Question

A block of mass $1\text{kg}$ is placed on a rough inclined plane at an angle $\theta ={60}^{\circ }$ with the horizontal and the block is connected with a string as shown in figure. If coefficient of friction between the block and inclined surface, ${\mu }_s=3/4$, find the tension in the string. [Take $g=10{\text{m/s}}^2$]

• A. $0.66\text{N}$
• B. $4.9\text{N}$
• C. $5\text{N}$
• D. $0\text{N}$

Answer 1

The correct option is B $4.9\text{N}$


As block is placed on an inclined plane, block has a tendency to slide, therefore static friction comes into play and opposes the tendency of motion. The component of $mg$ along the plane will be balanced by friction and tension force.

From the FBD,
$N=mg\cos \theta$
$\Rightarrow f_{max}={\mu }_{s}N={\mu }_{s}mg\cos \theta$
$=\frac{3}{4}\times 1\times 10\times \frac{1}{2}$
$=3.75\text{N}$
and $mg\sin \theta =1\times 10\times \frac{\sqrt{3}}{2}=5\sqrt{3}\text{N}$

Therefore the value of tension will be the difference of weight component and the maximum frictional force
$mg\sin \theta =f_{max}+T$
$\Rightarrow 5\sqrt{3}=3.75+T$
$\Rightarrow T=5\sqrt{3}-3.75$
$=4.9\text{N}$