Question

A 0.31 kg cart on a horizontal frictionless track is attached to a string. The string passes over a disk-shaped pulley of mass 0.08 kg and radius 0.012 m and moves without slipping. The string is pulled vertically downward with a constant force of 1.1 N. Find (a) the tension in the string between the pulley and the cart.

Answer 1

To solve this problem it is necessary to apply the concepts related to Newton's second law and its derived expressions for angular and linear movements.

Our values are given by,

$M_{cart} = 0.31kg\\m_{pulley} = 0.08kg\\r_{pulley} = 0.012m\\F = 1.1N$

If we carry out summation of Torques on the pulley we will have to,

$F_2*d-F_1*d = I \alpha$

Where,

I = Inertia moment

$\alpha =$Angular acceleration, which is equal in linear terms to a/r (acceleration and radius)

The moment of inertia for this object is given as

$I = \frac{1}{2} mr^2$

Replacing this equations we have know that

$(F_2 - F_1)d = (\frac{1}{2}(m_{pulley})r^2) (\frac{a}{r})$

$F_2 - F_1 = \frac{1}{2}m_{pulley} \frac{F_1}{M_{cart}}$

$F_2 = (1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))F_1$

Or

$F_1 = \frac{F_2}{(1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))}$

Replacing our values we have that

$F_1 = \frac{1.1}{(1 + (0.5)(\frac{0.08}{0.31}))}$

$F_1 = 0.974 N$

Therefore the tension in the string between the pulley and the cart is  0.974 N