Question

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