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Statics: Torque Equilibrium

Statics: Torque Equilibrium. T orque Equilibrium How to solve Example Whiteboards Torque and force Example Whiteboards. Clockwise torques are positive (+), anti-clockwise are negative (-)  = rFsin. Torque equilibrium - the sum of all torques is zero. How to set up torque equilibrium:

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Statics: Torque Equilibrium

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  1. Statics: Torque Equilibrium Torque Equilibrium How to solve Example Whiteboards Torque and force Example Whiteboards

  2. Clockwise torques are positive (+), anti-clockwise are negative (-) •  = rFsin Torque equilibrium - the sum of all torques is zero • How to set up torque equilibrium: • Pick a point to torque about. • Express all torques: • +rF +rF+rF… = 0 • + is CW, - is ACW • r is distance from pivot • Do math

  3. How to set up torque equilibrium: • Pick a point to torque about. • Express all torques: • +rF +rF+rF… = 0 • + is CW, - is ACW • r is distance from pivot • Do math 5.25 N F = ? 2.15 m 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.)

  4. Whiteboards Simple Torque Equilibrium 1 | 2 | 3 | 4

  5. Find the missing distance. Torque about the pivot point. 315 N 87.5 N 12 m r = ? (315 N)(12 m) - (87.5 N)r = 0 r = 43.2 m = 43 m W 43 m

  6. 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 6.2 N

  7. 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 360 N

  8. 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) W -15.3 N (down, not up as we guessed

  9. How to set up torque equilibrium: • Pick a point to torque about. • Express all torques • CW is + • ACW is - • Set <torques> = 0, solve if you can Find the tension in the cable 5.0 m 85 kg 58o 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 @58o

  10. Force Equilibrium: • Draw picture • Calculate weights • Draw arrows for forces. • (weights of beams act at their center of gravity) • Make components • Set up sum Fx = 0, sum Fy = 0 • Torque Equilibrium: • Pick a Pivot Point • (at location of unknown force) • Express all torques: • +rF +rF+rF… = 0 • + is CW, - is ACW • r is distance from pivot • Do Math

  11. F 2a. The hinge acts at r = 0, and so exerts no torque (torque = rF, r = 0) • How to set up torque equilibrium: • Pick a point to torque about. • Express all torques • CW is + • ACW is - • Set <torques> = 0, solve if you can Find the tension in the cable 5.0 m 85 kg 58o 1.0 m 350 kg 12.0 m

  12. 833 N 2b. The box weight (85*9.8) = 833 N, at a distance of 5.0 m torque = rF = (5.0 m)(833 N) = +4165 Nm (CW) Torques: Hinge = 0 Nm (torque = 0*F) Find the tension in the cable 5.0 m 85 kg 58o 1.0 m F 350 kg 12.0 m

  13. 3430 N 2c. The beam weight (350*9.8) = 3430 N, at its center of mass, 6.0 m from the left side torque = rF = (6.0 m)(3430 N) = +20,580 Nm (CW) Torques: Hinge = 0 Nm Box = +4165 Nm = (5.0 m)(833 N) Find the tension in the cable 5.0 m 85 kg 58o 1.0 m F 350 kg 833 N 12.0 m

  14. T Tsin(58o) 2d. The cable tension T, at 11.0 m, 58o angle torque = rFsin = (11.0 m)Tsin(58o) = -9.329T m (ACW) 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 5.0 m 85 kg 58o 1.0 m F 350 kg 833 N 3430 N 12.0 m

  15. Torques: Hinge = 0 Nm Box = +4165 Nm = (5.0 m)(833 N) Beam = +20,580 Nm = (6.0 m)(3430 N) Cable = - 9.329T m = (11.0 m)Tsin(58o) Find the tension in the cable T Tsin(58o) 5.0 m 85 kg 58o 1.0 m F 350 kg 833 N 3430 N 12.0 m 3. Set up your torque equation: 0 Nm + 4165 Nm + 20,580 Nm - 9.329T m = 0

  16. Torques: Hinge = 0 Nm Box = +4165 Nm = (5.0 m)(833 N) Beam = +20,580 Nm = (6.0 m)(3430 N) Cable = - 9.329T m = (11.0 m)Tsin(58o) Find the tension in the cable 5.0 m 85 kg 58o 1.0 m F 350 kg 833 N 12.0 m 4. Do Math: 0 Nm + 4165 Nm + 20,580 Nm - 9.329T Nm = 0 24,745 Nm = 9.329T Nm T = 2652.62 N = 2700 N

  17. Whiteboards: Torque Equilibrium 1a | 1b | 1c | 1d | 1e | 1f TOC

  18. 20.0m Step 1 - Let’s torque about the fulcrum 16.0m T 15 kg 35 kg 7.0m (Fulcrum) Find the tension in the cable W Blue, camels don’t need much water

  19. 20.0m 16.0m T 15 kg 35 kg 7.0m (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 W Green, because of their feet

  20. 20.0m r F Step 2a - Calculate the torque exerted by the beam itself. + = CW, - = ACW 16.0m T 15 kg 35 kg 7.0m (Fulcrum) Find the tension in the cable r = (20.0 m)/2 - 7.0 m = 3.0 m F = (35 kg)(9.8 N/kg) = 343 N torque = rF = (3.0 m)(343 N) = +1029 Nm (CW) W +1029 Nm (CW)

  21. 20.0m r F Step 2b - Calculate the torque exerted by the 15 kg box. + = CW, - = ACW 16.0m T 15 kg 35 kg 7.0m (Fulcrum) Find the tension in the cable r = 16.0 m - 7.0 m = 9.0 m F = (15 kg)(9.8 N/kg) = 147 N torque = rF = (9.0 m)(147 N) = +1323 Nm (CW) W +1323 Nm (CW)

  22. 20.0m T r Step 2c - Express the torque exerted by the cable. + = CW, - = ACW 16.0m 15 kg 35 kg 7.0m (Fulcrum) Find the tension in the cable r = 20.0 m - 7.0 m = 13.0 m F = T torque = rF = (13.0 m)T = -(13.0 m)T Nm (ACW) W -(13.0m)T m (ACW)

  23. 20.0m 16.0m T 15 kg 35 kg 7.0m (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 180 N

  24. Whiteboards: Torque and force 2a | 2b | 2c | 2d | 2e TOC

  25. 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. T1 T2 77 kg 52 kg Step 1 - Set up your vertical force equation T1 and T2 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) T1 + T2 -509.6 N - 754.6 N = 0 T1 + T2 -509.6 N - 754.6 N = 0

  26. 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. T1 T2 77 kg 52 kg 18.0 m 13.0 m 9.0 m 509.6 N 754.6 N Step 2 - Let’s torque about the left side Set up your torque equation: torque = rF T1 = 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) T2: T2 at 18.0 m = -(18.0 m)(T2) (ACW) Finally:+4586.4 Nm + 9809.8 Nm - (18.0 m)(T2) = 0 +4586.4 Nm + 9809.8 Nm - (18.0 m)(T2) = 0

  27. 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. T1 T2 77 kg 52 kg Step 3 - Math time. Solve these equations for T1 and T2: +4586.4 Nm + 9809.8 Nm - (18.0 m)(T2) = 0 T1 + T2 -509.6 N - 754.6 N = 0 +4586.4 Nm + 9809.8 Nm = (18.0 m)(T2), T2 = 799.8 N T1 + 799.8 N-509.6 N - 754.6 N = 0, T1 = 464.4 N T2 = 799.8 N T1 =464.4 N

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