To determine if an object is in equilibrium, there can be no resultant force or torque. This means the sum of all forces and sum of all torques must equal zero. To analyze equilibrium problems, one draws a free body diagram, chooses an axis of rotation where information is lacking, sums the torques and forces about that axis, setting each sum equal to zero to solve for unknowns. Common examples involve finding tensions in ropes or forces from supports of beams or structures.

1. 05B: Equilibrium

2. San Juanico Bridge

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

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

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

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

8. Choose axis of rotation at point where least information is given.

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

12. Finding the center of gravity 01. Plumb-line Method 02. Principle of Moments St(ccw) = St(cw)

13. Example 5:Find the center of gravity of the system shown below. Neglect the weight of the connecting rods. m3 = 6.00 kg m2 = 1.00 kg Since the system is 2-D: Choose now an axis of rotation/reference axis: 3.00 cm CG 2.00 cm m1 = 4.00 kg Trace in the Cartesian plane in a case that your axis of rotation is at the origin (-3.00 cm, 2.00 cm) (0, 2.00 cm) (0,0)

14. x 4 m 6 m 5 N 10 N 30 N Example 6:Find the center of gravity of the apparatus shown below. Neglect the weight of the connecting rods. C.G. Choose axis at left, then sum torques: x = 2 m

15. T 300 800.0 N T T Fx 300 300 3.00 m 5.00 m 2.00 m 200.0 N Fy 800.0 N 200.0 N 800.0 N Example 4: Find the tension in the rope and the force by the wall on the boom. The 10.00-mboom weighing 200.0 N. Rope is 2.00 m from right end. Since the boom is uniform, it’s weight is located at its center of gravity (geometric center)

16. Example 4 (Cont.) T T Fx 300 300 5.00 m 2.00 m 3.00 m Fy w1 w2 200.0 N 800.0 N r Choose axis of rotation at wall (least information) r1= 8.00 m St(ccw): r2= 5.00 m St(cw): r3= 10.00 m T = 2250 N

17. SF(up) = SF(down): SF(right) = SF(left): F = 1950 N, 356.3o

18. Example 3:Find the forces exerted by supports A and B. The weight of the 12-m boom is 200 N. B A 2 m 3 m 7 m 80 N 40 N 2 m 3 m 7 m Draw free-body diagram B A 40 N 80 N Rotational Equilibrium: Choose axis at point of unknown force. At A for example.

19. 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.

21. Choose axis of rotation at point where least information is given.

23. Solve for unknowns.Summary: Procedure