(25) Statics - Lecture 25 Statics Equilibrium A system is...

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Lecture 25 Statics
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Conditions for equilibrium: 0 and 0 F τ = = r r Equilibrium Equilibrium A system is in equilibrium if no part of it is moving. and CM is at rest Parts arenot moving about the CM. A very useful feature: Thesystem should not rotateabout ANY axis. We can takethe torques about any point we like .
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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 10.0-kg mass is placed 25.0 cm from the left side. What value does each scaleread? 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)
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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 t o solve f or (scales read in kg or lb). N g - 2 equations, 2 unknowns 0 τ = r 0 F = r 1 2 0 N N Mg mg + - - = 2 0 2 L LN dMg mg - - = Let us takethe torqueabout the left end:
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2 0 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 0 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 = + = ( 29 5.00 10.0 5.00 kg 10.0 kg = + - = 1 2 N N M m g g = + - On theleft On the right DEMO: Plank and scales Answer C
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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 scalethat can weigh a maximum of 2 lb. But here’s a good trick you can useto get an estimation… turkey Light rod scale d L
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