The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. If the angle of inclination of the plane is 60^ , then the coefficient of friction is

Step-by-step solution

According to the question, angle of inclination =60^ . Now, force of friction, f= N= m g and net retarding force, (F₁ )=m g +f Net accelerating force down the inclined plane is External force needed (up the inclined plane) to maintain sliding motion is (net retarding force) From Eqs. (i),(ii) and (iii) we get m g + m g =2(m g - m g ) or 3 m g =m g or =3 or =3 ( =60^ ) or 60^ =3 aligned 3 & =3 & = 1 3 aligned When inclination of plane is 60^ then the coefficient of friction, = 1 3 .