Mechanics
Module II: Equilibrium
Lesson 4: Equilibrium - I
1
Static Equilibrium
Figure 1:
A particle is said to be in equilibrium if the sum of all forces acting on
the particle is identically zero. A rigid body is said to be in equilibrium if
the equivalent force system at any point of the body or any extension of it
is identically zero. It can be easily shown that if the equivalent force system
at a point vanishes, then it vanishes at every other point. A particle or rigid
body is said to be in static equilibrium if it is in equilibrium and in the state
of rest. It may be noted that the conditions of equilibrium do not preclude,
for example, uniform rectilinear motion of a particle.
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Free Body Diagram
A system of connected or interacting rigid bodies is in static equilibrium
when, not only the complete system, but also each individual rigid body is in
static equilibrium. In order to analyze any part of such a system of interacting
rigid bodies, we need to separate out the part and replace each interaction
point by appropriate (internal) forces and/or moments as experienced by
the part at that point due to interaction. This diagram, known as the free
body diagram (FBD), is an extremely important part of visualization and
calculations in equilibrium problems. The interaction/reaction forces and/or
moments introduced at an interaction point opened-up for separation must
be able to exactly impose the motion restrictions imposed by the interaction
at that point.
The interaction forces and/or moments thus assigned to the two interacting bodies at the interaction point (that is now opened-up) must follow
the Newton’s third law. Some examples of connections/interactions, and the
corresponding forces/moments are shown in Figs. 2, 3 and 4.
2
W
W
Ideal pulley
W
Figure 2:
3
W
Figure 3:
4
Figure 4:
3
Equations of Equilibrium
For the equilibrium of a particle, there are three (two, for planar problems)
equations of equilibrium, which follow from vanishing of the three components
of the resultant force on the particle, as shown in Fig. 5(a).
In the case of a planar rigid body, as shown in Fig. 5(b), the equivalent force system at a point must vanish. This comprises three equations,
namely the vanishing of the two components of the resultant force Fx = 0
and Fy = 0, and the vanishing of the resultant moment MA = 0, as shown in
Fig. 5(c). It is, however, possible to replace this set of equations, depending
on convenience, by either
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Figure 5:
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• Two moment equations about two points, and a force equation along
these two points, as shown in Fig. 5(d), OR
• Three moment equations about three noncollinear points, as shown in
Fig. 5(e).
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What do we calculate in equilibrium problems?
The following types of calculations are possible.
1. Determining constraint or internal forces developed in members and connections. This type of calculations is required for design of members and
joint elements, and also for determining/estimating deflection of members when a small amount of flexibility is introduced.
2. Determining equilibrium configuration of a system subjected to a given
loading.
Since the equations of equilibrium are linear in the forces and moments, the
former type of problems is linear, while the determination of the equilibrium
configuration is usually a nonlinear problem. In planar equilibrium problems,
we can solve at most three unknown quantities.
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Problems
Problem 1
A uniform slab of mass 25000 kg is being raised as shown in Fig. 6. Determine
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Figure 6:
Figure 7:
all the forces on the slab.
Solution
Consider the Free Body Diagram (FBD) of the slab, as shown in Fig. 7. In
order to use only one equation to determine T , we locate D about which the
other two unknowns P and N do not produce any moment.
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From △ABD,
6m
a
=
sin 120◦
sin 30◦
⇒
√
a = 2 3 m.
From moment equilibrium about D,
X
MD = 0 ⇒ aT − 4 cos 60◦W = 0
√
√
⇒ T = W/ 3 = 25 × 103 × 9.81/ 3 = 141.6 kN.
Force equilibrium in the direction b̂
X
Fb = 0 ⇒ N cos 30◦ − T cos 30◦ = 0
√
⇒ N = W + T / 3 = 245.3 kN
Also,
X
Fx = 0 ⇒ T − P cos 30◦ = 0
⇒ P = 163.5 kN
Notice that the force equilibrium equations are along two non-orthogonal directions.
Problem 2
A 300 kg walk-way is supported as shown in Fig. 8. Calculate all the reaction
forces at A and B on the walkway.
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Figure 8:
Figure 9:
Solution
From the FBD of the walk-way shown in Fig. 9,
X
MB = 0 ⇒ −(8 cos 30◦)RA + (4 cos 30◦)W = 0
⇒ RA = W/2 = 1471.5 N
X
MC = 0 ⇒ −(4 cos 30◦)W + (8 cos 30◦)RB cos 30◦ = 0
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A
θ
45
D
600 mm
C
B
20 kg
Figure 10:
√
⇒ RB = W/ 3 = 1699 N
X
FCB = 0 ⇒ T − RB sin 30◦ = 0
⇒ T = 849.5 N
Problem 3
A load of 20 kg is supported as shown in Fig. 10. Assuming the pulley
to be ideal, determine the length of CD required to realize the equilibrium
configuration shown. Also, determine the tension in the cable CD.
Solution
This problem is equivalent to the equilibrium of a particle in a plane (since
rotation of the pulley is irrelevant and the moment equilibrium is vacuously
satisfied). From the FBD of the pulley shown in Fig. 11
X
Fx = 0 ⇒ FCD cos θ − 20g cos 45◦ = 0
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Figure 11:
X
Fy = 0 ⇒ FCD sin θ + 20g sin 45◦ − 20g = 0
√
Solving the above two equations, tan θ = 2−1 ⇒ θ = 22.5◦, and FCD =
p
20g cos2 45◦ + (1 − sin 45◦)2 = 150.2 N. Further, CD sin θ = 600 mm ⇒
CD = 1568 mm.
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