Question

A force of $100 \,N$ is applied on a block of mass $3\, kg$ as shown in the figure. The coefficient of friction between the surface and the block is $\mu=\frac{1}{\sqrt{3}}$ The frictional force acting on the block is

• A. $15\, N$ downwards
• B. $25\,N$ upwards
• C. $20\, N$ downwards
• D. $30\,N$ upwards

Answer 1

Answer: (C)

$N'=100\,cos\,30^{\circ}=100 \cdot \frac{\sqrt{3}}{2}=50\sqrt{3}N$
Net driving force $=\left(100\times\frac{1}{2}-30\right)$
$= 20\, N$ (upward)
$f_{\text{limiting}} \frac{1}{\sqrt{3}}\left(50\sqrt{3}\right)=50\,N$
$\therefore $ friction $= 20 \,N$ (downward)