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

3. 2 DEFINITION OF TORQUE Direction: The torque is positive when the force tends to produce a counterclockwise rotation about the axis, and negative when the force tends to produce a clockwise rotation. SI Unit of Torque: newton · meter (N · m)

4.  F F sin  If  =0, then the torque is zero  Therefore, torques (force times length) are responsible for rotational motion

5. 4 The line of action is an extended line drawn colinear with the force. The lever arm is the distance between the line of action and the axis of rotation, measured on a line that is perpendicular to both. The torque is represented by the symbol  (Greek letter tau), and its magnitude is defined as the magnitude of the force times the lever arm

6. 5 Forces of the same magnitude can produce different torques, depending on the value of the lever arm.

7. 6 Example 1. Different Lever Arms, Different Torques

8. 7 (a) (b) (c) In part c the line of action of F passes through the axis of rotation (the hinge). Hence, the lever arm is zero, and the torque is zero. A force whose magnitude is 55 N is applied to a door. However, the lever arms are different in the three parts of the drawing: (a) = 0.80 m, (b) = 0.60 m, and (c) = 0 m. Find the magnitude of the torque in each case.

9. 8 Example 2. The Achilles Tendon The ankle joint and the Achilles tendon attached to the heel at point P. The tendon exerts a force of magnitude F = 720 N. Determine the torque (magnitude and direction) of this force about the ankle joint, which is located 3.6 × 10 –2 m away from point P.

10. 9 = (3.6 × 10 –2 m) cos 55° The force F tends to produce a clockwise rotation about the ankle joint, so the torque is negative:

11. 10 Check Your Understanding 1 The drawing shows an overhead view of a horizontal bar that is free to rotate about an axis perpendicular to the page. Two forces act on the bar, and they have the same magnitude. However, one force is perpendicular to the bar, and the other makes an angle  with respect to it. The angle  can be 90°, 45°, or 0°. Rank the values of  according to the magnitude of the net torque (the sum of the torques) that the two forces produce, largest net torque first. 0°, 45°, 90°

12. Example 1: An 80-N force acts at the end of a 12-cm wrench as shown. Find the torque. Extend line of action, draw, calculate r.  = (80 N)(0.104 m) = 8.31 N m r = 12 cm sin 60 0 = 10.4 cm

13. Alternate: An 80-N force acts at the end of a 12-cm wrench as shown. Find the torque. Resolve 80-N force into components as shown. Note from figure: r x = 0 and r y = 12 cm  = (80N x Sin60)(0.12 m)  = 8.31 N m as before positive 12 cm

14. Calculating Resultant Torque Read, draw, and label a rough figure.Read, draw, and label a rough figure. Draw free-body diagram showing all forces, distances, and axis of rotation.Draw free-body diagram showing all forces, distances, and axis of rotation. Extend lines of action for each force.Extend lines of action for each force. Calculate moment arms if necessary.Calculate moment arms if necessary. Calculate torques due to EACH individual force affixing proper sign. CCW (+) and CW (-).Calculate torques due to EACH individual force affixing proper sign. CCW (+) and CW (-). Resultant torque is sum of individual torques.Resultant torque is sum of individual torques. Read, draw, and label a rough figure.Read, draw, and label a rough figure. Draw free-body diagram showing all forces, distances, and axis of rotation.Draw free-body diagram showing all forces, distances, and axis of rotation. Extend lines of action for each force.Extend lines of action for each force. Calculate moment arms if necessary.Calculate moment arms if necessary. Calculate torques due to EACH individual force affixing proper sign. CCW (+) and CW (-).Calculate torques due to EACH individual force affixing proper sign. CCW (+) and CW (-). Resultant torque is sum of individual torques.Resultant torque is sum of individual torques.

15. Example 2: Find resultant torque about axis A for the arrangement shown below: 30 0 6 m 2 m 4 m 20 N 30 N 40 N A Find  due to each force. Consider 20-N force first: r = (4 m) sin 30 0 = 2.00 m  = Fr = (20 N)(2 m) = 40 N m, cw The torque about A is clockwise and negative.  20 = -40 N m r negative

16. Example 2 (Cont.): Next we find torque due to 30-N force about same axis A. 30 0 6 m 2 m 4 m 20 N 30 N 40 N A Find  due to each force. Consider 30-N force next. r = (8 m) sin 30 0 = 4.00 m  = Fr = (30 N)(4 m) = 120 N m, cw The torque about A is clockwise and negative.  30 = -120 N m r negative

17. Example 2 (Cont.): Finally, we consider the torque due to the 40-N force. Find  due to each force. Consider 40-N force next: r = (2 m) sin 90 0 = 2.00 m  = Fr = (40 N)(2 m) = 80 N m, ccw The torque about A is CCW and positive.  40 = +80 N m 30 0 6 m 2 m 4 m 20 N 30 N 40 N A r positive

18. Example 2 (Conclusion): Find resultant torque about axis A for the arrangement shown below: 30 0 6 m 2 m 4 m 20 N 30 N 40 N A Resultant torque is the sum of individual torques.  R = - 80 N m Clockwise  R =  20 +  20 +  20 = -40 N m -120 N m + 80 N m

19. To Be Continued… 18