ch06-p029

# ch06-p029 - 29. First, we check to see if the bodies start...

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f s ,max = μ s F N = (0.56) (78 N) = 44 N. Since the magnitude f of the force of friction that holds the bodies motionless is less than f s ,max the bodies remain at rest. The acceleration is zero. 29. First, we check to see if the bodies start to move. We assume they remain at rest and compute the force of (static) friction which holds them there, and compare its magnitude with the maximum value s F N . The free-body diagrams are shown below. T is the magnitude of the tension force of the string, f is the magnitude of the force of friction on body A , F N is the magnitude of the normal force of the plane on body A , mg A G is the force of gravity on body A (with magnitude W A = 102 N), and B G is the force of gravity on body B (with magnitude W B = 32 N). θ = 40° is the angle of incline. We are told the direction of G f but we assume it is downhill. If we obtain a negative result for f , then we know the force is actually up the plane. (a) For A we take the + x to be uphill and + y to be in the direction of the normal force. The x and y

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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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ch06-p029 - 29. First, we check to see if the bodies start...

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