Two bodies of masses `m_1 and m_2` are connected by a light string going over a smooth lilght pulley at the end of an incline. The `mass_1` lies on the incline `m_2` hangs vertically. The system is t rest. Find the angle of the incline and the fore exerted by the incline on the body of mass `m_1.

Options

A. `sintheta=2m_2/m_1` , `N=m_1g cos theta`
B. `sintheta=m_2/m_1` , `N=m_1g cos theta`
C. `sintheta=m_2/m_1` , `N=m_1g sin theta`
D. `sintheta=m_2/m_1` , `N=4m_1g cos theta`

Answer

B

Explanation

ure shows the situation with the forces onn `m_1 and m_2` shown. Take the body of mass `m_2` as the system. The forces acting on it are

i. `m_2` g vertically downward (by the earth),
ii. T along the string up the incline (by the string),
N normal to the incline (by the incline).
As the string and the pulley are all light and smooth, the tension in the string is uniform everwhere. Hence, same T is used for the equations of `m_1 and m_2`.As the system is in equilibrium, these forces should add to zero.
Take components parallel to the incline,
`T=m_1 gcos (pi/2-theta)=m_1 g sin theta `.........ii
Taking comonents along the normal to the incline,
Taking components along the normal to the incline,
`N=m_1g cos theta` ............iii
ElimiN/Ating T form i and ii
`m_2g=m_1gsin theta`
or, `sintheta=m_2/m_1`
`giving ` theta= sin^-1 (m_2/m_1)`
From iii `N=m_1gsqrt(1- (m_2/m_1)2)`.