Chapter 5B
Rotational Equilibrium
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
©
2007
The Golden Gate
Bridge provides
an excellent
example of
balanced forces
and torques.
Engineers must
design such
structures so that
rotational and
translational
equilibrium is
maintained.
Photo © EP 101 Photodisk/Getty
Objectives: After completing this
module, you should be able to:
• State and describe with examples your
understanding of the first and second
conditions for equilibrium.
• Write and apply the first and second
conditions for equilibrium to the solution
of physical problems similar to those in
this module.
Translational Equilibrium
Car at rest
a  0;
 F  0;
Constant speed
No change in v
The linear speed is not changing with time.
There is no resultant force and therefore zero
acceleration. Translational equilibrium exists.
Rotational Equilibrium
Wheel at rest
Constant rotation
  0; No change in rotation.
The angular speed is not changing with
time. There is no resultant torque and,
therefore, zero change in rotational
velocity. Rotational equilibrium exists.
Equilibrium
• An object is said to be in equilibrium if
and only if there is no resultant force
and no resultant torque.
First
Condition:
F
Second
Condition:
  0
x
 0;
F
y
0
Does Equilibrium Exist?
T
300
IsYES!
the system
at left in
Observation
equilibrium
bothof
shows
that no part
and
thetranslationally
system is changing
rotationally?
its state
of motion.
A sky diver moments after the jump? Yes or
NoNo?
A sky diver who reaches terminal speed?
A fixed pulley rotating at constant speed?
Yes
Yes
Statics or Total Equilibrium
Statics is the physics that treats objects at
rest or objects in constant motion.
In this module, we will review the first
condition for equilibrium (treated in Part 5A
of these modules); then we will extend our
treatment by working with the second
condition for equilibrium. Both conditions
must be satisfied for true equilibrium.
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:
• Construct free-body diagram.
• Sum forces and set to zero:
Fx= 0;
Fy= 0
• Solve for unknowns.
Review: Free-body Diagrams
• Read problem; draw and label sketch.
• Construct force diagram for each object,
vectors at origin of x,y axes.
• Dot in rectangles and label x and y components opposite and adjacent to angles.
• Label all components; choose positive
direction.
Example 1. Find the tension in ropes A and B.
Free-body Diagram:
600
A
B
A
B
600
By
Bx
80 N
80 N
• Read problem; draw sketch; construct a freebody diagram, indicating components.
• Choose x-axis horizontal and choose right
direction as positive (+). There is no motion.
Example 1 (Continued). Find A and B.
Free-body Diagram:
600
A
B
A
B
600
By
Bx
80 N
80 N
Note: The components Bx and By can be
found from right triangle trigonometry:
Bx = B cos 600;
By = B sin 600
Example 1 (Cont.). Find tension in ropes A and B.
Free-body Diagram:
A
B
600
By
Bx
Fx = 0
Fy = 0
By
B sin 600
Bx
A
B cos 60o
80 N
80 N
• Apply the first condition for equilibrium.
F
x
 0;
F
y
 0;
Example 2. Find tension in ropes A and B.
A
350
550
B
500 N
Ay
A
B
By
350
550
Ax
Bx
W
Recall: Fx = Fy = 0
Fx = Bx - Ax = 0
W = 500 N
Fy = By + Ay – 500 N = 0
Example 2 (Cont.) Simplify by rotating axes:
x
y
A
Wx
350
Wy
B
550
W
Recall that W = 500 N
Fx = B - Wx = 0
B = Wx = (500 N) cos 350
B = 410 N
Fy = A - Wy = 0
A = Wx = (500 N) sin 350
A = 287 N
Total Equilibrium
In general, there are six degrees of freedom
(right, left, up, down, ccw, and cw):
ccw (+)
cw (-)
Fx= 0
Right = left
Fx= 0
Up = down
0
  (ccw)=   (ccw)
General Procedure:
• Draw free-body diagram and label.
• Choose axis of rotation at point where least
information is given.
• Extend line of action for forces, find moment
arms, and sum torques about chosen axis:
  1 + 2 + 3 + . . .  0
• Sum forces and set to zero: Fx= 0;
• Solve for unknowns.
Fy= 0
Example 3: Find the forces exerted by
supports A and B. Neglect the weight
of the 10-m boom.
Draw free-body
diagram
2m
40 N
Rotational Equilibrium:
Choose axis at point
of unknown force.
At A for example.
2m
40 N
7m
A
A
3m
B
80 N
7m
B
3m
80 N
Example 3 (Cont.)
Note: When applying
(ccw) = (cw)
we need only the
absolute (positive)
magnitudes of each
torque.
 (+) =  (-)
2m
A
40 N
7m
B
3m
80 N
Torques about axis ccw
are equal to those cw.
ccw (+)
cw (-)
Essentially, we are saying that the torques
are balanced about a chosen axis.
Example 3: (Cont.)
Rotational Equilibrium:
  1 + 2 + 3 + 4  0
or
(ccw) = (cw)
40 N
2m
With respect to Axis A:
7m
2m
A
A
40 N
3m
B
80 N
B
7m
3m
80 N
CCW Torques: Forces B and 40 N.
CW Torques:
80 N force.
Force A is ignored:
Neither ccw nor cw
Example 3 (Cont.)
First: (ccw)
1 = B (10 m)
2 = (40 N) (2 m)
= 80 Nm
Next: (cw)
3 = (80 N) (7 m)
= 560 Nm
7m
2m
A
40 N
2m
40 N
3m
A
B
80 N
B
7m
3m
80 N
(ccw) = (cw)
B(10 m) + 80 Nm = 560 Nm
B = 48.0 N
Example 3 (Cont.)
Translational
Equilibrium
Fx= 0; Fy= 0
F(up) = F(down)
A + B = 40 N + 80 N
A + B = 120 N
7m
2m
A
40 N
2m
A
3m
B
80 N
7m
40 N
B
3m
80 N
Recall that B = 48.0 N
A + 48 N = 120 N
A = 72.0 N
Example 3 (Cont.)
Check answer by
summing torques
about right end to
verify A = 72.0 N
(ccw) = (cw)
7m
2m
40 N
2m
A
A
3m
B
80 N
7m
40 N
B
3m
80 N
(40 N)(12 m) + (80 N)(3 m) = A (10 m)
480 Nm + 240 Nm = A (10 m)
A = 72.0 N
Reminder on Signs:
Absolute values
apply for:
F(up) = F(down)
We used absolute (+)
values for both UP
and DOWN terms.
7m
2m
40 N
2m
40 N
A
A
3m
B
80 N
7m
B
3m
80 N
Instead of: Fy = A + B – 40 N - 80 N = 0
We wrote: A + B = 40 N + 90 N
Example 4: Find the tension in
the rope and the force by the
wall on the boom. The 10-m
boom weighing 200 N. Rope is
2 m from right end.
T
300
800 N
For purposes of summing torques, we consider
entire weight to act at center of board.
Fy
T
300
200 N
800 N
T
Fx
300
5m
3m
200 N
2m
800 N
T
Example 4 F r
y
(Cont.)
300
200 N
T
Fx
300
5m
3m
200 N
800 N
2m
800 N
Choose axis of rotation at wall (least information)
(ccw):
(cw):
Tr = T (8 m)sin 300 = (4 m)T
(200 N)(5 m) + (800 N)(10 m) = 9000 Nm
(4 m)T = 9000
Nm
T = 2250 N
T
T
Example 4 F
y
(Cont.)
Fx T x
300
200 N
5m
00
3030
3m
200 N
800 N
2m
800 N
Ty + Fy = 200 N + 800 N
F(up) = F(down):
Fy = 200 N + 800 N - Ty ;
Fy = 1000 N - T sin 300
Fy = 1000 N - (2250 N)sin 300
F(right) = F(left):
Fx = 1950 N
Ty
Fy = -125 N
Fx = Ty = (2250 N) cos 300
or
F = 1954 N, 356.30
Center of Gravity
The center of gravity of an object is the point
at which all the weight of an object might be
considered as acting for purposes of treating
forces and torques that affect the object.
The single support force has line of action that passes
through the c. g. in any orientation.
Examples of Center of Gravity
Note: C. of G. is not always inside material.
Example 5: Find the center of gravity of the
apparatus shown below. Neglect the weight
of the connecting rods.
C. of G. is point at
which a single upupward force F will
balance the system.
Choose axis at left,
then sum torques:
(ccw) = (cw)
Fx = (10 N)(4 m) + (5 N)(10 m)
Fx = 90.0 Nm
F
x
4m
30 N
6m
10 N
5N
F(up) = F(down):
F = 30 N + 10 N + 5 N
(45 N) x = 90 N
x = 2.00 m
Summary
Conditions for Equilibrium:
An object is said to
be in equilibrium if
and only if there is
no resultant force
and no resultant
torque.
Fx  0
Fy  0
  0
Summary: Procedure
• Draw free-body diagram and label.
• Choose axis of rotation at point where least
information is given.
• Extend line of action for forces, find moment
arms, and sum torques about chosen axis:
  1 + 2 + 3 + . . .  0
• Sum forces and set to zero: Fx= 0;
• Solve for unknowns.
Fy= 0
CONCLUSION: Chapter 5B
Rotational Equilibrium