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STATICS: CE201
Chapter 3
Equilibrium of a Particle
Notes are prepared based on: Engineering Mechanics, Statics by R. C. Hibbeler, 12E Pearson
Dr M. Touahmia & Dr M. Boukendakdji
Civil Engineering Department, University of Hail
(2012/2013)
3.
Equilibrium of a Particle
________________________________________________________________________________________________________________________________________________
Chapter Goals:
1.
2.
Introduce the concept of the free-body diagram (FBD)
for an object modeled as a particle.
Show how to solve particle equilibrium problems using
the equations of equilibrium.
Contents:
1
2
3
4
Condition for the Equilibrium of a Particle
The Free-Body Diagram (FBD)
Coplanar Force Systems
Three-Dimensional Force Systems
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Condition for the Equilibrium of a Particle

A particle is said to be in equilibrium when the resultant
of all forces acting on it is zero.

Considering Newton’s first law of motion, equilibrium
can mean that the particle is either at rest or moving at
constant velocity.

To maintain equilibrium it is necessary and sufficient
that the resultant force acting on a particle be equal to
zero.
Equilibrium of a Particle

In terms of Newton`s first law of motion, this can be
expressed mathematically as:
F  0
Equation of Equilibrium
where  F is the vector sum of all the forces acting on
the particle.

For Equilibrium:
F
x
F
y
0
0
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The Free-Body Diagram (FBD)

To apply the equation of equilibrium, we must account
for all the known and unknown forces which act on the
particle.

In order to account for all the forces that act on a
particle, it is necessary to draw its Free-Body Diagram
(FBD)

The free-body diagram is simply a sketch which shows
the particle “free” from its surroundings with all the
forces that act on it.
Types of Connections

Two types of connections often encountered in particle
equilibrium: Springs and Cable and Pulleys.

Springs: The magnitude of force exerted on a linearly
elastic spring is:
F  ks
which has a stiffness k and is deformed
(elongated or compressed) a distance s,
measured from its unloaded position.
s  l  lo
where lo the original length and l the final length
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Types of Connections

Cables and Pulleys: All cables are assumed to have
negligible weight and they cannot stretch.

A cable can support only a tension or “pulling” force,
and this force always act in the direction of the cable.

The tension in a cable is constant throughout its length.
Cable in tension
Procedure for Drawing a Free-Body Diagram:

Draw Outlined Shape: Imagine the particle to be isolated
or cut “free” from its surroundings with all the forces
outlined shape.

Show All Forces: Indicate on this sketch all the forces
that act on the particle.

Identify Each Force: The forces which are known should
be labeled with their proper magnitudes and directions.
Letters are used to represent the magnitudes and
directions of forces that are unknown.
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Example 1:

The sphere has a mass of 6 kg and is supported as
shown. Draw a free-body diagram of the sphere, the
cord CE, and the knot at C.
Solution 1:

Sphere: There are two forces acting on the sphere:


◦ The weight of the sphere W, W  6kg 9.81m/s 2  58.9 N
◦ The force FCE of the cord CE acting on the sphere

The free body diagram:
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Solution 1:

Cord CE: When the cord CE is isolated from its
surroundings, its free-body diagram shows only two
forces acting on it:
◦ The force of the sphere and FCE.
◦ The force of the knot FEC.
FCE = FEC
Solution 1:
Knot: The knot at C is subjected to three forces. They are
caused by the cords CBA and CE and the spring CD:
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Coplanar Force Systems

If a particle is subjected to a system of coplanar forces
that lie in the x-y plane, then each force can be resolved
into i and j components.

For equilibrium, these forces must sum to produce a
zero resultant. Hence:
F  0
F i  F
x
y
F
x
j0
0
F  0
y
Example 2:

Determine the tension in cables BA and BC necessary to
support the 60-kg cylinder.
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Solution 2:

Due to equilibrium, the weight of the
cylinder causes the tension in cable BD to be:
TBD  609.81  588.6 N

The forces in cable BA and BC can be determined by
investigating the equilibrium of the ring B. Its free body
diagram is then:

The magnitudes of TA and TC are
unknown, but their directions are
known.
Solution 2:

Equations of equilibrium along the x and y axes:
F  0
x
 4
T cos 45   T  0
5
o
C
A
 3
 F  0 TC sin 45o   TA  588.6  0
y
5
TA  0.8839 TC
 3
TC sin 45o   (0.8839TC )  588.6  0
5
TC  475.66 N
TA  420 N
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Example 3:
The 200-kg box is suspended using the ropes AB and AC.
Each rope can withstand a maximum force of 10 kN before
it breaks. If AB always remains horizontal, determine the
smallest angle θ to which the box can be suspended before
one of the ropes breaks.
Solution 3:

Free-Body Diagram of the ring: There are three forces
acting on the ring: FC , FB , FD

The magnitude of FD is equal to the weight of the box.
FD = 200 (9.81) N = 1962 N < 10 kN
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Solution 3:

Equations of Equilibrium:
FC sin   1962  0
 FC 
 FC cos   FB  0
FB
cos 
FC  10 kN
 
10 10 3 sin   1962  0
  sin 1 0.1962  11.31

The force developed in rope AB:
 
FB  FC cos 11.31  10 10 3 cos 11.31
FB  9.81 kN
Three-Dimensional Force Systems

The necessary and sufficient condition for particle
equilibrium is:
F  0

We can resolve the forces into their respective i, j, k
components , so that:
F
x

i   Fy j   Fz k  0
To satisfy this equation we require:
F
F
x
0
y
0
F
z
0
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Example 5:

If the mass of cylinder C is 40 kg, determine the mass of
cylinder A in order to hold the assembly in the position
shown.
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