Question

A solid cylinder of mass m is kept in balance on a fixed incline of angle $\alpha ={37}^{\circ }$ with the help of a thread fastened to its jacket. The cylinder does not slip.
In what direction should the thread be pulled to a minimize the force required to hold the cylinder? What is the magnitude of this force?

• A. $\frac{3mg}{10}$ parallel to incline
• B. vertical $\frac{3mg}{8}$
• C. horizontal $\frac{3mg}{10}$
• D. None of these

Answer 1

The correct option is B vertical $\frac{3mg}{8}$

Rotational equilibrium:$Fr=fr⟹F=f⟹\left(1\right)$

Translational equilibrium:Horizontal direction

$Nsin\alpha =fcos\alpha ⟹N=fcos\alpha sin\alpha ⟹\left(2\right)$

Vertical direction:$F+Ncos\alpha +fsin\alpha =mg$

$F+\frac{Fcos2\alpha }{sin\alpha }+Fsin\alpha =mg$

Solving we get$F(1+\frac{1}{sin\alpha })=mg.$where $sin\alpha =\frac{3}{5}$

$\therefore F(1+\frac{5}{3})=mg⟹F=\frac{3mg}{8}$