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This is a three part question..Any help would be appreciated! A brick, with mass

ID: 1750082 • Letter: T

Question

This is a three part question..Any help would be appreciated!

A brick, with mass m = 30 kg, is suspended by a light string from apulley with mass M = 60 kg. Treat the pulley as a uniformdisk of radius .1 m. The brick is released from rest andfalls to the floor below. Assume that the massless stringdoes not slip on the pulley.

A) As the brick is falling, what is its downward acceleration?
10, 7.5, 5, 2.5, or 0

B) As the brick is falling, what is the tension in the string?
600, 300, 150, 75, 0

C) What is the magnitude of the torque acting on the pulley?
60, 6, 30, 3, 1.5

Explanation / Answer

You have to use sum of forces on the brick: .             mg - T = ma . And sum of torques on the pulley: .            Tr =   I             where    I = (1/2) M r2  and    = a / r .    So you get, for thepulley:       Tr = (1/2)M r2 * a /r          or .           T = M a / 2 . Now put that into the first equationand...          mg - Ma/2 = ma .           a = mg / (m + M/2) =   30 * 9.8 /(30 + 60/2)   =   4.9m/s2 ˜    5 .      T = Ma/2 =   60 * 4.9 / 2   =    147Newtons    ˜   150 .     torque = Tr = 147 *0.1 =   14.7 N-m   ˜    15 They do not give   15   as an option forpart c, but my answer is correct. It looks like they made amistake in giving you the choices they offered. They do not give   15   as an option forpart c, but my answer is correct. It looks like they made amistake in giving you the choices they offered.
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