1. Chapter 5B Rotational Equilibrium
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
© 2007

2. Photo © EP 101 Photodisk/Getty
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

3. 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.

4. Translational Equilibrium
Car at rest
Constant speed
The linear speed is not changing with time. There is no resultant force and therefore zero acceleration. Translational equilibrium exists.

5. Rotational Equilibrium
Wheel at rest
Constant rotation
The angular speed is not changing with time. There is no resultant torque and, therefore, zero change in rotational velocity. Rotational equilibrium exists.

6. Equilibrium
An object is said to be in equilibrium if and only if there is no resultant force and no resultant torque.
First Condition:
Second Condition:

7. Does Equilibrium Exist?
300 T
Is the system at left in equilibrium both translationally and rotationally?
YES! Observation shows that no part of the system is changing its state of motion.
A sky diver moments after the jump?
Yes or No? No
A sky diver who reaches terminal speed?
Yes
A fixed pulley rotating at constant speed?
Yes

8. 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.

9. 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:
SFx= 0; SFy= 0
Solve for unknowns.
Construct free-body diagram.
Sum forces and set to zero:

10. 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 compo- nents opposite and adjacent to angles.
Label all components; choose positive direction.

11. Example 1. Find the tension in ropes A and B.
600
80 N A B
600
Free-body Diagram: By Bx
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.

12. Example 1 (Continued). Find A and B.
600
80 N A B
600
Free-body Diagram: By Bx
Note: The components Bx and By can be found from right triangle trigonometry:
Bx = B cos 600; By = B sin 600

13. Example 1 (Cont.). Find tension in ropes A and B.
600
Free-body Diagram: By Bx
80 N A
B sin 600
B cos 60o Bx By
SFx = 0
SFy = 0
Apply the first condition for equilibrium.

14. Example 2. Find tension in ropes A and B.W
350
550 Bx By Ax Ay
350
550 A B
500 N
Recall: SFx = SFy = 0
SFx = Bx - Ax = 0
W = 500 N
SFy = By + Ay – 500 N = 0

15. Example 2 (Cont.) Simplify by rotating axes:
SFx = B - Wx = 0
350
550 A B W Wx Wy x y
B = Wx = (500 N) cos 350
B = 410 N
SFy = A - Wy = 0
A = Wx = (500 N) sin 350
Recall that W = 500 N
A = 287 N

16. Total Equilibrium SFx= 0
In general, there are six degrees of freedom (right, left, up, down, ccw, and cw):
SFx= 0
Right = left
Up = down
ccw (+)
cw (-)
S t = 0
S t (ccw)= S t (ccw)

17. General Procedure: St = t1 + t2 + t3 + . . . = 0
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:
St = t1 + t2 + t = 0
Sum forces and set to zero: SFx= 0; SFy= 0
Solve for unknowns.

18. Example 3: Find the forces exerted by supports A and B
Example 3: Find the forces exerted by supports A and B. Neglect the weight of the 10-m boom.
40 N
80 N
2 m
3 m
7 m A B
Draw free-body diagram
Rotational Equilibrium:
40 N
80 N
2 m
3 m
7 m A B
Choose axis at point of unknown force.
At A for example.

19. Torques about axis ccw are equal to those cw.
Example 3 (Cont.)
40 N
80 N
2 m
3 m
7 m A B
Torques about axis ccw are equal to those cw.
ccw (+)
cw (-)
St(ccw) = St(cw)
Note: When applying
we need only the absolute (positive) magnitudes of each torque.
 (+) =  (-)
Essentially, we are saying that the torques are balanced about a chosen axis.

20. Rotational Equilibrium:
Example 3: (Cont.)
40 N
80 N
2 m
3 m
7 m A B
Rotational Equilibrium:
St = t1 + t2 + t3 + t4 = 0 or
St(ccw) = St(cw)
With respect to Axis A:
CCW Torques: Forces B and 40 N.
CW Torques: 80 N force.
Force A is ignored: Neither ccw nor cw

21. Example 3 (Cont.) A B First: St(ccw) t1 = B (10 m)
t2 = (40 N) (2 m) = 80 Nm
St(ccw) = St(cw)
Next: St(cw)
B(10 m) + 80 Nm = 560 Nm
t3 = (80 N) (7 m) = 560 Nm
B = 48.0 N

22. Translational Equilibrium
Example 3 (Cont.)
40 N
80 N
2 m
3 m
7 m A B
Translational Equilibrium
SFx= 0; SFy= 0
SF(up) = SF(down)
Recall that B = 48.0 N
A + B = 40 N + 80 N
A + 48 N = 120 N
A + B = 120 N
A = 72.0 N

23. Check answer by summing torques about right end to verify A = 72.0 N
Example 3 (Cont.)
40 N
80 N
2 m
3 m
7 m A B
Check answer by summing torques about right end to verify A = 72.0 N
St(ccw) = St(cw)
(40 N)(12 m) + (80 N)(3 m) = A (10 m)
480 Nm Nm = A (10 m)
A = 72.0 N

24. Absolute values apply for:
Reminder on Signs:
40 N
80 N
2 m
3 m
7 m A B
SF(up) = SF(down)
Absolute values apply for:
We used absolute (+) values for both UP and DOWN terms.
Instead of: SFy = A + B – 40 N - 80 N = 0
We wrote: A + B = 40 N + 90 N

25. 300 T
800 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.
For purposes of summing torques, we consider entire weight to act at center of board.
300 T
800 N
200 N
300
800 N
200 N T Fx Fy
2 m
3 m
5 m

26. Choose axis of rotation at wall (least information)
300 T
800 N
200 N Fx Fy
2 m
3 m
5 m
Example 4 (Cont.) r
Choose axis of rotation at wall (least information)
St(ccw):
Tr = T (8 m)sin 300 = (4 m)T
St(cw):
(200 N)(5 m) + (800 N)(10 m) = 9000 Nm
(4 m)T = 9000 Nm
T = 2250 N

27. Example 4 (Cont.) Fy = -125 N or Fx = 1950 N F = 1954 N, 356.30 T T
300 T
300 T
800 N
200 N Fx Fy
2 m
3 m
5 m
Example 4 (Cont.) Ty Tx
300
SF(up) = SF(down):
Ty + Fy = 200 N N
Fy = 200 N N - Ty ;
Fy = 1000 N - T sin 300
Fy = 1000 N - (2250 N)sin 300
Fy = -125 N
SF(right) = SF(left):
Fx = Ty = (2250 N) cos 300
Fx = 1950 N or
F = 1954 N,

28. 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.

29. Examples of Center of Gravity
Note: C. of G. is not always inside material.

30. Choose axis at left, then sum torques:
Example 5: Find the center of gravity of the apparatus shown below. Neglect the weight of the connecting rods. x F
C. of G. is point at which a single up- upward force F will balance the system.
30 N
10 N
5 N
4 m
6 m
Choose axis at left, then sum torques:
SF(up) = SF(down):
F = 30 N + 10 N + 5 N
St(ccw) = St(cw)
(45 N) x = 90 N
Fx = (10 N)(4 m) + (5 N)(10 m)
Fx = 90.0 Nm
x = 2.00 m

31. Conditions for Equilibrium:
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.

32. Summary: Procedure St = t1 + t2 + t3 + . . . = 0
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:
St = t1 + t2 + t = 0
Sum forces and set to zero: SFx= 0; SFy= 0
Solve for unknowns.

33. CONCLUSION: Chapter 5B Rotational Equilibrium