Answer:
1.022 x 103 N.m
Step-by-step 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