Question

A body of mass $400\text{g}$ slides on a rough horizontal surface. If the frictional force is $3.0\text{N}$, find the angle made by the contact force on the body with the vertical. Take $g=10{\text{m/s}}^2$

• A. ${\tan }^{-1}\left(\frac{4}{3}\right)$
• B. ${\tan }^{-1}\left(\frac{3}{4}\right)$
• C. ${\tan }^{-1}\left(\frac{3}{5}\right)$
• D. ${\tan }^{-1}\left(\frac{4}{5}\right)$

Answer 1

The correct option is B ${\tan }^{-1}\left(\frac{3}{4}\right)$

Let the contact force on the block by the surface be $F$ which makes an angle $\theta$ with the vertical.


The component of $F$ perpendicular to the contact surface is the normal force $N$ and the component of $F$ parallel to the surface is the frictional force $f$. As the surface is horizontal, $N$ is vertically upwards. For vertical equilibrium,
$N=Mg=(0.400\text{kg}\times (10{\text{m/s}}^2)=4.0\text{N}$
The frictional force is $f=3.0\text{N}$
$\tan \theta =\frac{f}{N}=\frac{3}{4}$
$\theta ={\tan }^{-1}\left(\frac{3}{4}\right)$.