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- 1. TORQUE
- 2. Think of these! <ul><li>Why is it easier to open the tight lid of a can of paint using a long screw driver than a 5-peso coin? </li></ul><ul><li>Why are door knobs placed at the other end of the door opposite the hinges and not at the middle? </li></ul><ul><li>Why is it easier to maneuver a bicycle with handle bar than one without? </li></ul>
- 3. <ul><li>The answer to all is “torque”. </li></ul>
- 4. OUR TARGET <ul><li>Define torque; </li></ul><ul><li>Illustrate by example the understanding of the term moment arm; </li></ul><ul><li>Calculate the resultant torque about any axis when given the magnitude and position of forces on an extended object; </li></ul>
- 5. Let’s recall: <ul><li>What does the first condition of equilibrium tell us about the condition of an object in equilibrium? </li></ul>
- 6. The First Condition of Equilibrium <ul><li>“ If all forces acting on a body intersect at a single point and their vector sum is zero, the system must be in equilibrium.” </li></ul><ul><li>Mathematically: Σ F = 0 </li></ul>
- 7. <ul><li>LET’S ANALYZE THE FOLLOWING SITUATIONS </li></ul>
- 8. F F N S F F Does equilibrium exist if you exert a pair of equal opposing forces F to the right and to the left of the lug wrench? Situation 1
- 9. F F E W F F If the same two forces are applied as shown in the figure, does equilibrium exist? Situation 2
- 10. Questions <ul><li>How would you compare the vector sum of the forces in the two situations? </li></ul><ul><li>How would you compare the effect of the pair of equal forces applied on the lug wrench in the two situations? </li></ul><ul><li>What is the difference in the application of the two equal forces in situation 1 and 2? </li></ul>
- 11. N S F F E W F F VECTOR DIAGRAMS SITUATION 1 SITUATION 2 How would you compare the vector sum of the forces in the two situations?
- 12. N S F F N W F F VECTOR DIAGRAMS SITUATION 1 SITUATION 2 How would you compare the effect of the pair of equal forces applied on the lug wrench in the two situations?
- 13. N S F F N W F F VECTOR DIAGRAMS SITUATION 1 SITUATION 2 What is the difference in the application of the two equal forces in situations 1 and 2?
- 14. THE LINE OF ACTION <ul><li>The line of action of a force is an imaginary line extended indefinitely along the vector in both directions. </li></ul>F F F F S Λ
- 15. THE MOMENT ARM <ul><li>The moment arm of a force is the perpendicular distance from the line of action of the force to the axis of rotation . </li></ul>W Axis of rotation Line of action
- 16. THE MOMENT ARM <ul><li>The moment arm of a force is the perpendicular distance from the line of action of the force to the axis of rotation. </li></ul>F Line of action r = L sin θ Axis of rotation Length L
- 17. <ul><li>LET’S ANALYZE THE MEANING OF </li></ul><ul><li>MOMENT ARM </li></ul>
- 18. A B C F F F Given three unbalanced forces A, B and C 1. What is the moment arm of A? B? C? 2. How would you compare the rotational effect of A, B and C? 3. Which of the three unbalanced forces A, B, and C is the most effective in producing rotational motion? Why? Least effective? Why?
- 19. A B C F F F 1. What is the moment arm of A? B? C? 2. How would you compare the rotational effect of A, B and C? 3. Which of the three unbalanced forces A, B, and C is the most effective in producing rotational motion? Why? Least effective? Why?
- 20. TORQUE is the tendency to produce a change in rotational motion. It is also called moment of force.
- 21. <ul><li>Rotational motion is affected by both the magnitude of force F and its moment arm (also called torque arm ) r . </li></ul><ul><li>Thus we define torque as the product of a force and its moment arm. </li></ul>
- 22. Mathematically: <ul><li>τ = Fr </li></ul>Where: τ (tau) = torque( in Newton-meter, N-m or pound-foot, lb-ft) F = force (in Newton, N or pound, lb) r = moment arm (in meter, m or foot, ft) Torque = Force x moment arm
- 23. Sign Convention <ul><li>The directions of torque depends on whether it tends to produce clockwise or counterclockwise rotation about an axis. </li></ul><ul><li>We take counterclockwise rotation positive and </li></ul><ul><li>clockwise rotation negative . </li></ul>
- 24. F F (-) (+) (+) (-) F Clockwise torque Counterclockwise torque
- 25. Illustrative Example <ul><li>1. Draw and label the moment arm of force F about an axis at point A and point B in the figure below? What is the magnitude of the moment arm? What is the resultant torque about axis A and B? </li></ul>A F B 2 m 3 m
- 26. Solution <ul><li>The moment arm about point A is the perpendicular distance from point A to the line of action of the force F </li></ul>A F B 2 m 3 m The moment arm about point B is the perpendicular distance from point B to the line of action of the force F
- 27. Illustrative Example <ul><li>2. Draw and label the moment arm of force F about an axis at point A in the figure below? What is the magnitude of the moment arm? </li></ul>A F B 2 m 3 m 30 °
- 28. 2. Solution <ul><li>The moment arm about point A is the perpendicular distance from point A to the line of action of force F. </li></ul>A F B 2 m 3 m 30 ° Its magnitude is : r = L sin 30 ° = 2m (0.500) = 1.0 m
- 29. Illustrative Example <ul><li>3. Find the moment arm about axis B. </li></ul>A F B 2 m 3 m 30 °
- 30. 3. Solution <ul><li>The moment arm about point B is the perpendicular distance from point B to the line of action of force F. </li></ul>A F B 2 m 3 m 30 ° Its magnitude is : r = L sin 30 ° = 3m (0.500) = 1.5 m
- 31. Illustrative Examples <ul><li>4. If force F is 100 N, what is the resultant torque about an axis A? What is the resultant torque about axis B? </li></ul>A F B 2 m 3 m 30 °
- 32. 4. Solution A F B 2 m 3 m 30 ° Since r A = 1.0 m and F = 100 N T A = Fr A = 100 N (1.0 m) = 100N-m r = 1.0 m -
- 33. 4. Solution Since r B = 1.5 m and F = 100 N T B = Fr B = 100 N (1.5 m) = 150N-m A F B 2 m 3 m 30 ° r = 1.5 m
- 34. Illustrative Example <ul><li>5. What if force F is applied directly towards point B, what would be the resultant torque about axis A and B? </li></ul>A F B 2 m 3 m
- 35. Sum it up. <ul><li>If you are to explain torque to someone who does not study physics, how would you explain it so that she/he can readily understand it? </li></ul><ul><li>How would you explain to him/her the importance of the moment arm in producing torque? </li></ul>
- 36. Check your understanding. <ul><li>Why is it easier to open a tight lid of a can of paint using a long screw driver than with a coin? </li></ul><ul><li>Why is the door knob always located at the opposite end of the door hinge? Why is it not placed at middle of the door? </li></ul>
- 37. Which of the following will produce the greatest torque? D= 1000 N C=600 N A= 1000 N B=150 N 2.0 m 3.5 m 2.5 m 8.0 m
- 38. Which of the following will produce the greatest torque? D= 1000 N C=600 N A= 1000 N B=450 N 2.0 m 3.5 m 2.5 m 8.0 m 20 ° 60 ° T = 1000N (2.0m)sin60 ° = 1732 N-m T = 450N (5.50 m) = 2475 N-m T = 600N (5.50m)sin 20° = 1129 N-m T = 1000N (0.0m) = 0
- 39. Assignment <ul><li>A. Answer these in your assignment notebook. </li></ul><ul><li>What is the 2 nd Condition of Equilibrium? </li></ul><ul><li>How is it different from the 1st condition of Equilibrium? </li></ul>
- 40. <ul><li>B. Figure 1 shows an object with several forces acting on it. The pivot point is at O. </li></ul>If F 1 = 10 N, and is at a distance of 0.25 m from O, where Θ 1 = 30 ° F 2 = 7.0 N, acting perpendicular to the object, at a distance of 1.25 m from O F 3 = 12 N, is 0.60 m from O, and acts at Θ 3 = 40 ° from the horizontal Find the total (net) torque on the object.
- 41. <ul><li>Torque is greater with longer moment arm. How could this idea be applied to you when you are in trouble? Or when someone is in need? </li></ul>Did MCS ever extend a helping hand to you when you are in need? At your end, how do you support the school in helping others in need? Think about this.
- 42. <ul><li>THE END </li></ul>
- 50. Figure 1 object with several torques. The net total torque on the object is: A . -1.7 N m B . -15.8 N m C . -6.6 N m D . 18 N m
- 51. Solution <ul><li>Τ 1 = F 1 r 1 = </li></ul><ul><li>Τ 2 = F 2 r 2 = </li></ul><ul><li>Τ 3 = F 3 r 3 = </li></ul>
- 52. Calculate for moment arm and torque
- 54. r 1 r 2 r 3

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