Lecture #5
Translational Equilibrium
(2D Coplaner Force Systems,
Reference Chapter 3, section 3)
1
R. Michael PE 8/14/2012
Newton’s First Law of Motion
A body at rest will stay at rest and a
body in motion will stay in motion
unless acted upon by an unbalanced
force.
 Therefore, sum of all forces must be
zero:

F=0
2
Newton’s Second Law of Motion

The acceleration a body undergoes
when experiencing an unbalanced
force is proportional to the magnitude
of the unbalanced force, in the
direction of the unbalanced force, and
inversely proportional to the mass of
the object – Basis of Dynamics
 F = ma
3
Motion Definitions
Rectilinear Motion - Motion in a
straight line.
 Translational Motion - Motion where
the body does not rotate
 Again, this is studied in dynamics!

4
2D Particle Static Equilibrium (i.e. Section 2.3,
Coplanar Force Systems)
When a body is in this state there are
no unbalanced forces acting on it.
 The Resultant Force has a magnitude
of zero.

In general, for a particle in equilibrium,
 F = 0 or
 Fx i +  Fy j = 0 = 0 i + 0 j
(a vector equation)
Or, written in a scalar form,
Fx = 0 and  Fy = 0
These are two scalar equations of equilibrium (E-of-E). They can be
used to solve for up to two unknowns.
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3D Particle Static Equilibrium

Because the resultant force is
balanced, the following situation
occurs
ΣFx = 0
 ΣFy = 0
 ΣFz = 0

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Summary: 2D Static
Equilibrium:

F
F
Using Components:
0
x
 0;
F
y
 0;
F
z
0
F1

F
Graphically:
 0  F1  F2  F3
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F2
F3
Static Equilibrium Problems

In 2D, have 2 equations, so can solve for 2
unknowns
Find magnitudes of two forces with known
directions
 Find magnitude and direction of one force,
knowing magnitude and direction of other
force(s)


In 3D have 3 equations, so can solve for 3
unknowns
Do simple 2D
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2D COPLANAR FORCE SYSTEMS (Section 3.3)
This is an example of a 2-D or
coplanar force system.
If the whole assembly is in equilibrium,
then particle A is also in equilibrium.
To determine the tensions in the
cables for a given weight of the
cylinder, you need to learn how to
draw a free body diagram and apply
equations of equilibrium.
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THE WHAT, WHY AND HOW OF A FREE BODY
DIAGRAM (FBD)
Free Body Diagrams are one of the most important things for you to know
how to draw and use.
What ? - It is a drawing that shows all external forces acting on the particle.
Why ? - It is key to being able to write the equations of equilibrium—
which are used to solve for the unknowns (usually forces or angles).
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How ?
1. Imagine the particle to be isolated or cut free from its surroundings.
2. Show all the forces that act on the particle.
Active forces: They want to move the particle. Reactive
forces: They tend to resist the motion.
3. Identify each force and show all known magnitudes and directions. Show
all unknown magnitudes and / or directions as variables .
y
FBD at A
FD A
Note : Cylinder mass = 40 Kg
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A
FB
30˚
x
FC = 392.4 N (What is
this?)
EQUATIONS OF 2-D EQUILIBRIUM
y
FBD at A
A
FD A
FB
30˚
x
Since particle A is in equilibrium, the net force
at A is zero.
So FB + FC + FD = 0
or  F = 0
A
FC = 392.4 N
FBD at A
In general, for a particle in equilibrium,
 F = 0 or
 Fx i +  Fy j = 0 = 0 i + 0 j
(a vector equation)
Or, written in a scalar form,
Fx = 0 and  Fy = 0
These are two scalar equations of equilibrium (E-of-E). They can be
used to solve for up to two unknowns.
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EXAMPLE
y
FBD at A
FB
FD A
A
30˚
x
FC = 392.4 N
Note : Cylinder mass = 40
Kg Write the scalar E-of-E:
+   Fx = FB cos 30º – FD = 0
+   Fy = FB sin 30º – 392.4 N = 0
Solving the second equation gives: FB = 785 N →
From the first equation, we get: FD = 680 N ←
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SPRINGS, CABLES, AND PULLEYS
T1
T1
T2
Spring Force = spring constant *
deformation, or
F=k*s
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With a frictionless pulley,
T 1 = T 2= T
See HO
EXAMPLE
Given: Cylinder E weighs 30 lb
and the geometry is as
shown.
Find: Forces in the cables and
weight of cylinder F.
Plan:
1. Draw a FBD for Point C.
2. Apply E-of-E at Point C to solve for the unknowns (FCB & FCD).
3. Knowing FCB , repeat this process at point B.
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EXAMPLE (continued)
y
FCD
x
30
15
A FBD at C should look like the one at the left.
Note the assumed directions for the two cable
tensions.
FBC
30 lb
The scalar E-of-E are:
+   Fx = FBC cos 15º – FCD cos 30º = 0
+
 Fy = FCD sin 30º – FBC sin 15º – 30 = 0
Solving these two simultaneous equations for the two
unknowns FBC and FCD yields:
FBC = 100.4 lb
FCD = 112.0 lb
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EXAMPLE (continued)
y
FBC =100.4 lb
FBA
45
15
x
Now move on to ring B.
A FBD for B should look like the
one to the left.
WF
The scalar E-of-E are:
   Fx = FBA cos 45 – 100.4 cos 15 = 0
   Fy = FBA sin 45 + 100.4 sin 15 – WF = 0
Solving the first equation and then the second yields
FBA = 137 lb
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and
WF = 123 lb
GROUP PROBLEM SOLVING
Given: The box weighs 550 lb and
geometry is as shown.
Find:
The forces in the ropes AB and AC.
Plan:
1. Draw a FBD for point A.
2. Apply the E-of-E to solve for the forces in
ropes AB and AC.
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GROUP PROBLEM SOLVING (continued)
y
FB
30
˚
FC
5
3
4
A
x
FD = 550 lb
Applying the scalar E-of-E at A, we get;
+   F x = FB cos 30° – FC (4/5) = 0
+   F y = FB sin 30° + FC (3/5) - 550 lb = 0
Solving the above equations, we get;
FB = 478 lb and FC = 518 lb
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FBD at point A
ATTENTION QUIZ
1. Select the correct FBD of particle A.
30
A
40
100 lb
A
A)
B)
100
lb
F
C)
30°
F1
F2
30
40°
A
D)
F2
40°
F1
30°
A
A
100 lb
20
100 lb
ATTENTION QUIZ
2. Using this FBD of Point C, the sum of forces in the
F2
x-direction ( FX) is ___ . Use a sign convention
of +  .
A) F2 sin 50° – 20 = 0
B) F2 cos 50° – 20 = 0
C) F2 sin 50° – F1 = 0
D) F2 cos 50° + 20 = 0
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20 lb
50°
C
F1
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23
24
25
26
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READING QUIZ
1) When a particle is in equilibrium, the sum of forces acting
on it equals ___ . (Choose the most appropriate answer)
A) A constant
D) A negative number
B) A positive number
E) An integer
C) Zero
2) For a frictionless pulley and cable, tensions in the cable
(T1 and T2) are related as _____ .
A) T1 > T2
B) T1 = T2
C) T1 < T2
D) T1 = T2 sin 
T1
T2
Determine the forces in cables AC and AB needed to hold the 20-kg
ball D in equilibrium. Take F = 300 N and d = 1 m
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Practice Problem 1
Reference 3-18/19.
Determine the tensions developed in wires CD, CB and BA and the
angle  required for equilibrium of the 30 lb cylinder E and 60 lb
cylinder F.
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Practice Problem 2
Reference 3-26.
Determine the stretch in springs AC and AB for equilibrium of the 2kg block. The springs are shown in the equilibrium position.
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Practice Problem 3
Reference 3-14.