Eighth
Edition
Vector Mechanics for Engineers: Statics
CE 102 Statics
Chapter 4
Equilibrium of Rigid Bodies
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4-1
Eighth
Edition
Vector Mechanics for Engineers: Statics
Contents
Introduction
Free-Body Diagram
Equilibrium of a Rigid Body in Three
Dimensions
Reactions at Supports and Connections
for a Two-Dimensional Structure
Reactions at Supports and Connections for a
Three-Dimensional Structure
Equilibrium of a Rigid Body in Two
Dimensions
Sample Problem 4.5
Statically Indeterminate Reactions
Sample Problem 4.1
Sample Problem 4.2
Sample Problem 4.3
Equilibrium of a Two-Force Body
Equilibrium of a Three-Force Body
Sample Problem 4.4
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4-2
Eighth
Edition
Vector Mechanics for Engineers: Statics
Introduction
• For a rigid body in static equilibrium, the external forces and
moments are balanced and will impart no translational or rotational
motion to the body.
• The necessary and sufficient condition for the static equilibrium of a
body are that the resultant force and couple from all external forces
form a system equivalent to zero,


 
 F  0  M O   r  F   0
• Resolving each force and moment into its rectangular components
leads to 6 scalar equations which also express the conditions for static
equilibrium,
 Fx  0  Fy  0  Fz  0
Mx  0 My  0 Mz  0
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4-3
Eighth
Edition
Vector Mechanics for Engineers: Statics
Free-Body Diagram
First step in the static equilibrium analysis of a rigid
body is identification of all forces acting on the
body with a free-body diagram.
• Select the extent of the free-body and detach it
from the ground and all other bodies.
• Indicate point of application, magnitude, and
direction of external forces, including the rigid
body weight.
• Indicate point of application and assumed
direction of unknown applied forces. These
usually consist of reactions through which the
ground and other bodies oppose the possible
motion of the rigid body.
• Include the dimensions necessary to compute
the moments of the forces.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4-4
Eighth
Edition
Vector Mechanics for Engineers: Statics
Reactions at Supports and Connections for a TwoDimensional Structure
• Reactions equivalent to a
force with known line of
action.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4-5
Eighth
Edition
Vector Mechanics for Engineers: Statics
Reactions at Supports and Connections for a TwoDimensional Structure
• Reactions equivalent to a
force of unknown direction
and magnitude.
• Reactions equivalent to a
force of unknown
direction and magnitude
and a couple.of unknown
magnitude
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4-6
Eighth
Edition
Vector Mechanics for Engineers: Statics
Equilibrium of a Rigid Body in Two Dimensions
• For all forces and moments acting on a twodimensional structure,
Fz  0 M x  M y  0 M z  M O
• Equations of equilibrium become
 Fx  0  Fy  0  M A  0
where A is any point in the plane of the
structure.
• The 3 equations can be solved for no more
than 3 unknowns.
• The 3 equations can not be augmented with
additional equations, but they can be replaced
 Fx  0  M A  0  M B  0
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4-7
Eighth
Edition
Vector Mechanics for Engineers: Statics
Statically Indeterminate Reactions
• More unknowns than
equations
• Fewer unknowns than • Equal number unknowns
and equations but
equations, partially
improperly constrained
constrained
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4-8
Eighth
Edition
Vector Mechanics for Engineers: Statics
Sample Problem 4.1
SOLUTION:
• Create a free-body diagram for the crane.
• Determine B by solving the equation for
the sum of the moments of all forces
about A. Note there will be no
contribution from the unknown
reactions at A.
A fixed crane has a mass of 1000 kg
and is used to lift a 2400 kg crate. It
is held in place by a pin at A and a
rocker at B. The center of gravity of
the crane is located at G.
Determine the components of the
reactions at A and B.
• Determine the reactions at A by
solving the equations for the sum of
all horizontal force components and
all vertical force components.
• Check the values obtained for the
reactions by verifying that the sum of
the moments about B of all forces is
zero.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4-9
Eighth
Edition
Vector Mechanics for Engineers: Statics
Sample Problem 4.1
• Determine B by solving the equation for the
sum of the moments of all forces about A.
 M A  0 :  B1.5m   9.81 kN2m 
 23.5 kN6m   0
B  107.1 kN
• Create the free-body diagram.
• Determine the reactions at A by solving the
equations for the sum of all horizontal forces
and all vertical forces.
 Fx  0 : Ax  B  0
Ax  107.1kN
 Fy  0 : Ay  9.81kN  23.5 kN  0
Ay  33.3 kN
• Check the values obtained.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 10
Eighth
Edition
Vector Mechanics for Engineers: Statics
Sample Problem 4.2
SOLUTION:
• Create a free-body diagram for the car
with the coordinate system aligned
with the track.
• Determine the reactions at the wheels
by solving equations for the sum of
moments about points above each axle.
• Determine the cable tension by
A loading car is at rest on an inclined
solving the equation for the sum of
track. The gross weight of the car and
force components parallel to the track.
its load is 5500 lb, and it is applied at
at G. The cart is held in position by
• Check the values obtained by verifying
the cable.
that the sum of force components
perpendicular to the track are zero.
Determine the tension in the cable and
the reaction at each pair of wheels.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 11
Eighth
Edition
Vector Mechanics for Engineers: Statics
Sample Problem 4.2
• Determine the reactions at the wheels.
 M A  0 :  2320 lb 25in.  4980 lb 6in.
 R2 50in.  0
R2  1758 lb
 M B  0 :  2320 lb 25in.  4980 lb 6in.
 R1 50in.  0
R1  562 lb
• Create a free-body diagram
W x  5500 lb cos 25
 4980 lb
W y  5500 lb sin 25
• Determine the cable tension.
 Fx  0 :  4980 lb  T  0
T  4980 lb
 2320 lb
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 12
Eighth
Edition
Vector Mechanics for Engineers: Statics
Sample Problem 4.3
SOLUTION:
• Create a free-body diagram for the
frame and cable.
• Solve 3 equilibrium equations for the
reaction force components and
couple at E.
The frame supports part of the roof of
a small building. The tension in the
cable is 150 kN.
Determine the reaction at the fixed
end E.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 13
Eighth
Edition
Vector Mechanics for Engineers: Statics
Sample Problem 4.3
• Solve 3 equilibrium equations for the
reaction force components and couple.
4.5
150 kN   0
F

0
:
E

 x
x
7.5
E x  90.0 kN
 Fy  0 : E y  420 kN  
6
150 kN   0
7.5
E y  200 kN
• Create a free-body diagram for
the frame and cable.
 M E  0 :  20 kN 7.2 m   20 kN 5.4 m 
 20 kN 3.6 m   20 kN 1.8 m 

6
150 kN 4.5 m  M E  0
7.5
M E  180.0 kN  m
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 14
Eighth
Edition
Vector Mechanics for Engineers: Statics
Equilibrium of a Two-Force Body
• Consider a plate subjected to two forces F1 and F2
• For static equilibrium, the sum of moments about A
must be zero. The moment of F2 must be zero. It
follows that the line of action of F2 must pass
through A.
• Similarly, the line of action of F1 must pass
through B for the sum of moments about B to be
zero.
• Requiring that the sum of forces in any direction be
zero leads to the conclusion that F1 and F2 must
have equal magnitude but opposite sense.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 15
Eighth
Edition
Vector Mechanics for Engineers: Statics
Equilibrium of a Three-Force Body
• Consider a rigid body subjected to forces acting at
only 3 points.
• Assuming that their lines of action intersect, the
moment of F1 and F2 about the point of intersection
represented by D is zero.
• Since the rigid body is in equilibrium, the sum of the
moments of F1, F2, and F3 about any axis must be
zero. It follows that the moment of F3 about D must
be zero as well and that the line of action of F3 must
pass through D.
• The lines of action of the three forces must be
concurrent or parallel.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 16
Eighth
Edition
Vector Mechanics for Engineers: Statics
Sample Problem 4.4
SOLUTION:
• Create a free-body diagram of the joist.
Note that the joist is a 3 force body acted
upon by the rope, its weight, and the
reaction at A.
A man raises a 10 kg joist, of
length 4 m, by pulling on a rope.
Find the tension in the rope and
the reaction at A.
• The three forces must be concurrent for
static equilibrium. Therefore, the reaction
R must pass through the intersection of the
lines of action of the weight and rope
forces. Determine the direction of the
reaction force R.
• Utilize a force triangle to determine the
magnitude of the reaction force R.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 17
Eighth
Edition
Vector Mechanics for Engineers: Statics
Sample Problem 4.4
• Create a free-body diagram of the joist.
• Determine the direction of the reaction
force R.
AF  AB cos 45  4 m cos 45  2.828 m
CD  AE  12 AF  1.414 m
BD  CD cot(45  25)  1.414 m tan 20  0.515 m
CE  BF  BD  2.828  0.515 m  2.313 m
tan 
CE 2.313

 1.636
AE 1.414
  58.6
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 18
Eighth
Edition
Vector Mechanics for Engineers: Statics
Sample Problem 4.4
• Determine the magnitude of the reaction
force R.
T
sin 31.4

R
sin 110

98.1 N
sin 38.6
T  81.9 N
R  147.8 N
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 19
Eighth
Edition
Vector Mechanics for Engineers: Statics
Equilibrium of a Rigid Body in Three Dimensions
• Six scalar equations are required to express the
conditions for the equilibrium of a rigid body in the
general three dimensional case.
 Fx  0  Fy  0  Fz  0
Mx  0 My  0 Mz  0
• These equations can be solved for no more than 6
unknowns which generally represent reactions at supports
or connections.
• The scalar equations are conveniently obtained by applying the
vector forms of the conditions for equilibrium,


 
 F  0  M O   r  F   0
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 20
Eighth
Edition
Vector Mechanics for Engineers: Statics
Reactions at Supports and Connections for a ThreeDimensional Structure
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 21
Eighth
Edition
Vector Mechanics for Engineers: Statics
Reactions at Supports and Connections for a ThreeDimensional Structure
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 22
Eighth
Edition
Vector Mechanics for Engineers: Statics
Sample Problem 4.5
SOLUTION:
• Create a free-body diagram for the sign.
• Apply the conditions for static
equilibrium to develop equations for
the unknown reactions.
A sign of uniform density weighs 270
lb and is supported by a ball-andsocket joint at A and by two cables.
Determine the tension in each cable
and the reaction at A.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 23
Eighth
Edition
Vector Mechanics for Engineers: Statics
Sample Problem 4.5

TBD  TBD
 TBD
 TBD

TEC  TEC
• Create a free-body diagram for the
sign.
 TEC
Since there are only 5 unknowns,
the sign is partially constrain. It is
free to rotate about the x axis. It is,
however, in equilibrium for the
given loading.
 TEC
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.


rD  rB


rD  rB



 8i  4 j  8k
12
 1 2
2
 3i  3 j  3k
 
rC  rE
 
rC  rE



 6i  3 j  2 k
7
 3 2
6
7i 7 j 7k




4 - 24
Eighth
Edition
Vector Mechanics for Engineers: Statics
Sample Problem 4.5
• Apply the conditions for
static equilibrium to
develop equations for the
unknown reactions.

F 

i:

j:

k:

MA

j:

k:
 


A  TBD  TEC  270 lb j  0
Ax  23 TBD  76 TEC  0
Ay  13 TBD  73 TEC  270 lb  0
Az  23 TBD  72 TEC  0



 

 rB  TBD  rE  TEC  4 ft i   270 lb j  0
5.333TBD  1.714TEC  0
2.667TBD  2.571TEC  1080 lb  0
Solve the 5 equations for the 5 unknowns,
TBD  101.3 lb TEC  315 lb




A  338 lbi  101.2 lb j  22.5 lbk
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 25
Problem 4.6

P
A
45o
B
O
D
45o
C
The semicircular rod ABCD is
maintained in equilibrium by
the small wheel at D and the
rollers at B and C. Knowing
that  = 45o, determine the
reactions at B , C , and D.
26
Problem 4.6

P
A
45
B
Solving Problems on Your Own
O
o
D
45
o
C
The semicircular rod ABCD is
maintained in equilibrium by
the small wheel at D and the
rollers at B and C. Knowing
that  = 45o, determine the
reactions at B , C , and D.
1. Draw a free-body diagram of the body. This diagram shows
the body and all the forces acting on it.
27
Problem 4.6

P
A
45
B
Solving Problems on Your Own
O
o
D
45
o
C
The semicircular rod ABCD is
maintained in equilibrium by
the small wheel at D and the
rollers at B and C. Knowing
that  = 45o, determine the
reactions at B , C , and D.
2. Write equilibrium equations and solve for the unknowns.
For two-dimensional structure the three equations might be:
SFx = 0 SFy = 0
SMO = 0
where O is an arbitrary point in the plane of the structure
or
SFx = 0 SMA = 0 SMB = 0
where point B is such that line AB is not parallel to the y axis
or
SMA = 0 SMB = 0 SMC = 0
where the points A, B , and C do not lie in a straight line. 28
Problem 4.6 Solution

P
A
45
B
O
o
D
45
o
Draw a free-body diagram
of the body.
C
P sin
P
A
P cos
O
45o
B
D
45o
C
R
B/ 2
B
D
B/ 2
C/ 2
C
C/ 2
29
Problem 4.6 Solution
P sin
P
A
P cos
O
45o
B
Write three equilibrium
equations and solve
for the unknowns.
D
45o
C
R
B/ 2
B
D
C/ 2
C
B/ 2
C/ 2
+ S MO = 0: (P sin) R _ D (R) = 0
+ S Fx = 0:
P cos + B/ 2
+ S Fy = 0:
_
_
1
D = P sin
C/ 2=0
(2)
P sin + B/ 2 + C / 2 _ P sin = 0
_
2P sin + B/ 2 + C / 2 = 0
(3)
30
P sin
P
A
P cos
O
45o
C
R
B/ 2
(2) + (3)
D
45o
B
B
Problem 4.6 Solution
D
B/ 2
C/ 2
C
C/ 2
P(cos _ 2sin) + 2 B/ 2 = 0
2
B=
(2sin _ cos) P
2
(2) _ (3)
(4)
P(cos + 2sin) _ 2 C/ 2 = 0
2
C=
(2sin + cos) P
2
(5)
31
P sin
P
A
P cos
O
45o
B
B
EQ. (4) :
EQ. (5) :
EQ. (1) :
D
45o
C
R
B/ 2
B/ 2
Problem 4.6 Solution
D
For  = 45o
C/ 2
C
sin = cos = 1/ 2
C/ 2
2 2
B=
(
2
2
2 2
C=
(
2
2
D = P/ 2
_
_
1
1
)P= P;
2
2
1
3
)P= P;
2
2
1
B= P
2
C= 3 P
2
45o
45o
D = P/ 2
32
Problem 4.7
4 in 4 in
20 lb
A
C
q
40 lb
B
2 in
3 in
D
3 in
E
The T-shaped bracket shown is
supported by a small wheel at E
and pegs at C and D. Neglecting
the effect of friction, determine
the reactions at C , D , and E
when q = 30o.
33
4 in 4 in
20 lb
A
C
q
40 lb
B
2 in
3 in
D
3 in
E
Problem 4.7
Solving Problems on Your Own
The T-shaped bracket shown is
supported by a small wheel at E
and pegs at C and D. Neglecting
the effect of friction, determine
the reactions at C , D , and E
when q = 30o.
1. Draw a free-body diagram of the body. This diagram shows
the body and all the forces acting on it.
34
4 in 4 in
20 lb
A
C
q
40 lb
B
2 in
3 in
D
3 in
E
Problem 4.7
Solving Problems on Your Own
The T-shaped bracket shown is
supported by a small wheel at E
and pegs at C and D. Neglecting
the effect of friction, determine
the reactions at C , D , and E
when q = 30o.
2. Write equilibrium equations and solve for the unknowns.
For two-dimensional structure the three equations might be:
SFx = 0 SFy = 0
SMO = 0
where O is an arbitrary point in the plane of the structure
or
SFx = 0 SMA = 0 SMB = 0
where point B is such that line AB is not parallel to the y axis
or
SMA = 0 SMB = 0 SMC = 0
where the points A, B , and C do not lie in a straight line. 35
Problem 4.7 Solution
4 in 4 in
20 lb
A
C
q
40 lb
B
2 in
3 in
D
3 in
E
Draw a free-body diagram of the body.
4 in 4 in
20 lb
A
C
40 lb
B
2 in
C
D
D
E
3 in
3 in
E
30o
36
Problem 4.7 Solution
4 in 4 in
20 lb
A
C
40 lb
B
2 in
C
D
D
E
3 in
3 in
Write equilibrium equations and
solve for the unknowns.
E
30o
 S Fy = 0:
E cos 30o _ 20 _ 40 = 0
60 lb
E=
o = 69.28 lb
cos 30
E = 69.3 lb
60o
37
Problem 4.7 Solution
4 in 4 in
20 lb
A
C
40 lb
B
2 in
C
D
D
E
3 in
3 in
E
30o
 S MD = 0:
( 20 lb)( 4 in) _ ( 40 lb)( 4 in) _ C ( 3 in) + E sin 30o ( 3 in) = 0
_
80 _ 3C + 69.28 ( 0.5 )( 3 ) = 0
C = 7.974 lb
C = 7.97 lb
38
Problem 4.7 Solution
4 in 4 in
20 lb
A
C
40 lb
B
2 in
C
D
D
E
3 in
3 in
+
S Fx= 0:
E sin 30o + C _ D = 0
E
( 69.28 lb )( 0.5 ) + 7.974 lb _ D = 0
30o
D = 42.6 lb
39
Problem 4.8
y 1.2 m
E
1.2 m
D
1.5 m
z
C
A
B
5 kN
2m
f
1m
x
A 3-m pole is supported by a
ball-and-socket joint at A and
by the cables CD and CE.
Knowing that the line of action
of the 5-kN force forms an
angle f=30o with the vertical
xy plane, determine (a) the
tension in cables CD and CE,
(b) the reaction at A.
40
Problem 4.8
y 1.2 m
Solving Problems on Your Own
E
1.2 m
D
1.5 m
z
C
A
B
5 kN
2m
f
1m
x
A 3-m pole is supported by a
ball-and-socket joint at A and
by the cables CD and CE.
Knowing that the line of action
of the 5-kN force forms an
angle f=30o with the vertical
xy plane, determine (a) the
tension in cables CD and CE,
(b) the reaction at A.
1. Draw a free-body diagram of the body. This diagram shows
the body and all the forces acting on it.
41
y 1.2 m
Problem 4.8
E
Solving Problems on Your Own
1.2 m
A 3-m pole is supported by a balland-socket joint at A and by the
cables CD and CE. Knowing that
D
C the line of actionof the 5-kN force
1.5 m
x
A
B
forms an angle f=30o with the
5 kN
vertical xy plane, determine (a)
z
f
the tension in cables CD and CE,
1m
2m
(b) the reaction at A.
2. Write equilibrium equations and solve for the unknowns.
For three-dimensional body the six scalar equations
SFx = 0 SFy = 0 SFz = 0
SMx = 0 SMy = 0 SMz = 0
should be used and solved for six unknowns. These equations
can also be written as
SF = 0
SMO = S (r x F ) = 0
where F are the forces and r are position vectors.
42
Problem 4.8 Solution
y 1.2 m
E
1.2 m
Draw a free-body diagram
of the body.
D
1.5 m
z
C
A
B
5 kN
2m
x
y 1.2 m
f
E
1.2 m
1m
D
TCD
Ax i
1.5 m
z
Az k
TCE
C
B
A
x
30o
Ay j 5 kN
2m
1m
43
Problem 4.8 Solution
y 1.2 m
Write equilibrium equations
and solve for the unknowns.
E
1.2 m
D
TCD
Ax i
1.5 m
z
Az k
TCE
C
B
A
x
30o
Ay j 5 kN
2m
5 unknowns and 6 equations of
equilibrium, but equilibrium is
maintained, S MAC = 0 .
rB/A = 2 i
rC/A = 3 i
1m
Load at B, FB = _ ( 5 cos 30o ) j + ( 5 sin 30o ) k = _ 4.33 j + 2.5 k
CD = _ 3 i+ 1.5 j + 1.2 k
CD = 3.562 m
T
TCD = TCD CD = CD (_ 3 i + 1.5 j + 1.2 k)
CD 3.562
TCE = TCE
CE = TCD (_ 3 i + 1.5 j _ 1.2 k)
CE
3.562
44
Problem 4.8 Solution
y 1.2 m
E
1.2 m
D
TCD
Ax i
1.5 m
z
Az k
TCE
C
B
A
x
30o
Ay j 5 kN
2m
1m
SMA = 0: rC/A x TCD + rC/A x TCE + rB/A x FB = 0
i j k
i j k
i j k
TCD
TCE
3 0 0
+ 2 0 0 =0
+ 3 0 0
3.562
3.562
_
_
_
3 1.5 1.2
3 1.5 1.2
0 _4.33 2.5
45
Problem 4.8 Solution
y 1.2 m
E
1.2 m
D
TCD
Ax i
1.5 m
z
Az k
TCE
B
A
TCD
TCE _
j: 3.6
+ 3.6
5=0
3.562
3.562
C
_
_
3.6
T
+3.6
T
(1)
x
CD
CE 17.81 = 0
30o
Ay j 5 kN
2m
(2) + 1.25 (1):
Eq. (1):
Equate coefficients of unit
vectors to zero.
_
1m
9TCE
_
TCD
TCE _
k: 4.5
+ 4.5
8.66 = 0
3.562
3.562
4.5 TCD+4.5 TCE = 30.85
(2)
_
53.11 = 0 ;
3.6TCD + 3.6 (5.902) _ 17.81 = 0
TCE = 5.90 kN
TCD = 0.954 kN
46
Problem 4.8 Solution
y 1.2 m
E
1.2 m
D
TCD
Ax i
1.5 m
z
Az k
SF = 0: A + TCD + TCE + FB = 0
TCE
C
B
A
i: Ax +
x
0.954 _
5.902 _
( 3) +
( 3) = 0
3.562
3.562
Ax = 5.77 kN
30o
Ay j 5 kN
2m
1m
0.954
5.902
j: Ay +
(1.5) +
(1.5) _ 4.33 = 0
3.562
3.562
0.954
5.902 _
k: Az +
(1.2) +
( 1.2) + 2.5 = 0
3.562
3.562
Ay = 1.443 kN
Az = _ 0.833 kN
A = ( 5.77 kN) i + ( 1.443 kN ) j - ( 0.833 kN ) k
47
Problem 4.9
A
a
B
C
60 lb
10 in
20 in
Rod AC is supported by a pin and
bracket at A and rests against a
peg at B. Neglecting the effect of
friction, determine (a) the reactions
at A and B when a = 8 in., (b) the
distance a for which the reaction
at A is horizontal and the
corresponding magnitudes of the
reactions at A and B.
48
Problem 4.9
A
Solving Problems on Your Own
a
B
C
60 lb
10 in
20 in
Rod AC is supported by a pin and
bracket at A and rests against a
peg at B. Neglecting the effect of
friction, determine (a) the reactions
at A and B when a = 8 in., (b) the
distance a for which the reaction
at A is horizontal and the
corresponding magnitudes of the
reactions at A and B.
1. Draw a free-body diagram of the body. This diagram shows
the body and all the forces acting on it.
49
Problem 4.9
Solving Problems on Your Own
A
a
B
C
60 lb
10 in
20 in
Rod AC is supported by a pin and
bracket at A and rests against a
peg at B. Neglecting the effect of
friction, determine (a) the reactions
at A and B when a = 8 in., (b) the
distance a for which the reaction
at A is horizontal and the
corresponding magnitudes of the
reactions at A and B.
2. For a three-force body, solution can be obtained by
constructing a force triangle. The resultants of the three forces
must be concurrent or parallel. To solve a problem involving a
three-force body with concurrent forces, draw the free-body
diagram showing that the three forces pass through the same
point. Complete the solution by using a force triangle.
50
Problem 4.9 Solution
A
(a) a = 8 in
a
B
Draw a free-body diagram
of the body.
20 in
10 in
C
A
60 lb
10 in
F
 = 26.57
A
D
G
tan  = 21
B
o
2
C
60 lb
g
E
8 in

B
12 in
1
10 in
51
Problem 4.9 Solution
10 in
Construct a force triangle.
A
F
A
D
G
B
2
C
60 lb
g
E
3 - FORCE BODY
8 in

B
12 in
Reaction at A passes through D
where B and 60-lb load intersect
AE =
1
1
EB =
(8) = 4 in.
2
2
1
EF = BG = 10 _ 4 = 6 in
10 in
DG = 2 BG = 2 (6) = 3 in.
1
1
FD = FG _ DG = 8 _ 3 = 5 in.
FD
5
Tan g = AF =
;
10
g = 26.57o
52
Problem 4.9 Solution
10 in
A
A
D
G
2
60 lb
8 in
 = 26.57o
30 lb
60 lb
A
B
C
g
E
F
FORCE TRIANGLE
1
10 in

B
12 in
B
 = 26.57o
30 lb
30 lb
A=B=
o = 67.08 lb
sin 26.57
A = 67.1 lb
26.6o
B = 67.1 lb
26.6o
53
Problem 4.9 Solution
A
(b) For A horizontal
a
B
20 in
Draw a free-body diagram
of the body.
10 in
C
60 lb
10 in
A
F
G


B
2
C
a
A

B
 = 26.57o
1
60 lb
54
Problem 4.9 Solution
10 in
A
F
G


2
C
60 lb
1
a
A
B
Construct a force triangle.

B
 = 26.57o
D ABF : BF = AF cos 
D BFG : FG = BF sin 
a = FG = AF cos  sin 
a = (10 in.) cos 26.57o sin 26.57o
a = 4.00 in.
55
Problem 4.9 Solution
10 in
FORCE TRIANGLE
A
F
G


B
2
C
a
A

B
B
 = 26.57o
60 lb
 = 26.57o
1
A
60 lb
A = 60 lb = 120 lb
tan 
B = 60 lb = 134.16 lb
sin 
A = 120.0 lb
B = 134.2 lb
26.6o
56
Problem 4.10
a
a
a

A
P
B
30o
30o
C
D
Rod AD supports a vertical load
P and is attached to collars B
and C, which may slide freely on
the rods shown. Knowing that
the wire attached at D forms an
angle  = 30o with the vertical,
determine (a) the tension in the
wire, (b) the reactions at B and C.
57
Problem 4.10
Solving Problems on Your Own
a
a
a

A
P
B
30o
30o
C
D
Rod AD supports a vertical load
P and is attached to collars B
and C, which may slide freely on
the rods shown. Knowing that
the wire attached at D forms an
angle  = 30o with the vertical,
determine (a) the tension in the
wire, (b) the reactions at B and C.
1. Draw a free-body diagram of the body. This diagram shows
the body and all the forces acting on it.
58
Problem 4.10
Solving Problems on Your Own
a
a
a
Rod AD supports a vertical load
P and is attached to collars B

and C, which may slide freely on
A
B
C
D
the rods shown. Knowing that
o
o
30
30
the wire attached at D forms an
P
angle  = 30o with the vertical,
determine (a) the tension in the
wire, (b) the reactions at B and C.
2. Write equilibrium equations and solve for the unknowns.
For two-dimensional structure the three equations might be:
SFx = 0 SFy = 0 SMO = 0
where O is an arbitrary point in the plane of the structure
or
SFx = 0 SMA = 0 SMB = 0
where point B is such that line AB is not parallel to the y axis
or
SMA = 0 SMB = 0 SMC = 0
where the points A, B , and C do not lie in a straight line. 59
Problem 4.10 Solution
a
a
a
Draw a free-body diagram
of the body.

A
P
B
30o
C
D
30o
30o
30o
o
30o 30o 30
C
A
B
C
T
D
30o
B
P
a
a
a
60
30o
30o
30
C
A
B
C
30
o
30
o
30o
T
D
o
B
P
a
a
Write equilibrium equations
and solve for the unknowns.
a
S F = 0:
30o
Problem 4.10 Solution
_
P cos 30o + T cos 60o = 0
cos 30o
3/2
T= P
T=
o = P 1/2
cos 60
+ S MB = 0: P a _ (C sin 30o) a + T cos 30o (2a) = 0
Pa_( 1 C)a+ 3P(
3 ) 2a = 0
2
1
C + (1 + 3) P = 0;
2
C=8P
2
_
C=8P
3 P
30o61
30o
30o
30
C
A
B
C
30
o
30
o
Problem 4.10 Solution
30o
T
D
o
B
P
a
a
+ S F = 0:
a
_
_
B cos 30o + C cos 30o _ T sin 30o = 0
3
3
B 2 +8P 2
B= 7P
_
1
3 P ( 2 ) = 0;
B=7P
30o
62