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Unformatted text preview: 60° 6 kg 100 N Solution: Free body diagram: f F v F h mg N a F v = 100 sin 60° = 100 sin 60° F h = 100 cos 60° Vertical: F v = mg + N Solving for N gives N = F v mg Horizontal: F h – f = ma but f = µ k N = k (F v – mg) Substituting expressions for F v into equation for f , followed by substitution of f and F v into equation for horizontal motion gives: 100 cos 60°  k ( 100 sin 60° – mg) = ma Solving for k gives: mg ma k − − = o o 60 sin 100 60 cos 100 = 0.28...
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This note was uploaded on 09/25/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, Force, Work

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