Question

. A block is placed on an incline. The coefficient of static friction between the block and the plane is 0.59. What is the maximum value for θ, such that the block remains in equilibrium?

Answer 1

Answer:

$\theta = 30.5 degree$

Explanation:

If the block remains in equilibrium on the inclined plane

then we can say that component of weight along the inclined plane is counter balanced by the static friction force.

Here we know that maximum static friction force on the block is given by

$f_s = \mu_s F_n$

here we have

$F_n = mgcos\theta$

now the component of weight along the inclined is balance by maximum static friction force

$\mu_s (mgcos\theta) = mg sin\theta$

$tan\theta = \mu_s$

$tan\theta = 0.59$

$\theta = tan^{-1}(0.59)$

$\theta = 30.5 degree$