1111ProblemSet07_solns.pdf

# 1111ProblemSet07_solns.pdf - Physics 1111 Fall 2017 PS#7...

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Physics 1111 Fall 2017 PS #7 Solutions Page 1 of 7 Physics 1111 Problem Set #7 Solutions (1) Atwood Machine: (a) The textbook example mentioned gives a handy formula for the acceleration of the masses in an Atwood’s Machine: a = m 1 - m 2 m 1 + m 2 g. Let’s say that both masses are increased by the same factor f ; that is, m 1 fm 1 and m 2 fm 2 . Then the acceleration for this new pair of masses is a 0 = fm 1 - fm 2 fm 1 + fm 2 g = f ( m 1 - m 2 ) f ( m 1 + m 2 ) g = m 1 - m 2 m 1 + m 2 g = a. So the acceleration is unchanged. (b) The same textbook example provides another formula for the tension in the rope: T = m 1 g + m 1 a = m 1 ( g + a ) . In part (a) we showed that the acceleration doesn’t change when the masses are in- creased by the same factor. However, the above expression shows that the tension is proportional to the mass m 1 , so the tension increases. (Actually, the tension could also be thought of as proportional to m 2 .) (c) Now instead we add the same amount m to each mass: m 1 m 1 + m and m 2 m 2 + m . In this case, the acceleration becomes a 0 = m 1 + m - ( m 2 + m ) m 1 + m + m 2 + m g = m 1 - m 2 m 1 + m 2 + 2 m g. Because the denominator is larger, this new acceleration a 0 is smaller than the original. So the acceleration decreases. (2) Pulley Problem with Friction: We start this problem the same way as any other, by identifying the forces acting on each object. The hang- ing block ( m 2 ) has only gravitational and tension forces acting on it, both vertical. The block on the table feels tension from the rope, friction, gravity, a normal force, and the pulling force. Let’s choose coordinate systems for the two blocks that “follow the rope”, so that the accelerations of m 1 and m 2 are both in the same ( - x ) direction. Copyright c 2017 University of Georgia. Unauthorized duplication or distribution prohibited.

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Physics 1111 Fall 2017 PS #7 Solutions Page 2 of 7 N1 F g m 1 F ’ T T F g 2 m F θ x y y x f k From our free-body diagrams, we can write down the force components in each direction for each block: X F 1 x = - F cos θ + F T + f k 1 X F 1 y = F sin θ + F N 1 - m 1 g X F 2 x = m 2 g - F 0 T X F 2 y = 0 . Now let’s apply Newton’s Second Law. There’s no ac- celeration in what we’re calling the y direction for either block, so the second equation above becomes X F 1 y = 0 = F N 1 = m 1 g - F sin θ. We need this normal force because the force of kinetic friction uses it.
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