Question

In the arrangement shown $\theta$ is varied between $0$ to $\frac{\pi }{2}.$ ( If $f_c:$ contact force, $N$ : normal Reaction, $f$ : friction), then which of the following variations are correctly shown

• A.
• B.
• C.
• D.

Answer 1

Answer: (B), (D)

Till $\theta$ reaches the angle of repose, friction force is equal to applied external forece.

$f=mg\sin \theta ,N=mg\cos \theta$

$f_c=\sqrt{f^2+N^2}$

$f_c=mg$

For $\theta$ exceeding angle of repose, kinetic friction starts acting on the block, which is given by

$f_k={\mu }_{k}N$

$f=\mu mg\cos \theta ,N=mg\cos \theta$

$f_c=\left(\sqrt{1+{\mu }^2}\right)mg\cos \theta$

Hence, option $\left(A\right)$ and $\left(D\right)$ are the correct choices.