What is the acceleration of a particle sliding down an inclined plane?

[Lavace]

Homework Statement

A tile on a roof becomes loose and slides from rest down the roof. The roof is modeled as a plane inclined at 30 degrees to the horizontal. The co-efficent of friction between the tile and the roof is 0.4. The tile is modeled as a particle of mass 'm'kg.

a) Find the acceleration of the particle as it slides down the roof.

Homework Equations

Fr(Friction Force)=uR
Reaction force from the plane = mgcos(angle)
Fr= 0.4mgcos(angle)
The force of the particle sliding down the plane x = mgsin(angle)
Total force going down the plane is x - Fr=ma

The Attempt at a Solution

R = mgcos(30)
F = uR
Fr= 0.4mgcos(30).

Total force down the plane as it is sldiing down is:
mgsin(30) - 0.4mgcos(30) = ma

I can't solve this, as there are TWO unknowns, then there's the fact I can't find m. I've tried looking thorugh textbooks but no avail.

This is my first post here =) I am attempting to complete as many practice examination questions throughout March and started early instead.

I believe once I answer a), the rest will be easy as well (a needs to be answered for b and c).

Thanks to all!

 

[Doc Al]

Total force down the plane as it is sldiing down is:
mgsin(30) - 0.4mgcos(30) = ma

I can't solve this, as there are TWO unknowns, then there's the fact I can't find m. I've tried looking thorugh textbooks but no avail.

Hint: Divide by m.

 

[radou]

mgsin(30) - 0.4mgcos(30) = ma

I can't solve this, as there are TWO unknowns, then there's the fact I can't find m. I've tried looking thorugh textbooks but no avail.

Are you sure you need to find m?

Edit: too late, and too similar.

 

[Lavace]

Dear god, I feel embarressed.

I have worked it out, THANK YOU.
mgsin30 - 0.4mgcos(30) = ma

All over m

gsin30 - 0.4gcos(30) = a
1.50 ... = a
1.5ms^-2 = a

Thank you all, I can resume studying now :)
Such a speed reply!