Question

a body of mass M starts sliding down an inclined plane with inclination theta ,such that tan theta = 1/3 the force acting on the body down the plane in this position is

Answer 1

tan theta=1/3 so

sin theta=1/$\sqrt{10}$

cos theta=3/$\sqrt{10}$

then total downward force=Mgsin theta

that is M*g*1/$\sqrt{10}$

Answer 2

Answer:  Force acting on the body is $Force = mg/\sqrt{10}$

Given:

          $tan \Theta = 1/3$

Solution:

A body carrying a mass will be sliding down an inclined plane based on the inclination angle and the friction. But we don’t have any friction value here. Therefore,

On applying Pythagoras theorem, we get the other side value as \sqrt10

IMAGE

           $Sin \Theta = opposite/ hypotenuse = 1/\sqrt {10}$

           $Force\quad F = mg sin\Theta = mg (1/\sqrt {10})$

           $Force = mg/\sqrt{10}$