Answer to Question #148784 in Mechanics | Relativity for althea alcorin

Question

A wooden incline plane is being built, along which wooden crates of merchandise are to be slid down into the basement of a store.

a.what angle with the horizontal should the incline plane be made if the crates are to slide down at constant speed?

b.With what force must a 100 kg crate be pushed in order to start it sliding down the incline plane if the angle of the incline plane is that the one found in (a)?

Answer 1

a. For crates sliding down at constant speed along the x axis (see figure), the friction force F should be balanced by the projection of the gravity force on the x axis, that equal $mgsin\alpha$.

Friction force is $F=\mu N$,

where $N=mgcos\alpha$ (reaction of the incline plane),

$\mu$ - coefficient of sliding friction, wich for sliding of wood on wood is 0.48.

$mgsin\alpha=\mu mgcos\alpha,$

$tan\alpha=\mu$,

$\alpha=arctan\mu = arctan0.48=26 \space degrees.$

b. To start a 100 kg crate sliding down the incline plane the pushing force should reach the value of static friction force, that is $\mu_sN,$

where $\mu_s$ - coefficient of static friction, wich for wood on wood is 0.62.

So,

$F_p=\mu_s mgcos\alpha=0.62\cdot100\cdot9.81\cdot \cos26\degree=546.7 N$

Answer: a: 26 deg, b: 546.7 N.