Particle on a slope with friction question

[Jackelkes95]

Hi guys! I am really struggling with this poorly written question. Any help would be appreciated!
a 3.5 kg box is on a slope. What is the minimum angle which will cause the box to slip if μ (friction) =0.6? If the plane is tilted to an angle of θ above this slipping value what is the acceleration on the box when θ is 10 degrees and -10 degrees (below). How long will it take the box to stop if θ = -10 degrees with an initial velocity of 2 m/s and with θ = 10 degrees how far will the box slide in that time starting from rest?

 

[PeroK]

Can you try the first part about finding the minimum angle?

 

[Svein]

This is standard mechanics. m*g is straight down, the friction is dependent on the force normal to the slope and the accelerating force is parallel to the slope.

 

[Jackelkes95]

If i was given the angle i can easily calculate the forces parallel and perpendicular to the slope ie mgcosθ for perpendicular resultant force. I just haven't a clue how to go about this question! Dynamics are not my forte.

 

[Svein]

ie mgcosθ for perpendicular resultant force.

Yes - just continue...

 

[haruspex]

mgcosθ for perpendicular resultant

Be careful about reusing symbols with different meanings. I assume here you mean theta as the angle of an arbitrary slope, but in the question it is the additional angle of the slope beyond a critical angle.

 

[dean barry]

I can get you started.
The point at which it breaks stasis is when the forces up and down the incline are equal.
So:
Gravitational force = friction force
( m * g * sine ( incline angle ) ) = ( m * g * cosine ( incline angle ) * friction coefficient )
Transpose for friction coefficient