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Unformatted text preview: Lecture 25 Statics Conditions for equilibrium: a n d F τ = = ∑ ∑ G G Equilibrium Equilibrium A system is in equilibrium if no part of it is moving. and CM is at rest Parts are not moving about the CM. A very useful feature: The system should not rotate about ANY axis. We can take the torques about any point we like . Example: Board on scales A 1.00 m long board with a mass of 5.00 kg is supported at each end by a scale. A 10.0kg mass is placed 25.0 cm from the left side. What value does each scale read? L = 1.00 m M = 10.0 kg m = 5.00 kg d = 25.0 cm A. 15.0 kg (L), 0 kg (R) B. 12.5 kg (L), 2.50 kg (R) C. 10.0 kg (L), 5.00 kg (R) D. 7.50 kg (L), 7.50 kg (R) E. 5.00 kg (L), 10.0 kg (R) d = 25.0 cm L = 1.00 m M = 10.0 kg m = 5.00 kg N 1 N 2 mg Mg L /2 We want to solve for (scales read in kg or lb). N g − 2 equations, 2 unknowns 0 τ = ∑ G 0 F = ∑ G 1 2 N N Mg mg + − − = 2 2 L LN dMg mg − − = Let us take the torque about the left end: 2 2 L LN dMg mg − − = d = 25.0 cm L = 1.00 m M = 10.0 kg m = 5.00 kg N 1 N 2 mg Mg L /2 1 2 N N Mg mg + − − = 2 2 N d m M g L ⇓ = + 5.00 kg 25.0 cm (10.0 kg) 100 cm 2 5.00 kg = + = ( ) 5.00 10.0 5.00 kg 10.0 kg = + − = 1 2 N N M m g g ⇓ = + − On the left On the right DEMO: Plank and scales Answer C ACT: How to weight a turkey on a tiny kitchen scale You bought a frozen turkey and forgot how many pounds it was. All you have is a tiny kitchen scale that can weigh it was....
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This note was uploaded on 04/16/2008 for the course PHYSICS 221 taught by Professor Johnson during the Fall '06 term at Iowa State.
 Fall '06
 Johnson
 Physics

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