1. 1 Rotational Dynamics The Action of Forces and Torques on Rigid Objects Chapter 9 Lesson 2 (a) Translation (b) Combined translation and rotation

2. 2 CONCEPTS AT A GLANCE If a rigid body is in equilibrium, neither its linear motion nor its rotational motion changes. This lack of change leads to certain equations that apply for rigid-body equilibrium. Rigid Objects in Equilibrium

3. 3 EQUILIBRIUM OF A RIGID BODY: A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. In equilibrium, the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero:

4. 4 Example 3. A Diving Board

5. 5 A woman whose weight is 530 N is poised at the right end of a diving board with a length of 3.90 m. The board has negligible weight and is bolted down at the left end, while being supported 1.40 m away by a fulcrum. Find the forces F 1 and F 2 that the bolt and the fulcrum, respectively, exert on the board W = 530 N.

6. 6 Example 4. Fighting a Fire

7. 7 An 8.00-m ladder of weight W L = 355 N leans against a smooth vertical wall. The term “smooth” means that the wall can exert only a normal force directed perpendicular to the wall and cannot exert a frictional force parallel to it. A firefighter, whose weight is W F = 875 N, stands 6.30 m from the bottom of the ladder. Assume that the ladder’s weight acts at the ladder’s center and neglect the hose’s weight. Find the forces that the wall and the ground exert on the ladder.

8. 8 Force Lever Arm Torque W L = 355 N L = (4.00 m) cos 50.0° –W L L W F = 875 N F = (6.30 m) cos 50.0° –W F F P P = (8.00 m) sin 50.0° +P P

9. 9 P = 727 N.

10. 10 Check Your Understanding 2 Three forces act on each of the thin, square sheets shown in the drawing. In parts A and B of the drawing, the force labeled 2F acts at the center of the sheet. The forces can have different magnitudes (F or 2F) and can be applied at different points on an object. In which drawing is (a) the translational acceleration equal to zero, but the angular acceleration is not equal to zero, (b) the translational acceleration is not equal to zero, but the angular acceleration is equal to zero, and (c) the object in equilibrium? (a) C, (b) A, (c) B

11. 11 DEFINITION OF CENTER OF GRAVITY: The center of gravity of a rigid body is the point at which its weight can be considered to act when the torque due to the weight is being calculated. Center of Gravity

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13. Problem-Solving Strategy – Equilibrium Problems, 2 Establish a convenient coordinate system Find the components of the forces along the two axes Apply the first condition for equilibrium (  F=0) Be careful of signs

14. Problem-Solving Strategy – Equilibrium Problems, 3 Choose a convenient axis for calculating the net torque on the object Remember the choice of the axis is arbitrary Choose an origin that simplifies the calculations as much as possible A force that acts along a line passing through the origin produces a zero torque

15. Problem-Solving Strategy – Equilibrium Problems, 4 Apply the second condition for equilibrium (  = 0) The two conditions of equilibrium will give a system of equations Solve the equations simultaneously If the solution gives a negative for a force, it is in the opposite direction than what you drew in the free body diagram

16. Weighted Hand Example Model the forearm as a rigid bar The weight of the forearm is ignored There are no forces in the x-direction Apply the first condition for equilibrium (  F y = 0)

17. Weighted Hand Example, cont Apply the second condition for equilibrium using the joint O as the axis of rotation (  =0) Generate the equilibrium conditions from the free-body diagram Solve for the unknown forces (F and R)

18. Horizontal Beam Example The beam is uniform So the center of gravity is at the geometric center of the beam The person is standing on the beam What are the tension in the cable and the force exerted by the wall on the beam?

19. Horizontal Beam Example, 2 Draw a free-body diagram Use the pivot in the problem (at the wall) as the pivot This will generally be easiest Note there are three unknowns (T, R,  )

20. The forces can be resolved into components in the free-body diagram Apply the two conditions of equilibrium to obtain three equations Solve for the unknowns Horizontal Beam Example, 3

21. Ladder Example The ladder is uniform So the weight of the ladder acts through its geometric center (its center of gravity) There is static friction between the ladder and the ground

22. Ladder Example, 2 Draw a free-body diagram for the ladder The frictional force is ƒ = µn Let O be the axis of rotation Apply the equations for the two conditions of equilibrium Solve the equations

23. Add a person of mass M at a distance d from the base of the ladder The higher the person climbs, the larger the angle at the base needs to be Eventually, the ladder may slip Ladder Example, Extended

24. To Be Continued… 24