Mastering Torque Equilibrium in Statics with Practical Examples
Explore the concept of torque equilibrium in statics through detailed explanations, step-by-step solutions, and real-world examples. Learn how to set up torque equilibrium, calculate missing forces and distances, and solve complex problems effectively. Enhance your understanding of torque and force relationships using whiteboard illustrations and solve practice scenarios.
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Presentation Transcript
Statics: Torque Equilibrium 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 (-) (Arbitrary) = rFsin 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
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 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.)
Whiteboards Simple Torque Equilibrium 1-4
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 43 m
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 6.2 N
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 = ? 360 N
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)
Example: The uniform beam is 6.00 m long. The box is 2.00 m from the left side, the person is 1.00 m from the right side. What does F have to be to support the beam if it is exerted 4.10 m from the left side? 71.0kg 16.0kg 42.0kg F = ? 1230 N
Whiteboards Less Simple Torque Equilibrium 1-3
The uniform beam is 8.00 m long. The person is 2.10 m from the right side. What distance must the force of 1900 N be exerted from the left side to hold up the beam? 56.0 kg 38.0kg F = 1900. N 2.49 m from the left side
A 7.00 m long uniform beam has a mass of 43.0 kg. A 64.0 kg person is standing 2.00 m from the left side, and a 52.0 kg person is standing 1.00 m from the right side. What is the tension in the cable on the right side? Pretend that the beam is all at its middle. (3.50 m from the side) 64.0 kg 52.0 kg T = ? 7.00 m 43.0 kg solution 828 N
The 14.0 kg beam is uniform and 4.20 m long. The 11.0 kg beam is 3.30 m long, and the 18.0 kg box is 1.20 m wide. How far from the left side must the 263 N supporting force be exerted? 18 kg 11.0 kg 14.0 kg x = ? 263 N solution 3.15 m
72.0 kg Dad sits 1.20 m from the center of the seesaw, and Keenan sits 3.40 m from the center to balance. What is Keenan s mass? 72.0 kg m = 1.20 m 3.40 m solution 25.4 kg
= ____ cats? = ____ 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)
