1. Calculating the amount of torque (A confusing task)

2. A door Hinge Knob

3. Door (Top View) Hinge knob

4. How much torque? (Zero, a Little, or Large?)

9. How to calculate torque Method One (“Adjust the force”) F

10. F Draw a vector from the axis to the point of application of force. d

11. F Extend the vector to make an angle between the force and d.  d

12. F Complete the triangle to find the component of F that is perpendicular to d. This is the only part of F that makes the torque.  F  = Fsin  d

13. F  = d·F   = d·Fsin   F  = Fsin  d

14. F How to calculate torque Method Two {“Adjust the distance”} (more handy than you might think)

15. F Draw the usual distance vector. d 

16. F Extend the Force Vector to see the “Line of Force” d  

17. F d  How far is the Line of Force from the pivot? (We call this distance the ‘Lever Arm’ or “Moment Arm’) d  = dsin(  ) 

18. F d    = F·d   = F·dsin   = d Fsin  dsin 

19. Numerical Examples. For each, the force is 20 N, and the dimensions of the door are: 0.8 meters 0.3 m Note: The diagonal distance is 0.854 meters.

20. Numerical Example A (Method One - Adjust Force) F 40˚ d = 0.8 m

21. Numerical Example A (Method One - Adjust Force) 50˚ 40˚ d = 0.8 m F = 20 N  = 50˚

22. Numerical Example A (Method One - Adjust Force) 50˚ 40˚ d = 0.8 m F  = 20 N·sin( 50˚ ) = 15.3N FF

23. Numerical Example A (Method One - Adjust Force) 50˚ 40˚ d = 0.8 m F  = 20 N·sin( 50˚ ) = 15.3N FF  = d·F   = (0.8 m)·(15.3 N) = 12.3 Nm

24. Numerical Example A (Method Two - Adjust Distance) F 40˚ 50˚ 40˚ 90˚ d = 0.8 m d  = ?

25. Numerical Example A (Method Two - Adjust Distance) F 40˚ 50˚ 40˚ d  = (0.8 m)sin(  ˚) = 0.61 m d = 0.8 m

26. Numerical Example A (Method Two - Adjust Distance) F 40˚ 50˚ 40˚ d  = (0.8 m)sin(  ˚) = 0.61 m d = 0.8 m  = d  ·F  = (0.61 m)·(20 N) = 12.3 Nm

27. Numerical Example B Will this be: More, Less, or the Same amount of Torque? d = 0.8 m F = 20 N

28. Numerical Example B Method One - Adjust Force 90˚ Why is F  = 20 N?

29. Numerical Example B Method One - Adjust Force 90˚  = d·F   = (0.8 m)·(20 N) = 16 Nm Is this more, less, or the same?

30. Numerical Example B Method Two - Adjust Distance d = 0.8 m F = 20 N

31. Numerical Example B Method Two - Adjust Distance d = 0.8 m F = 20 N Why is d  = 0.8 m?

32. Numerical Example B Method Two - Adjust Distance d = 0.8 m F = 20 N  = d  ·F  = (0.8 m)·(20 N) = 16 Nm

33. Numerical Example C (shows why Method 2 is useful)

34. Numerical Example C (Method One - Adjust Force) WARNING: This is a bad method for this problem. 20 N 0.8 m 0.3 m Just sit back and see how much work this method is for this problem, and then how easy the other method is for the same problem.

35. Numerical Example C In Method One, we find F . Draw in the component of F that is  to d. d D = √(0.8 2 + 0.3 2 ) = 0.854 m

36. Extend d. Find F . d  

37. d FF    = InvTan(Opp/Adj)  = tan -1 (0.8m/0.3m) = 69.4˚

38. Find F . d FF   F  = (20 N)sin(69.4˚) = 18.7 N

39.  = d·F   = (0.854m)·(18.7N)  = 16 Nm d FF   This was a tough way to get the answer.

40. Numerical Example C (Method Two - Adjust Distance) 20 N 0.8 m 0.3 m

41.  = F·d  So, find the distance from the ‘Line of force’ to the pivot. 20 N 0.8 m 0.3 m

42. dd 20 N 0.8 m 0.3 m

43.  = F·d   = (20 N)·(0.8 m)  = 16 Nm 20 N 0.8 m 0.3 m That was much better.

44. Addendum (skippable) A closer look at Method 1 in Example C In the next slide notice that the distance (d) comes from the width (W) of the door and the thickness (T) of the door:

45. d  W T d = square root of …

46. In the previous slide, notice that The sine of the angle is Opp/Hyp So

47. Calculate the torque and see the simplification to the old result  = dFsin 