Question

A mass of $1kg$ is just able to slide down the slope of an inclined rough surface when the angle of inclination is ${60}^o$ . The minimum force necessary to pull the mass up the inclined plane $(g=10m{s}^{-2})$ is

• A. 14.14 N
• B. 17.32 N
• C. 10 N
• D. 16.66 N
• E. 0.866 N

Answer 1

Answer: (B)

The component of weight mg of block along the inclined plane $=mgsin\theta$ .
The minimum frictional force to be overcome is also
$mgsin$ $\theta$ . To make the block just move up the plane the minimum force applied must overcome the component $mgsin\theta$ of gravitational force as well as the frictional force
$mgsin\theta =2mg$ $\sin \theta$
$\therefore$ Force needed $=2mgsin\theta$
$=2\times 1\times 10\times \sin {60}^o$
$=20\times \frac{\sqrt{3}}{2}$ $=17.32N$