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Unformatted text preview: get x:T A f = 0 y: N m A g = 0 For block B, the translational equilibrium condition yields T B m B g = 0 For the joint we have x: Tsin T A = 0 y: Tcos T B = 0 b) What are the tension in the cord attached to block B and that attached to block A ? Solving block Bs balance of forces equation for T B , we get T B = m B g = 58.8 N Solving the two balance of forces equation for the joint simultaneously, we obtain T = T B / cos T A = Tsin = (T B / cos)sin = T B tan = 27.4 N c) What is the coefficient of static friction between block A and the surface below it? We solve the balance of forces equation for block A to obtain the normal force and the maximum static friction. N = m A g = 78.4 N f = T A = 27.4 N By definition, the coefficient of static friction is the ratio of the maximum static friction to the normal force, thus we have s = f/N = 0.349...
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This note was uploaded on 01/15/2012 for the course PHY 2048 taught by Professor Field during the Fall '08 term at University of Florida.
 Fall '08
 Field
 Physics, Work

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