Question

A body of mass $10\text{kg}$ is lying on a rough plane inclined at an angle of ${30}^{\circ }$ to the horizontal and the co-efficient of friction is $0.5$. The minimum force required to pull the body up the plane is ($g=9.8{\text{m/s}}^2$)

• A. $914\text{N}$
• B. $91.4\text{N}$
• C. $9.14\text{N}$
• D. $0.914\text{N}$

Answer 1

The correct option is B $91.4\text{N}$

By the free body diagram of the block,

$N$ is the normal force and $f$ is the frictional force.
$f=\mu N=0.5\times 10\times 9.8\times \frac{\sqrt{3}}{2}=24.5\sqrt{3}$, where $N=10g\cos {30}^{\text{o}}$.

Now, for the body to move up the incline plane the friction force will also act downward, as the tendency of motion of the body will be up the incline plane.

Hence the minimum force to pull the body up the incline plane will be,
$F_{\text{min}}=(mg\sin \theta +f)$
$\Rightarrow F_{\text{min}}=10\times 9.8\times \frac{1}{2}+24.5\sqrt{3}$
$=91.4\text{N}$

Hence option B is the correct answer.