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 = (1)/(2) mr^2`

Replacing this equations we have know that

`(F_2 - F_1)d = ((1)/(2)(m_(pulley))r^2) ((a)/(r))`

`F_2 - F_1 = (1)/(2)m_(pulley) (F_1)/(M_(cart))`

`F_2 = (1+(1)/(2)((m_(pulley))/(M_(cart))))F_1`

Or

`F_1 = (F_2)/((1+(1)/(2)((m_(pulley))/(M_(cart)))))`

Replacing our values we have that

`F_1 = (1.1)/((1 + (0.5)((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