# What is the angular range for theta, for which the masses will not move in the following diagram?

Nov 21, 2016

First considering the situation when the block A tends to move downward along the inclined plane.In this situation the force of friction as well as weight of block B together will balance the downward force on block A.
Taking acceleration due to gravity $g = 9.8 m {s}^{-} 2$

So

$30 g \times \sin \theta - {\mu}_{s} \times 30 g \times \cos \theta = 20 g$

$\implies 30 \sin \theta - {\mu}_{s} 30 \cos \theta = 20$

$\implies 30 \sin \theta - 0.3 \times 30 \cos \theta = 20$

$\implies 10 \sin \theta - 3 \cos \theta = \frac{20}{3}$

$\implies \frac{10}{\sqrt{109}} \sin \theta - \frac{3}{\sqrt{109}} \cos \theta = \frac{20}{3 \sqrt{109}}$

Let 10/sqrt109=cosalpha and 3/sqrt109=sinalpha#

This means $\tan \alpha = 0.3 \implies \alpha = {\tan}^{-} 1 \left(0.3\right) = {16.7}^{\circ}$

The above equation becomes

$\implies \cos \alpha \sin \theta - \sin \alpha \cos \theta = \frac{20}{3 \sqrt{109}}$

$\implies \sin \left(\theta - \alpha\right) = \frac{20}{3 \sqrt{109}}$

$\implies \left(\theta - \alpha\right) = {\sin}^{-} 1 \left(\frac{20}{3 \sqrt{109}}\right) \approx {39.68}^{\circ}$

$\theta = 39.68 + \alpha = 39.68 + 16.7 = {56.38}^{\circ}$

Again considering the situation when the block A tends to move upward along the inclined plane.In this situation the force of friction as well as downward force on block A together will balance the downward force on block B.

So

$30 g \times \sin \theta + {\mu}_{s} \times 30 g \times \cos \theta = 20 g$

$\implies 30 \sin \theta + {\mu}_{s} 30 \cos \theta = 20$

$\implies 30 \sin \theta + 0.3 \times 30 \cos \theta = 20$

Similar manner we get

$\implies \left(\theta + \alpha\right) = {\sin}^{-} 1 \left(\frac{20}{3 \sqrt{109}}\right) \approx {39.68}^{\circ}$

Here

$\theta = 39.68 - 16.7 = {22.98}^{\circ}$

So when $\text{ } {22.98}^{\circ} \le \theta \le {56.38}^{\circ}$ the combined sytem of blocks will not move.