A block of mass M is pulled on a smooth horizontal table by a string making an angle `theta` with the horizontal as shown in figure. If the acceleration of the block is a find the force applied by the string and by the table N on the block.

Options

A. `T=(Ma)/(tantheta)` , `N=(Mg-Ma tan theta)`
B. `T=(Mg)/(costheta)` , `N=(Mg-Ma tan theta)`
C. `T=(Ma)/(costheta)` , `N=(Mg-Ma cos theta)`
D. `T=(Ma)/(costheta)` , `N=(Mg-Ma tan theta)`

Answer

D

Explanation

Let us consider the block as the system.
The forces on the block are
a. pull of the earth, Mg, vertically downward,
contact force by the table , vertically upward,
pull of the string T, along the string.
The free body diagram for the block is shownin ure.
The acceleration of the block is horizontal and towards the right. Take this direction as the X-axis and vertically upward direction as the Y-axis. We have,


component of Mg along the X-axis =0
component of N along the X-axis =0
component of LT along the X-axis = T `costheta`
Hence the total force along the X-axis = `T cos theta`
Using Newton's law, `T costheta=Ma`
Component of Mg along the Y-axis =-Mg
component of `N along the Y-axis =N`
Component of T along the Y-axis `=T sin theta`
Total force along the Y-axis `= N+T sin theta-Mg`
Using Newton's law `N+T sintheta- Mg=0` ...........ii
From equation i. `T=(Ma)/(costheta)`. Putting this in equation ii. `N=(Mg-Ma tan theta)`