Translational Equilibrium
Physics
Montwood High School
R. Casao
• Concurrent forces are those forces that are applied
to or act on the same point, as shown at the point
where T1, T2, and T3 meet.
• When two or more forces act at the same point, the
resultant force is the sum of the forces applied at
that point.
• The resultant force is the
single force that has the same
effect as the two or more forces
that act together.
• When forces act in the same direction
or opposite directions, the total force
can be found by adding the forces that
act in one direction and subtracting the
forces that act in the opposite direction.
• Equilibrium is the state of a body in which there is
no change in its motion.
• A body is in equilibrium when the net force acting
on it is zero; there is no acceleration (a = 0 m/s2),
the body is either at rest or is moving at a constant
velocity.
• The study of objects in equilibrium is called statics.
• When two or more forces are acting at the same
point, the equilibrant force is the force that when
applied at that same point produces equilibrium.
• The equilibrant force is equal in magnitude to that
of the resultant force but acts in the opposite
direction.
• If an object is in equilibrium in one dimension, the
forces in one direction must equal the forces in the
opposite direction.
ΣF+ direction = ΣF- direction
• If an object is in equilibrium in two dimensions, the
net force acting on it must be zero.
– For the net force to be zero, the sum of the x-components
must be zero and the sum of the y-components must be
zero.
A

B

C

0
x
x
x
– Conditions for equilibrium:
Ay  By  Cy  0
• In general, to solve equilibrium problems:
– Draw a free-body diagram from the point at which the
unknown forces act.
– Find the x- and y-components of each force.
Fx  0
– Substitute the components in the equations:
– Solve for the unknowns. This may involve Fy  0
two simultaneous equations.
Two Conditions for Equilibrium
• An object is said to be in equilibrium if and only if
there is no resultant force and no resultant torque
(rotation).
– First condition (translational equilibrium):
ΣFx = 0; ΣFy = 0
– The linear speed is not changing with time.
Car at rest
a  0;
Constant speed
 F  0;
No change in v
– Second condition (rotational equilibrium)
Σtorque = 0; the object does not rotate, or rotates at a
constant number of turns per unit time.
– Torque is the tendency of a force to rotate an object
around an axis; torque measures how hard something is
twisted.
Wheel at rest
Constant rotation
  0; No change in rotation.
Translational Equilibrium Only
• If all forces act at the same point, then there is no
torque to consider and one need only apply the
first condition for equilibrium: ΣFx = 0; ΣFy = 0
• Example: Find the tension in ropes A and B.
Free-body Diagram:
600
B
B
A
A
By
600
Bx
80 N
80 N
• Read problem; draw sketch; construct a free-body
diagram, indicating components.
• Choose x-axis horizontal and choose right
direction as positive (+). There is no motion.
• The components Bx and By can be found from right
triangle trigonometry.
R
y
q
x
y
sin q  ; y  R  sin q
R
x
cos q  ; x  R  cos q
R
2
2
2
R x y
Free-body Diagram:
600
B
B
A
A
By
600
Bx
80 N
80 N
Bx = B cos 600
By = B sin 600
Free-body Diagram:
B
A
By
600
Fx = 0
Fy = 0
By
B sin 600
Bx
A
Bx
80 N
• ΣFx = 0; Bx – A = 0; Bx = A
B cos 60o
B·cos 60 = A; two unknowns
80 N
proceed to ΣFy = 0
• ΣFy = 0; By – 80 N = 0; By = 80 N; B·sin 60 = 80
N
80 N
B
 92.38 N
sin 60
• Determine A:
Problem Solving Strategy
1. Draw a sketch and label all information.
2. Draw a free-body diagram.
3. Find components of all forces (+ and -).
4. Apply First Condition for Equilibrium:
ΣFx = 0; ΣFy = 0
5. Solve for unknown forces or angles.
Example: Find Tension in Ropes A and B.
300
A
600
B
B
600
300
Ay
300
Ax
400 N
1. Draw free-body diagram.
2. Determine angles.
3. Draw/label components.
By
A
600
Bx
400 N
Next we will find
components of
each vector.
Example: Find the tension in ropes A and B.
First Condition for Equilibrium:
B
Ay
Fx= 0 ;
Fy= 0
By
A
600
300
Ax
Bx
W 400 N
4. Apply 1st Condition for Equilibrium:
Fx = Bx - Ax = 0
Fy = By + Ay - W = 0
Bx = Ax
By + Ay = W
Example: Find the tension in ropes A and B.
Ax = A cos 300; Ay = A sin 300
B
Bx = B cos 600
By = B sin 600
Wx = 0; Wy = -400 N
Ay
By
A
300
Ax
600
Bx
W 400 N
Using Trigonometry, the first condition yields:
Bx = Ax
By + Ay = W
B cos 600 = A cos 300
A sin 300 + B sin 600 = 400 N
Example: Find the tension in A and B.
B cos 600 = A cos 300
B
Ay
By
A
300
Ax
W 400 N
600
A sin 300 + B sin 600 = 400 N
Bx
We now solve for A and B: Two
Equations and Two Unknowns.
We will first solve the horizontal equation for B in
terms of the unknown A:
A  cos 30
B
 1.73  A
cos 60
B = 1.732· A
Example: Find Tensions in A and B.
B = 1.732 A
B
Ay
By
A
300
Ax
600
Bx
400 N
B = 1.732·A
Now apply Trig to:
Ay + By = 400 N
A sin 300 + B sin 600 = 400 N
A·sin 300 + B·sin 600 = 400 N
A·sin 300 + (1.732 A)·sin 600 = 400 N
0.500·A + 1.50·A = 400 N
2·A = 400 N
A = 200 N
Example: Find B with A = 200 N.
A = 200 N
B
Ay
By
A
300
600
Ax
Bx
B = 1.732·A
B = 1.732·(200 N)
W 400 N
Rope tensions are:
A = 200 N
B = 346 N
B = 346 N
• You may need to find the tension or
compression in part of a structure, such as
in a beam or a cable.
• Tension is a stretching force produced by
forces pulling outward on the ends of an
object.
• Compression is a force produced by forces
pushing inward on the ends of an object.