Question

A 250 kg crate is placed on an adjustable inclined plane. If the crate slides down the incline with an acceleration of 0.7 m/s2 when the incline angle is 25ø, then what should the incline angle be for the crate to slide down the plane at constant speed?

Answer 1

Answer:

21.2 degrees

Explanation:

Let gravitational acceleration constant g = 9.81 m/s2 directed downward. We can calculate the g component that is parallel to the 25 degree incline:

$gsin25^0 = 9.81sin25^0 = 4.15 m/s^2$

This parallel component would produce an acceleration of the block. But since the net acceleration is only 0.7 m/s2, there's a friction acceleration that hinders g parallel. This acceleration can be calculated by

$4.15 - a_f = a = 0.7$

$a_f = 4.15 - 0.7 = 3.45 m/s^2$

The force of friction would be

$F_f = a_fm = 250 * 3.45 = 816.5 N$

Friction is a product of normal force and its coefficient

$F_f = \mu N = \mu mgcos25^0 = \mu 250*9.81*cos25^0 = 2222.72 \mu$

$\mu = F / 2222.72 = 816.5 / 2222.72 = 0.388$

For the crate to slide down at constant speed, the net acceleration must be 0. In other words, the parallel g must be the same as acceleration caused by friction. Let this incline angle be α

$g sin\alpha = a_f = F_f/m = \frac{\mu N}{m} = \frac{\mu mgcos\alpha}{m} = \mu g cos\alpha$

$sin\alpha = \mu cos\alpha$

$\frac{sin \alpha}{cos \alpha} = \mu$

$tan\alpha = \mu = 0.388$

$\alpha = tan^{-1} 0.388 = 0.37 rad = 0.37*180/\pi \approx 21.2^0$