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

2. Torque equilibrium - the sum of all torques is zero How to set up torque equilibrium: 1.Pick a point to torque about. 2.Express all torques: 3.+rF +rF+rF… = 0 + is CW, - is ACW r is distance from pivot 4.Do math Clockwise torques are positive (+), anti-clockwise are negative (-)  = rFsin 

3. How to set up torque equilibrium: 1.Pick a point to torque about. 2.Express all torques: 3.+rF +rF+rF… = 0 + is CW, - is ACW r is distance from pivot 4.Do math 5.25 N F = ? 2.15 m5.82 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. 315 N 87.5 N 12 mr = ? Find the missing distance. Torque about the pivot point. 43 m (315 N)(12 m) - (87.5 N)r = 0 r = 43.2 m = 43 m W

6. 34 N F = ? 1.5 m6.7 m Find the missing force. Torque about the pivot point. (Be careful of the way the distances are marked) 6.2 N (34 N)(1.5 m) - (8.2 N)F = 0 F = 6.2 N W

7. 512 N F = ? 2.0 m4.5 m Find the missing Force. Torque about the pivot point. 360 N -(512 N)(2.0 m) - (481 N)(5.1 m) + F(9.6 m) = 0 F = 362 N = 360 N W 481 N 3.1 m

8. F = ? 12.2 N 35.0 cm 186 cm Find the missing force. Torque about the pivot point. -15.3 N (down, not up as we guessed -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 102 cm 27.5 N

9. How to set up torque equilibrium: 1.Pick a point to torque about. 2.Express all torques CW is + ACW is - 3.Set = 0, solve if you can 350 kg 85 kg Find the tension in the cable 58 o 12.0 m 1.0 m 5.0 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

10. Force Equilibrium: 1.Draw picture 2.Calculate weights 3.Draw arrows for 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.+rF +rF+rF… = 0 + is CW, - is ACW r is distance from pivot Do Math

11. 350 kg 85 kg Find the tension in the cable 58 o 12.0 m 1.0 m 5.0 m 2a. The hinge acts at r = 0, and so exerts no torque (torque = rF, r = 0) F How to set up torque equilibrium: 1.Pick a point to torque about. 2.Express all torques CW is + ACW is - 3.Set = 0, solve if you can

12. 350 kg 85 kg Find the tension in the cable 58 o 12.0 m 1.0 m 5.0 m F Torques: Hinge = 0 Nm (torque = 0*F) 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) 833 N

13. 350 kg 85 kg Find the tension in the cable 58 o 12.0 m 1.0 m 5.0 m F Torques: Hinge = 0 Nm Box = +4165 Nm = (5.0 m)(833 N) 833 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) 3430 N

14. 350 kg 85 kg Find the tension in the cable 58 o 12.0 m 1.0 m 5.0 m F Torques: Hinge = 0 Nm Box = +4165 Nm = (5.0 m)(833 N) Beam = +20,580 Nm = (6.0 m)(3430 N) 833 N 3430 N 2d. The cable tension T, at 11.0 m, 58 o angle torque = rFsin  = (11.0 m)Tsin(58 o ) = -9.329T m (ACW) T Tsin(58 o )

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

16. 350 kg 85 kg Find the tension in the cable 58 o 12.0 m 1.0 m 5.0 m F 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 833 N 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(58 o )

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

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

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

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

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

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

23. 180 N W 35 kg 15 kg 7.0m 16.0m 20.0m (Fulcrum) Find the tension in the cable T 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

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

25. T 1 + T 2 -509.6 N - 754.6 N = 0 Step 1 - Set up your vertical force equation 52 kg 77 kg T1T1 T2T2 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. 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

26. +4586.4 Nm + 9809.8 Nm - (18.0 m)(T 2 ) = 0 Step 2 - Let’s torque about the left side Set up your torque equation: torque = rF 52 kg 77 kg T1T1 T2T2 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. 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 9.0 m 13.0 m 18.0 m 509.6 N 754.6 N

27. T 2 = 799.8 N T 1 =464.4 N 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 52 kg 77 kg T1T1 T2T2 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. +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