1 / 27

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:

Download Presentation

Statics: Torque Equilibrium

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related