Physics: A0.31 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.

A. Set A is showing the same motion in both graphs!...

A0.31 kg cart on a horizontal frictionless track

Ответ:

uhhgray

10.01.2022

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

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

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

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

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

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

0,0(0 оценок)

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Physics

Ask an AI advisor a question