Question

A block of mass $m$ is placed over rough inclined plane having inclination $30^{\circ }$ . The co-efficient of friction between the block and the inclined plane is $0.75$ . The contact force on the block is

• A. $\frac{3}{4}mg$
• B. $2mg$
• C. $\frac{5}{4}mg$
• D. $mg$

Answer 1

Answer: (D)

$mgsin\theta =\frac{\sqrt{3}}{2}mg=f$
$f_l=\mu mgcos\theta$
$=\frac{3}{4}mg\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{8}mg$
$f<f_r$
$f=mgsin\theta$
$N=mgcos\theta$
$f_{net}=mg\sqrt{si{n}^2\theta +co{s}^2\theta }$
$=mg$