Question

a block of mass 1kg is stationery with respect to conveyor belt that is accelerating with 1m/s^2 upwards at an angle 30 degree
what is the frictional force on the block

Answer 1

Question 32553

If block is stationary with respect to conveyor which is accelerating with $a = 1\frac{m}{s^2}$ , then in laboratory frame of reference the block is moving upwards with the same acceleration. In this case, the friction force is causing the object to accelerate.

For motion over inclined plane, projection on axis, which is perpendicular to the surface of the plane gives: $N = mg\cos \varphi$ , $\varphi$ is the angle of inclination.

Projection over axis, parallel to the plane, gives:

$m a = F_{f} - mg\sin \varphi$ . $F_{f}$ is the friction force, which is calculated as $F_{f} = \mu N = \mu mg\cos \varphi$ ( $\mu$ is the friction coefficient one needs to find).

Hence, $m a = \mu m g \cos \varphi - m g \sin \varphi \Rightarrow \mu = m \frac{(a + g \sin \varphi)}{m g \cos \varphi} = \frac{a + g \sin \varphi}{g \cos \varphi} = 0.68$ .