1. 1. How is torque calculated?

2. Torque = Force X length of torque arm T = F x l

3. 2. What are the two requirements for equilibrium?

4. The net force must be zero, and the net torque must be zero. (The clockwise torques must be equal to the counterclockwise torques.)

5. 3. If you push the knob of a heavy door with a force of 230 N, and the knob is 0.82 meters from the hinge, what is the torque applied?

6. T = F x l T = 230 x 0.82 T = 189 Nm

7. 4. What force is needed to provide the same torque on the door in problem three if you apply this force halfway between the knob and the hinge?

8. The length of the torque arm l is cut in half. If the torque must stay the same and T = F x l, then F must be doubled. 230 x 2 = 460 N

9. 5. A 50 kg child sits 3 meters from the fulcrum of a seesaw. Where must the fat physics teacher (150 kg) sit to balance the seesaw?

10. Clockwise torque must balance counterclockwise torque. CWCCW 50 x 10 x 3 = 150 x 10 x l l = 1 m

11. 6. Two workers lift a heavy beam (800 N, 4 meters long) by the ends. The beam is not uniform; the center of gravity is 1.5 meters from the heavy end. What force must each worker apply to lift the beam?

12. Clockwise torque must balance counterclockwise torque. CWCCW 800 x 1.5 = F l x 4 F L = 300N F H = 800 – 300 = 500 N

13. 7. A gymnast (mass of 45 kg) is on the balance beam (5 meters, 30 kg). The beam is held by two support bars, attached 1 meter from each end. If the gymnast is standing 2 meters from the left end, what is the upward force supplied by each support?

14. If we pick the pivot point as the support bar closest to the gymnast, the gymnast has a weight of 450 N and is 1 m from the pivot point. The beam has a weight of 300 N and is 1.5 m from the pivot. So the total clockwise torques are: 450 x 1 + 300 x 1.5 = 900 Nm. (continued)

15. The counterclockwise torque is the force applied by the other support (F R ) multiplied by the distance from the pivot, which is 3 m. This CCW torque must be equal to the CW torque, so: 900 Nm = F R x 3. F R = 300 N Since the total weight of the gymnast and the beam is 750 N, the force supplied by the other support F L is: 750 = F R + 300 F R = 450 N

16. 8. A uniform horizontal beam with a length of 6 meters and a weight of 120 N is attached at one end to a wall by a pin so the beam can rotate. The opposite end of the beam is supported by a cable attached to the wall above the pin. The cable makes an angle of 60º with the beam. A) What torque is produced by the weight of the beam?

17. A: The torque is the weight of the bar (120 N) X the length of the torque arm. The length of the torque arm is the distance from the pivot (pin) to the center of gravity (3 m). T = F x l T = 120 N x 3 m = 360 Nm

18. B) What torque is produced by the tension in the cable? B: The bar is not changing its rotation, so the torques must be balanced. So the torque produced by the cable must be the same as the torque produced by the weight, 360 Nm.

19. C) What is the vertical component of the tension in the cable that produces equilibrium? C: The torque is 360 Nm and the cable is attached 6 m from the pivot. T = F x l 360 Nm = F x 6 m F = 60 N

20. D) What is the tension in the cable needed to keep the beam in equilibrium? If 60 N is the opposite of the angle, the tension T is the hypotenuse. Sin 60º = 60/T T = 69N

21. 9. What are the six simple machines? What are the two basic types?

22. Lever, pulley, wheel and axle, inclined plane, wedge, screw. The two basic types are inclined plane and lever.

23. 10. How do most simple machines multiply force?

24. All machines that multiply force do so by increasing the distance the force is applied to the machine. You must apply your force over a longer distance to produce larger force over a smaller distance. You trade force for distance.

25. 11. What is the law of machines?

26. Work output cannot be larger than work input.

27. 12. An inclined plane is 6 meters long and is used to raise a box 2 meters above the ground. What is the ideal mechanical advantage of the inclined plane?

28. IMA = d in / d out IMA = 6 m / 2 m IMA = 3

29. 13. If the box lifted in problem 12 has a weight of 1200 N, what is the work output?

30. W.O. = F out x d out W.O. = 1200 N x 2 m W.O. = 2400 J

31. 14. If there is no friction on the inclined plane described above, what force must be applied to the box to push it up the plane? What is the work input in this case?

32. Mechanical advantage is a measure of how much force is multiplied. Since the IMA is 3, it only takes a 400 N force to produce a 1200 N force. W.I. = F in x d in W.I. = 400 N x 6 m W.I. = 2400 J

33. 15. Due to friction, the actual force applied to the box described above to push it up the plane is 600 N. What is the actual mechanical advantage of the inclined plane? What is the work input in this case?

34. AMA = F out / F in AMA = 1200 N / 600 N AMA = 2 W. I. = F in X d in W. I. = 600 N X 6 m W. I. = 3600 J

35. 16. Including friction, what is the efficiency of the inclined plane described in problems 12 through 15?

36. Eff = (W.O. / W. I.) x 100 Eff = (2400 J / 3600 J) x 100 Eff = 67%

37. 17. A fellow student tells you he has a machine that will produce 1000 N of force over a distance of 3 meters if you apply a force of 500 N over 4 meters. Is this possible? What law does this situation violate?

38. No, it is not possible. The work output, 3000 J, is larger than the work input, 2000 J. This violates the law of machines and the law of conservation of energy.