qz7sol_6839sp10 - get x:T A f = 0 y: N m A g = 0 For block...

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TA: Tomoyuki Nakayama Tuesday, March 23 rd , 2010 PHY 2048: Physic 1, Discussion Section 6839 Quiz 7 (Homework Set # 10) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ In the figure below right, block A (mass 8.00 kg) is in equilibrium, but it would slip if block B (mass 6.00 kg) were any heavier. Angle θ is 25.0º. a) Draw all the forces exerted on the block A , block B and joint of the three cords, and write down the translational equilibrium condition or conditions (balance of forces equations) for each of them. As shown in the figure, four forces are exerted on block A, two forces are exerted on block B and three forces are exerted on the joint. Applying translational equilibrium conditions to block A, we
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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.

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