Download Engineering Mechanics CHAPTER 4

4.1 Introduction
This chapter studies one of the most important section of
statics, where a body remains in equilibrium under the action
of a system of forces and couples.
Upon completion of this chapter, the student will be able
1. Draw complete free-body diagrams for the coplanar force
systems.
2. Apply the three equations of equilibrium :
Fx = 0
Fy = 0
M = 0
4.2 Equilibrium
Newton’s First Law states that “ Every body remains at rest or
maintains a constant velocity in a straight line unless an
unbalanced force acts upon it.
A body is said to be in equilibrium if the forces and moment
acting on it are balanced, and the body remains at rest or, if it
is in motion, there is no tendency for it to change its speed or
direction.
Hence for static equilibrium to occur:
1. Vertical forces must be balance;
Fy = 0
2. Horizontal forces must be balance;
Fx = 0
3. Moments must balance ie. ccw = cw;
M = 0
4.3 Free-Body Diagrams (F.B.D.)
Draw the object showing all the forces acting on it.
1. Start by drawing the body floating freely in space without
any supports or connections like rollers, pinned joints etc.
2. Draw in all the known external applied forces on the body,
and include the weight if given, acting vertically
downwards from the centre of gravity (C.G) of the body.
3. Replace the supports or connections with the type of
reaction forces on the body at these points (according to
table 3.1 in the next slide)
4. Finally fill in information on slope angles,lengths or
distances, etc for the analysis
Table 3.1: Reactions at Various
Types of Supports and Connections
1. Rollers
F.B.D.




Frictionless
Surface
Rc
Single reaction force at 90 0 to
supporting surface
2. Cable & Short Link
F.B.D.


T
Short Rope/
Cable

Short Link
Single force following
direction of rope or link
3. Sliders Or Slot
F.B.D.
R

Slider on
Frictionless Rod
90o


Frictionless Pin in
Slot
Single force at 90o to the
guide (direction assumed)
4. Hinged Joints
F.B.D.
or


Fx
Fy
Frictionless pin
or hinged joint


R

Rough surface
Two components resolved
into x and y axis or
a single reaction of
unknown direction
Example 4.1
Draw the free-body diagram for the lever ABC supported by
roller at B and hinged at A as shown in the figure below.
A
10 kN
B
C
Solution 4.1
1. Draw the body ABC
(Remove all the supports)
2. Draw all applied loads
F.B.D.
Ay
A
3. Draw the reaction forces at
the supports.
Ax
10 kN
A
10 kN
Bx
B
B
C
C
Example 4.2
Draw free-body diagrams of the bar ABC and BDE as shown in the
figure below. At A and D are pin joints, B is a slot and C is a smooth
wall. (This example shows 2 objects connected via a slot at B)
6
3
5
2
All dimensions in cm
A
8
B
E
D
4
C
30kN
y
x
Solution 4.2
1. Draw the body ABC
F.B.D.
2. Draw the forces at the
supports.
6
3
5
12/3 = 4

B
E
D
4
C
9/3 = 3
2
A
8
Ay
Rc
30kN
B

C
A
Ax
3

4
Rb
1. Draw the body BDE
2. Draw all applied loads
3. Draw the forces at the
supports.
F.B.D.
Rb
3
4
6
3
5
2
B
Dy
E
DDx
A
8
B
E
D
4
C
30kN
30 kN
4.4 Concurrent Force Systems
When a system of forces intersect at a point (in the
same plane) the system is known as the concurrent
force system.
To solve this we first draw a F.B.D.
Next we apply the equations
 Fx = 0
 Fy = 0
Example 4.3
Two forces are in a state of equilibrium with a load of mass M kg
as shown in the figure below. Determine the magnitude of force F
and the mass M.
F kN
5 kN
y
40o
A
M kg
60o
x
Solution 4.3
y
1. Draw F.B.D. of point A
2.
Fx = 0
5 kN

F
3.
Cos60o –
5
Cos40o
 Fy = 0
=0
F kN
60o
40o
(1)
A
L kN
(2)
F Sin60o + 5 Sin40o – M x 9.81 x 10-3 = 0
From (1), F = 7.66 kN
From (2), M x 9.81 x 10-3 = 7.66 Sin 60o + 5 Sin 40o
= 9.85 kN
M = (9.85/9.81) x 103
= 1004 kg
x
4.5 Equilibrium of A Two Force
Body
P
P
P
P
Consider the body in the first figure above under the action of 2
forces. The body is not in equilibrium because there is a net
moment.
For equilibrium, the two forces must be equal and opposite so
that the net force is zero and they must also act along the same
line of action so that the net moment is also zero, as shown by
the second figure.
Equilibrium will be covered by Intermediate Mechanics in
Level 2.
4.6 Equilibrium of A Three Force
Body
If a body is subjected to three forces at 3 points, for
equilibrium, the vector sum of the components and the
moments about any point must be zero.
Body With 3 Forces
F2
F2
F2
F1
D
D
D
F1
d
MD = 0,
Not In Equilibrium
F3
F3
F3
MD = 0, Fx = 0, Fy = 0,
In Equilibrium
Example 4.4
A slender rod BC of length 2 m and weight 200 N is held by a
cable AB and by a pin at C which slides in a vertical slot. Draw
the free-body diagram of the rod BC. If  is 30 0, determine the
reaction at C, the angle  and the tension in the cable AB.
A

Slot
B

C
Solution 4.4
F.B.D. of ABC
y
T
+ve

B
D
1m
x
1m
30o
E
200 N
Rc
C
Applying the moment equation at point B
 Mb = 0
(-200 x 1 x cos 30) + (Rc x 2 sin 30) = 0
Rc = 173.21 N
To find , apply the Fx = 0 equation:
T
 B
Fx = 0
Rc – T cos  = 0,
T cos  = 173.21 (1)
D
y
+ve
1m
1m
30o
E
200 N
x
Rc
C
Fy = 0
T sin  - 200 = 0,
T sin  = 200
(2)
Equation (2) / (1), we get tan  = 1.155
 = 49.1 0
T
y
And from (2) ,
T = 200/ sin 
= 264.6 N
 B
D
1m
+ve
1m
30o
E
200 N
x
Rc
C
4.7 Equilibrium of Non-concurrent
Multiple Force System
Steps for solving problems
1. Draw Free-body Diagram
2. Apply the three Equations of Equilibrium
a)
b)
c)
M  0
F  0
F  0
A
x
y
(Take moment at pin joint)
Example 4.5
A light bar ABCD is hinged at B, supported by smooth roller at D
is subjected to two forces as shown in the figure above. Find the
reaction at B and D.
4m
3m
1m
65 0
B
A
50 0
200N
480N
C
D
Solution 4.5
F.B.D. of ABCD
y
+ve
4m
3m
1m
x
65 0
B
A
50 0
200N
480N
Bx
By
C
D
Rd
4m
3m
1m
65 0
B
480N
A
Bx
50 0
200N
By
C
D
Rd
Apply  MB  0 at point B where there are 2
unknowns and note that the x-components of
480 N and 200 N passes through point B
(Rd x 5) – (480 cos 65 x 4) + (200 sin 50 x 3) = 0
(Rd x 5) = 811.43 – 459.63
Rd = 70.36 N
4m
3m
1m
65 0
B
480N
A
50 0
D
C
Bx
Fx = 0,
200N
By
Bx + (480 sin 65) – 200 cos 50 = 0
Bx = (128.56 - 435.03) = - 306.47 N
Bx = 306.47 N (
)
Applying Fy = 0,
By + Rd – (480 cos 65) – (200 sin 50) = 0
By = (202.86 + 153.21 – 70.36) = + 285.71 N
= 285.71 N (
Rd
)
To find the magnitude and direction of the single
reaction at B.
Use vector addition for the Bx and By components:
Rb =  ( Bx 2 + By 2) = 418.99 N
tan  = (By / Bx)
= (285.71 / 306.47) = 0.932
By
 = 43.0 0
Rb = 418.99 N
 = 43.0 0
43.0 0
Rb

Bx
Example 4.6
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 roller at B. The center
of gravity of the crane is located at G. Determine the components
of the reactions at A and B.
C
A
G
2400 kg
1.5m
B
2m
4m
Solution 4.6
F.B.D. of Crane
C
Ax
Ay
A
y
+ve
x
G
(2400 x g) N
1.5m
(1000 x g) N
RB
B
2m
4m
Apply
M
A
0
(RB x 1.5) – (1000 x 9.81 x 2) – (2400 x 9.81 x 6) = 0
RB = (160,884 / 1.5) = 107,256 N (
)
C
Ay
Fx = 0
Ax A
G
Ax + RB = 0
1.5m
Ax = - 107,256 N
RB
Ax = 107,256 N (
)
B
2m
Fy = 0
Ay – 1000 x 9.81 – 2400 x 9.81 = 0
Ay = 33,354 N (
)
(2400 x g) N
(1000 x g) N
4m
4.8 Pulleys
Draw the F.B.D. of the pulley shown in the diagram
below.
300 N
B
3
A
C
4
D
M kg
The tension in the cable is due to the 300 N force
applied at one end of the cable.
Since the cable has the same tension throughout its
length, the free-body diagram of the pulley would
have two forces of 300 N and the x and ycomponents at C.
300 N
300 N
3
Cx
4
Cy
If we apply the vector addition for the two 300 N
forces, we obtain a resultant R = 537 N.
For the system in equilibrium, the sum of moment
at C must be zero and hence the resultant must
pass through the point C.
Cx
Cx
R
R=537 N
Cy
Cy
We can then resolve R back to the two original
300 N forces but now acting on the centre of the
pulley, as shown below.
300 N
300 N
3
Cx
C
R
4
Cx
Cy
Cy
300 N
300 N
Cx
Cy
3
4
Original system
Hence for the example above, we could draw the
following free-body diagrams, both of which are
correct and the solution of either will yield the same
answers for the reaction forces. When the pulley is
drawn, the diameter of the pulley becomes
insignificant.
300N
1) With pulley
300 N
Ay
Ax
3
C
4
D
A
(M x 9.81)N
2) Without pulley
300N
300 N
3
Ay
4
Ax
A
C
D
(M x 9.81)N
Example 4.7
A light bar AD is hinged at A and a pulley of diameter 3 cm is
attached to the bar at C. It is held in horizontal position, with a
load attached at D, by a force F applied via a rope attached at B
and passing over pulley C. Draw the free-body diagram of the bar
AD and determine the magnitude of the force F and the
components of reaction at A.
F
B
40o
A
C
500 mm
D
700 mm
500 kg
Solution 4.7
F.B.D. of bar AD
y
F
F
C
A
0.5 m
x
40o
Ay
Ax
+ve
D
0.7 m
(500 x 9.81)N
Apply
M
A
0
(F x 0.5) + ( F sin 40 x 0.5) – (4905 x 1.2) = 0
F
0.5 F + 0.321 F = 5886
Ay
F = 7,169.31 N
40o
Fx = 0
Ax C
Ax +7169.31 Cos 40o = 0 A
Ax = 5492.01 N (
)
0.5 m
Fy = 0
Ay + Fsin 40o – 4905 + F = 0
Ay = 6872.65 N (
)
F
D
0.7 m
(500 x 9.81)N