Solving the Friction of a Crate on a 20° Plane

[laxboi33]

Homework Statement

A crate is on a 20 degree plane where the coefficients of static and kinetic friction are .45 and .35 respectively. What is the acceleration of the crate?

Homework Equations

Mk= .35
Ms= .45
F= sin20 * mg (I think this is the net force?)

The Attempt at a Solution

F - Fk= M*A

(sin20 * mg) - (.35 * mg) = m * a

a = sin20 * g - .35 * g

a= .34 * 9.8 - .35 * 9.8

a = -.098


The answer in the book says 0. As I was writing this problem I think I figured it out. If sin 20 is the net force, then that would make net force less than the static friction which would give it 0 acceleration. Maybe I just miscalculated the net force. If so could someone explain to me what net force would be for this equation? Thanks, guys.

 

[RoyalCat]

You presupposed that the crate slides down the incline.

You must first check to see if it starts moving at all!

The static friction force is a reaction force. It will always oppose the total sum of the forces parallel to the surface it acts from.

The only force in the direction of the incline, other than friction, is as you've pointed out, mg\sin{\theta}

Now, the force of static friction tries to oppose this. Can it? What is the maximum force that the static friction can oppose? Is the total force it has to oppose less than, or more than this limit?

 

[laxboi33]

thanks cat