Question

A 320-g ball and a 400-g ball are attached to the two ends of a string that goes over a pulley with a radius of 8.7 cm. Because of friction in its axle, the pulley does not begin to rotate. What is the magnitude of the frictional torque at the axle of the pulley if the system remains at rest when the balls are released?

Answer 1

To solve this problem, it is necessary to apply the concepts related to force described in Newton's second law, so that

F = ma

Where,

m = mass

a = Acceleration (Gravitational acceleration when there is action over the object of the earth)

Torque, as we know, is the force applied at a certain distance, that is,

$\tau = F*d$

Where

F= Force

d = Distance

Our values are given as,

$m_1 = 0.32Kg$

$m_2 = 0.4Kg$

$d = 8.7*10^{-2}m$

Since the system is in equilibrium the difference of the torques is the result of the total Torque applied, that is to say

$\tau = T_2-T_1$

$\tau = F_2*d-F_1*d$

$\tau = m_2g*d-m_1*g*d$

$\tau = (m_2-m_1)g*d$

$\tau = (0.4-0.32)(9.8)(8.7*10^{-2})$

$\tau = 0.068N\cdot m$

Therefore the magnitude of the frictional torque at the axle of the pulley if the system remains at rest when the balls are released is $\tau = 0.068N\cdot m$