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Torque & Equilibrium Review

Torque & Equilibrium Review. For this review, answer each question then continue to the next page for the correct answer. . What are the two factors that can affect the amount of torque acting on an object?. Force How hard you push on the lever

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Torque & Equilibrium Review

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  1. Torque & Equilibrium Review For this review, answer each question then continue to the next page for the correct answer.

  2. What are the two factors that can affect the amount of torque acting on an object? • Force • How hard you push on the lever • Could say mass because the force can be found by taking m x g • Lever Arm • (distance between force and axis of rotation)

  3. How can I increase the torque acting on an object? • Increase Force • Increase Mass

  4. What is torque? • A force that causes rotation

  5. Calculate the amount of torque acting on a door if you push on the door .75 m from the hinges with a force of 13.5 N. τ=r F τ = (.75)(13.5) τ= 10. Nm

  6. There are 18 Nm of torque acting on a 1.2 m long fishing rod that has a fish hanging off the end. What is the mass of the fish? τ=r F 18 = (1.2)(m x 9.8) m =1.5 kg **If you have an answer of 15, you’ve found the force acting on the fish, now calculate the mass (Fg=mg)

  7. If a teeter-totter is in equilibrium, what does this mean in terms of torque? • The total torques acting on one side of the teeter-totter MUST equal the total torque acting on the opposite side • Tside 1 = Tside 1

  8. A 12 kg teeter-totter is centered on a fulcrum. Pudgy Pete (m=45 kg) sits 2.3 m from the center. Little Lucy has a mass of 22 kg. Where does Lucy need to sit in order to balance Pudgy Pete? rF = rF (2.3)(45 x 9.8) = r (22 x 9.8) 1014.3 = 215.6 r r = 4.7 m from center

  9. A 20. kg teeter-totter is centered on a fulcrum. Tiny Tina (m=26 kg) sits 3.2 m from the center. Huge Hugh finds he must sit 1.8 meters from the center in order to balance Tina. What is Huge Hugh’s mass? rF = rF (3.2)(26 x 9.8) = r (22 x 9.8) 815.36 = (1.8)(m x 9.8) 452.97 = (m x 9.8) m = 46 kg

  10. Solve for the unknown forces in the mobile 2.38 + 3.0 + 5.20 = 10.58 N (.600)(10.58)=(.150)(3.0) + (.700)(F2) F2 = 8.43 N (.218)(5.20)=(.100)(3.0) + F1(.350) F1 = 2.38 N

  11. Solve for the unknown forces in the mobile 5.29 + 1.0 + 2.97 = 9.26 N F3 1.79 + 1.0 + 2.50 = 5.29 N (.550)(9.26)= (.150)(1.0 )+(.750)(F3) F3= 6.59 N (.150)(5.29)= (.050)(1.0 )+(.250)(F2) F2 = 2.97 N (.250)(2.50)=(.350)(F1) F1 = 1.79 N

  12. Congratulations! You’ve completed the review!

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