Discussion Forum

A wooden box lying at rest on an inclined surface of wet wood is held at static equilibrium by a constant force F applied perpendicular to the incline. If the mass of the box  is 1 kg , the angle of inclination is $30^{0}$  and the coefficient of static  friction between the box and the inclined plane is 0.2, the minimum magnitude of F is (Use g=10 m/s ² )

Answer:

Explanation:

 According to question , we can draw the following  diagram

 $mg \sin \theta= \mu (F+ mg \cos \theta)$

 $F= \frac{mg \sin \theta}{\mu}-mg \cos \theta$

 $F=mg\left[\frac{\sin \theta}{\mu}-\cos \theta\right]$

 Here,

     m=1kg ,g=10 m/s², $\theta=30^{0}$, $\mu=0.2$

 $F=1\times 10\left[\frac{\sin 30}{0.2}-\cos 30^{0}\right]$

     = $10\left[\frac{1}{2\times 0.2}-\frac{\sqrt{3}}{2}\right]$

  = $10\left[\frac{5}{2}-\frac{\sqrt{3}}{2}\right]=5[5-\sqrt{3}]$

   =$5[5-1.732]=16.34 N$