Question

In what direction should the thread be pulled to minimise the force required to hold the cylinder? What is the magnitude of this force?

• A. $3mg/10$ parallel to incline
• B. vertical $3mg/8$
• C. horizontal $mg/5$
• D. $3mg/5$ perpendicular to incline

Answer 1

The correct option is A $3mg/10$ parallel to incline

Let force is applied at angle $\theta$ as shown with the incline again $f=F$

$\therefore Fcos\theta +f=mgsin\alpha$

$Fcos\theta +f=\frac{3mg}{5}$

$⟹f=\frac{3mg}{5(1+cos\theta )}$

$f$ is minimum when $cos\theta =1$

$⟹\theta =0^o$

$f_{min}=\frac{3mg}{10}$ parallel to the incline