The masses attached to each side of an ideal Atwood\'s machine consist of a stac

Question

The masses attached to each side of an ideal Atwood's machine consist of a stack of five washers, each of mass m, as shown in the figure below. The tension in the light string is T0. When one of the washers is removed from the left side, the remaining washers accelerate and the tension decreases by 0.290 N. (Assume that the pulley is massless and frictionless.) (a) Find m. ? g (b) Find the new tension and the acceleration of each mass when a second washer is removed from the left side. new acceleration ? m/s2 new tension ? N

Explanation / Answer

(a):

T0 = 5mg N

let T1 be the Tension after washer is removed from left

therefore,

T1 = 4mg + 4 ma

T1 + 5ma = 5mg

therefore

4mg + 4ma + 5ma = 5mg

a = g/9

therefore

T1 = (40/9)mg

according to ques.

T0 - T1 = (3/10)

so,

5mg + (40/9)mg = (3/10)

m = 27 / (50*g)

m = 0.0551 kg

(b):

let T2 be tension when 2'nd washer is removed from left

T2 = 3mg + 3ma

T2 + 5ma = 5mg

therefore,

a = g/4 = 2.45 m/s^2

so,

T2 = 4mg = 4*0.0551*9.8 = 2.1599 N

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