Chapter 6 Hmwk. 4 Phy 101 2009

# Chapter 6 Hmwk. 4 Phy 101 2009 - Solutions to Homework#4...

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Solutions to Homework #4 Chapter 6 32. Picture the Problem: The forces exerted on the left and right blocks are depicted at right. Strategy: Because the pulley is ideal, the tension in the string is equal to the weight of the hanging block. This can be verified by Newton’s Second Law in the vertical direction for the hanging block: . Write Newton’s Second Law along the direction parallel to the incline for the 6.7 kg block and substitute into the resulting equation to find m. Solution: 1. Write Newton’s Second Law along the direction parallel to the incline: 2. Substitute into the resulting equation to find m. Insight: A larger m is required if the angle θ is increased. If it is increased all the way to θ =90°, the large mass will be hanging straight down and the mass m required to maintain equilibrium would be 6.7 kg. 40. Picture the Problem: Refer to the figure at right: Strategy: Write Newton’s Second Law for each block and add the equations to eliminate the unknown tension T . Solve the resulting equation for the acceleration a , and use the acceleration to find the tension. Let x be positive in the direction of each mass’s motion, m 1 be the mass on the table, and m 2 be the hanging mass. Solution: 1. (a) The tension in the string is less than the weight of the hanging mass. If it were equal to the weight, the hanging mass would not accelerate. 2. (b) Write Newton’s Second Law for each block and add the equations: 3. Solve the resulting equation for a: 4. (c) Use the first equation to find T:

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Insight: Note that the blocks move as if they were a single block of mass 6.30 kg under the influence of a force equal to The tension in the string would be zero if m 2 fell freely, 27.5 N if m 2 (and the entire system) were at rest. 46. Picture the Problem: The forces acting on the car are
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Chapter 6 Hmwk. 4 Phy 101 2009 - Solutions to Homework#4...

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