A 10 kg object rests on a 30 degrees...

Question

A 10 kg object rests on a 30 degrees ramp which has a coefficient of friction of 0.20.

a) What will the object do?

b) What would a 50 kg mass do?

Answer 1

Miguel,

I hope you've already found your answer. But incase you haven't, here it is.

 

Friction force is found by the equation F_f = μN, coefficient of friction times the normal force.

 

The normal force, in this case, is created by the vector portion of gravity which is perpendicular to the ramp surface.

We calculate that by the equation N = Wcosθ where Θ = 30º

 

The force that wants to make the object slide down the ramp is found by F_p = WsinΘ

 

When F_p exceeds F_f, the object will slide. Otherwise, it sits there.

 

a)  N = 10(9.81)cos30 = 84.96 N

     F_f = .2(84.96) = 17 N

     F_p = 10(9.81)sin30 = 49 N    ∴ It will slide down the ramp.

 

b)  N = 50(9.81)cos30 = 424.8 N

     F_f = .2(424.8) = 85 N

     F_p = 50(9.81)sin30 = 245 N   ∴ It will slide down the ramp.

 

If you look a little closer, you will see that the vector components of a vertical force on a 30º ramp will always have a normal/parallel ratio of √3 /1, or about 1.7/1.  But with μ = .2, that reverses the ratio at 1/5.  The mass of the object will never change the outcome. The angle and coefficient of friction are the controlling factors.