Finding the angle at which a box starts to slide down a ramp

[joe426]

Homework Statement

A box of mass   4.0 kg is placed at rest on a ramp. The coefficients of static and kinetic
friction are $&  0.35 and $%  0.22, respectively. As the ramp is slowly raised from a
horizontal orientation, find the critical angle above which the box starts to slide.

Homework Equations

ƩF=ma

The Attempt at a Solution

I break ƩF=ma down to

F(gravity) x sin∅ - μ(kinetic) x F(normal) = ma

This is as far as I get before I start to get confused. In order to solve for ∅ I need to find accerlation, which i can't think to do without knowing the angle, and the normal force. I can see that the normal force is equal to mgcos∅.


Thanks for the help

 

[TSny]

Hello, joe426

[ find the critical angle above which the box starts to slide.

Note that you are looking for the maximum angle before the block begins to slide. So, what would the acceleration be? Should the friction force in your equation be static or kinetic?

 

[joe426]

Hello, joe426

Note that you are looking for the maximum angle before the block begins to slide. So, what would the acceleration be? Should the friction force in your equation be static or kinetic?

The acceleration would be 0.

The equation would then be

ƩFx = ma = 0
mgsinθ - μmg = 0
θ = sin^-1 (.35)
θ = 20 degrees

Thank you!

 

[TSny]

The acceleration would be 0.

The equation would then be

ƩFx = ma = 0
mgsinθ - μmg = 0
θ = sin^-1 (.35)
θ = 20 degrees

Yes, a = 0. But, earlier you stated that the normal force would be mgcos∅. Now you have it as just mg.
Which is correct?

 

[joe426]

Yes, a = 0. But, earlier you stated that the normal force would be mgcos∅. Now you have it as just mg.
Which is correct?

I think mgcosθ is correct but now I don't know how I would solve for theta.

mgsinθ - μ_smgcosθ = 0

 

[lendav_rott]

mg( sinθ - μcosθ) = 0
sinθ - μcosθ = 0
μ = sinθ/cosθ = tanθ

θ = arctan μ