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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
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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
j0
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 609.81 588.6 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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0
