Statics Torque Equilibrium How to solve Example Whiteboards
Statics: Torque Equilibrium How to solve Example Whiteboards Torque and force Example Whiteboards
Torque equilibrium - the sum of all torques is zero • Clockwise torques are positive (+), anti-clockwise are negative (-) • = r. Fsin How to set up torque equilibrium: 1. Pick a point to torque about. 2. Express all torques: 3. +r. F+r. F… = 0 • + is CW, - is ACW • r is distance from pivot 4. Do math
How to set up torque equilibrium: 1. Pick a point to torque about. 2. 3. Express all torques: +r. F+r. F… = 0 • + is CW, - is ACW • r is distance from pivot 4. Do math 5. 25 N 2. 15 m F=? 5. 82 m 1. Torque about the pivot point 2 and 3. (2. 15 m)(5. 25 N) - (5. 82 m)F = 0 F = 1. 94 N (mech. adv. )
Whiteboards Simple Torque Equilibrium 1|2|3|4
Find the missing distance. Torque about the pivot point. 315 N 12 m 87. 5 N r=? (315 N)(12 m) - (87. 5 N)r = 0 r = 43. 2 m = 43 m W
Find the missing force. Torque about the pivot point. (Be careful of the way the distances are marked) F= ? 1. 5 m 6. 7 m 34 N (34 N)(1. 5 m) - (8. 2 N)F = 0 F = 6. 2 N W
Find the missing Force. Torque about the pivot point. 512 N 481 N 2. 0 m 3. 1 m 4. 5 m -(512 N)(2. 0 m) - (481 N)(5. 1 m) + F(9. 6 m) = 0 F = 362 N = 360 N F=? W
Find the missing force. Torque about the pivot point. 27. 5 N F=? 35. 0 cm 102 cm 186 cm 12. 2 N -F(. 35 m) - (27. 5 N)(1. 02 m) + (12. 2 N)(1. 86 m) = 0 F = -15. 3 N (it would be downward, not upward as we guessed) -15. 3 N (down, not up as we guessed W
How to set up torque equilibrium: 1. Pick a point to torque about. 2. Express all torques • CW is + • ACW is 3. Set <torques> = 0, solve if you can 5. 0 m 85 kg Find the tension in the cable 58 o 1. 0 m 350 kg 12. 0 m 1. Let’s choose the left side to torque about. Four forces - hinge (up? ), weight of box down , weight of beam down and the tension in the cable up @58 o
Force Equilibrium: 1. Draw picture 2. Calculate weights 3. Draw arrows forces. (weights of beams act at their center of gravity) 4. Make components 5. Set up sum Fx = 0, sum Fy = 0 Torque Equilibrium: 1. Pick a Pivot Point (at location of unknown force) 2. Express all torques: 3. +r. F+r. F… = 0 + is CW, - is ACW r is distance from pivot Do Math
How to set up torque equilibrium: 1. Pick a point to torque about. 2. Express all torques • CW is + • ACW is 3. Set <torques> = 0, solve if you can 5. 0 m F 85 kg Find the tension in the cable 58 o 350 kg 12. 0 m 2 a. The hinge acts at r = 0, and so exerts no torque (torque = r. F, r = 0) 1. 0 m
Torques: Hinge = 0 Nm (torque = 0*F) Find the tension in the cable 5. 0 m F 85 kg 833 N 58 o 1. 0 m 350 kg 12. 0 m 2 b. The box weight (85*9. 8) = 833 N, at a distance of 5. 0 m torque = r. F = (5. 0 m)(833 N) = +4165 Nm (CW)
Torques: Hinge = 0 Nm Box = +4165 Nm = (5. 0 m)(833 N) 5. 0 m F 85 kg 833 N Find the tension in the cable 58 o 1. 0 m 350 kg 3430 N 12. 0 m 2 c. The beam weight (350*9. 8) = 3430 N, at its center of mass, 6. 0 m from the left side torque = r. F = (6. 0 m)(3430 N) = +20, 580 Nm (CW)
Torques: Hinge = 0 Nm Box = +4165 Nm = (5. 0 m)(833 N) Beam = +20, 580 Nm = (6. 0 m)(3430 N) Find the tension in the cable Tsin(58 o) T 5. 0 m F 85 kg 833 N 58 o 350 kg 3430 N 12. 0 m 2 d. The cable tension T, at 11. 0 m, 58 o angle torque = r. Fsin = (11. 0 m)Tsin(58 o) = -9. 329 T m (ACW) 1. 0 m
Torques: Hinge = 0 Nm Box = +4165 Nm = (5. 0 m)(833 N) Beam = +20, 580 Nm = (6. 0 m)(3430 N) Cable = - 9. 329 T m = (11. 0 m)Tsin(58 o) Find the tension in the cable Tsin(58 o) T 5. 0 m F 85 kg 833 N 58 o 350 kg 3430 N 12. 0 m 3. Set up your torque equation: 0 Nm + 4165 Nm + 20, 580 Nm - 9. 329 T m = 0 1. 0 m
Torques: Hinge = 0 Nm Box = +4165 Nm = (5. 0 m)(833 N) Beam = +20, 580 Nm = (6. 0 m)(3430 N) Cable = - 9. 329 T m = (11. 0 m)Tsin(58 o) 5. 0 m F 85 kg 833 N Find the tension in the cable 58 o 350 kg 12. 0 m 4. Do Math: 0 Nm + 4165 Nm + 20, 580 Nm - 9. 329 T Nm = 0 24, 745 Nm = 9. 329 T Nm T = 2652. 62 N = 2700 N 1. 0 m
Whiteboards: Torque Equilibrium 1 a | 1 b | 1 c | 1 d | 1 e | 1 f TOC
20. 0 m 16. 0 m 15 kg T 35 kg 7. 0 m (Fulcrum) Find the tension in the cable Step 1 - Let’s torque about the fulcrum Blue, camels don’t need much water W
20. 0 m 16. 0 m 15 kg T 35 kg 7. 0 m (Fulcrum) Find the tension in the cable Step 2 - Express your torques: The fulcrum, the beam, the box, and the cable The fulcrum is r = 0 from the fulcrum, and so exerts no torque Green, because of their feet W
20. 0 m 16. 0 m 15 kg r T 35 kg 7. 0 m (Fulcrum) Find the tension in the cable F Step 2 a - Calculate the torque exerted by the beam itself. + = CW, - = ACW r = (20. 0 m)/2 - 7. 0 m = 3. 0 m F = (35 kg)(9. 8 N/kg) = 343 N torque = r. F = (3. 0 m)(343 N) = +1029 Nm (CW) W
20. 0 m 16. 0 m r 15 kg T 35 kg 7. 0 m (Fulcrum) Find the tension in the cable F Step 2 b - Calculate the torque exerted by the 15 kg box. + = CW, - = ACW r = 16. 0 m - 7. 0 m = 9. 0 m F = (15 kg)(9. 8 N/kg) = 147 N torque = r. F = (9. 0 m)(147 N) = +1323 Nm (CW) W
20. 0 m 16. 0 m 15 kg r T 35 kg 7. 0 m (Fulcrum) Find the tension in the cable Step 2 c - Express the torque exerted by the cable. + = CW, - = ACW r = 20. 0 m - 7. 0 m = 13. 0 m F=T torque = r. F = (13. 0 m)T = -(13. 0 m)T Nm (ACW) -(13. 0 m)T m (ACW) W
20. 0 m 16. 0 m 15 kg T 35 kg 7. 0 m (Fulcrum) Find the tension in the cable Step 3, and 4 - Set up your torque equilibrium, and solve for T: Beam = +1029 Nm Box = +1323 Nm Cable = -(13. 0 m)T Nm +1029 Nm + 1323 Nm - (13. 0 m)T = 0 T = (1029 Nm + 1323 Nm)/(13. 0 m) = 180. 9 N = 180 N W
Whiteboards: Torque and force 2 a | 2 b | 2 c | 2 d | 2 e TOC
T 1 Beam is 18. 0 m long, person is 5. 0 m from the right side. Find the two tensions in the cables at either end. 77 kg T 2 52 kg Step 1 - Set up your vertical force equation T 1 and T 2 are up, and the beam and person weights are down: Beam: -(52 kg)(9. 8 N/kg) = -509. 6 N (down) Person: -(77 kg)(9. 8 N/kg) = -754. 6 N (down) T 1 + T 2 -509. 6 N - 754. 6 N = 0
T 1 Beam is 18. 0 m long, person is 5. 0 m from the right side. Find the two tensions in the cables at either end. 52 kg 9. 0 m 509. 6 N 13. 0 m 77 kg 18. 0 m 754. 6 N Step 2 - Let’s torque about the left side Set up your torque equation: torque = r. F T 1 = 0 Nm torque, (r = 0) Beam: (9. 0 m)(509. 6 N) = +4586. 4 Nm (CW) Person: (13. 0 m)(754. 6 N) = +9809. 8 Nm (CW) T 2: T 2 at 18. 0 m = -(18. 0 m)(T 2) (ACW) Finally: +4586. 4 Nm + 9809. 8 Nm - (18. 0 m)(T 2) = 0 T 2
T 1 Beam is 18. 0 m long, person is 5. 0 m from the right side. Find the two tensions in the cables at either end. 77 kg T 2 52 kg Step 3 - Math time. Solve these equations for T 1 and T 2: +4586. 4 Nm + 9809. 8 Nm - (18. 0 m)(T 2) = 0 T 1 + T 2 -509. 6 N - 754. 6 N = 0 +4586. 4 Nm + 9809. 8 Nm = (18. 0 m)(T 2), T 2 = 799. 8 N T 1 + 799. 8 N -509. 6 N - 754. 6 N = 0, T 1 = 464. 4 N T 2 = 799. 8 N T 1 =464. 4 N
= ____ cats? How many cats in the “? ” box with the two bombs on it? (Hint – figure out the cat equivalent of the bomb and the bucket o’ chicken – neglect the masses of other things)
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