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

Question #148784
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)?
1
Expert's answer
2020-12-06T17:25:14-0500


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 mgsinαmgsin\alpha.

Friction force is F=μNF=\mu N,

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

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

mgsinα=μmgcosα,mgsin\alpha=\mu mgcos\alpha,

tanα=μtan\alpha=\mu,

α=arctanμ=arctan0.48=26 degrees.\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 μsN,\mu_sN,

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

So,

Fp=μsmgcosα=0.621009.81cos26°=546.7NF_p=\mu_s mgcos\alpha=0.62\cdot100\cdot9.81\cdot \cos26\degree=546.7 N

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



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment