80 lb is supported by a pin and
bracket at A and by cable BC. Find the reaction at A and the tension
in the cable.
SOLUTION
Since CB is a two–force member, the force it exerts on member AB is directed along CB.
Free–Body Diagram of AB: (Three–Force member)
Triangle CFD:
( )
( )
26 ft 1.50 ft
8
BG
DG AC
AB
DG
=
= =
Triangle EAD:
1.50
tan 6
14.0362
DG
AE
a
a
= =
=
A
Triangle EGB: Force triangle
80 lb
sin53.13 sin75.964 sin 50.906°
AT
= =
°°
82.5 lb=A
14.04°
100.0 lbT=
consent of McGraw–Hill Education.
PROBLEM 4.30
A T–shaped bracket supports a 300–N load as shown. Determine
the reactions at A and C when α = 45°.
SOLUTION
Free–Body Diagram:
(Three-force body)
The line of action of C must pass through E, where A and the 300-N force intersect.
SOLUTION Continued
Force Triangle
Law of sines:
300 N
sin32.905 sin135 sin12.0948
AC
= =
°° °
778 N=A
;
1012 N=C
77.9°
consent of McGraw–Hill Education.
PROBLEM 4.31
One end of rod AB rests in the corner A and the other end is attached to
cord BD. If the rod supports a 150–N load at its midpoint C, find the
reaction at A and the tension in the cord.
SOLUTION
Free–Body Diagram: (Three-force body) Dimensions in mm
The line of action of reaction at A must pass through E, where T and the 150–N load intersect.
Force triangle
460
tan 240
62.447
100
tan 240
22.620
EF
AF
EH
DH
a
a
β
β
= =
= °
= =
= °
150 N
sin67.380 sin 27.553 sin 85.067°
AT
= =
°°
139.0 N=A
62.4°
69.6 NT=
PROBLEM 4.32
Using the method of Section 4.2B, solve Problem 4.12.
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
Reaction at C must pass through E, where FAD and 500-N force intersect.
Since
250 mm,AC CD= =
triangle
ACD
is isosceles.
We have
90 30 120C= °+ °= °�
and
1(180 120 ) 30
2
AD= = °− ° = °��
Dimensions in mm
On the other hand, from triangle BCF:
( )sin30 200 sin 30 100 mm
250 100 150 mm
CF BC
FD CD CF
= °= °=
=−= −=
From triangle EFD, and since
30 :D= °�
( )tan30 150tan30 86.60 mmEF FD= °= °=
From triangle EFC:
100 mm
tan 86.60 mm
49.11
CF
EF
a
a
= =
= °
Force triangle
Law of sines
500 N
sin 49.11 sin 60 sin 70.89°
400 N, 458 N
AD
AD
FC
FC
= =
°°
= =
(a)
400 N
AD
F=
(b)
458 N=C
49.1°
Free–Body Diagram:
(Three-Force body)
consent of McGraw–Hill Education.
PROBLEM 4.33
Using the method of Section 4.2B, solve Problem 4.16.
PROBLEM 4.16 Determine the reactions at A and B when
(a)
0,h=
(b)
200 mm.h=
SOLUTION
(a) h = 0
Free–Body Diagram:
PROBLEM 4.34
A 40–lb roller, of diameter 8 in., which is to be used on a tile floor, is
resting directly on the subflooring as shown. Knowing that the thickness
of each tile is 0.3 in., determine the force P required to move the roller
onto the tiles if the roller is (a) pushed to the left, (b) pulled to the right.
SOLUTION
Force Triangle
tile,
13.7 in.
cos 4 in.
22.332
α
α
−
=
= °
(a) Roller pushed to left (three-force body):
Forces must pass through O.
Law of sines:
40 lb ; 24.87 lb
sin37.668 sin 22.332
PP= =
°°
24.9 lb=P
30.0°
PROBLEM 4.35
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
Reaction at B must pass through D.
7 in.
tan 12 in.
30.256
7 in.
tan 24 in.
16.26
a
a
β
β
=
= °
=
= °
Force Triangle
Law of sines:
72 lb
sin59.744 sin13.996 sin106.26
(sin13.996 ) ( 72 lb)(sin 59.744°)
(0.24185) ( 72)(0.86378)
TT B
TT
TT
−
= =
°°
°= −
= −
100.00 lbT=
100.0 lbT=
sin 106.26°
(100 lb) sin59.744
111.14 lb
B=°
=
111.1 lb
=B
30.3°
Free–Body Diagram: