Question

A block of mass $10kg$ is placed on rough inclined plane of variable angle $\theta$ and friction coefficient ${\mu }_s={\mu }_k=3/4$. When $\theta$ is ${37}^o$ net reaction force applied by inclined is $\overrightarrow{N_1}$ and when $\theta ={53}^o$ net reaction force applied by inclined is $\overrightarrow{N_2}$, find $\left|\overrightarrow{N_1}\right|-\left|\overrightarrow{N_2}\right|$.

Answer 1

$\mid N_1\mid =mg\cos {37}^{\circ }=mg.\frac{4}{5}$

$\mid N_2\mid =mg\cos {53}^{\circ }=mg.\frac{3}{5}$

$\mid N_1\mid -$$\mid N_2\mid =$$\frac{mg}{5}=\frac{10\times 10}{5}=20N$