A force of 1200 N is applied to a drum of radius 80 cm which has a 12 kg mass, m1, attached by a cord
wound on the drum. A second drum of radius 41 cm is bolted on the first and has a 39 kg mass, m2, attached
by a cord wound on it. What is the resulting torque applied to the system?

1.022 x 103 N.m

Explanation:

Solution

Given:

The weight of the block of mass m₂ is :

w₂ = m₂*g

Where

w₂ = 39 x 9.8 = 382.2 N

Then,

The weight of the block of mass m₁

w₁= m₁*g;

so,

w₁ = 12 x 9.8 = 117.5 N

Thus,

The tension wrapped in cord on drum (80 cm) T₁ = F - w₁

Now,

T₁ = 1200 - 117.5

T₁ = 1082.5 N

The tension  wrapped in the cord on drum (41 cm) T₂ = w₂;

T₂ = 382.2 N

Hence,

We calculate net torque on the center of the drum:

The net torque = T₁ x 0.8 + T₂ x 0.41;

= 1082.5 x 0.8 + 382.2 x 0.41;

= 1.022 x 103 N.m

Therefore, the resulting torque applied to the system is 1.022 x 103 N.m

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