# Calculating the coefficient of friction

1. Oct 28, 2009

### phyz

1. The problem statement, all variables and given/known data
A wedge slides down a frictionless inclined plane that makes angle with the horizontal. A small block is placed at the horizontal top side of the wedge. During the slide, the block does not move relative to the wedge. Find the minimum possible coefficient of static friction µs between the block and the wedge.

2. Relevant equations
us=fsmax/Fn uk=Fk/Fn
Fn=mg
Fnet=ma

3. The attempt at a solution
i cannot seem to figure out a way to start.
the only variable that would be given is gravity and i cant seem to derive an equation that would make me find another variable. I drew an fbd but it cannot be displayed here.

Given: g=9.8m/s

2. Oct 29, 2009

### rl.bhat

Hi phyz, welcome to PF.
If you draw the free body diagram, you can see that the force acting on the wedge and the block is
F = (M + m)*g*sinθ along the inclined plane. Here M is the mass of the wedge and m is the mass of the block.
F*cosθ pushes the block in the forward direction and F*sinθ pushes the block in the downward direction.
F*cosθ causes the frictional force between wedge and the block. To prevent the slipping of the block it should be equal to..........?

3. Oct 29, 2009

### ideasrule

The purpose of drawing a FBD is so that you can write out Newton's second law for both the x and y directions. Try doing that. Remember that the acceleration of the block must be the same as that of the wedge (and of the wedge-block system, since the wedge and block can be considered one object).

4. Oct 29, 2009

### phyz

thanks guys i appreciate the help.
rl.bhat i understand the equation u got and that would mean F*cosθ and F*sinθ are both components to equal fk ? and to prevent the slipping of the block it should be equal to fsmax?

how can u solve this equation tho without varaibles other then gravity?

5. Oct 29, 2009

### rl.bhat

Obviously in terms of masses of wedge, block and the angle of inclination. Write the answer in symbols.

6. Oct 30, 2009

### phyz

F=(mw+mb)*g*cosθ is the force going down the ramp. therefore for the wedge to move down the ramp and the block to not slip the F=(mw+mb)*g*cosθ must be equal to the Fsmax of the block.
so....
Fsmax = (mw+mb)*g*cosθ... since fsmax is equal to us*Fn we can sub it in...

us*Fn = (mw+mb)*g*cosθ... since Fn is equal to mb*g we can sub that in...

us*(mb*g)=(mw+mb)*g*cosθ...now we isolate for us and we get our answer...

[us*(mb*g)] / (mb*g) = [(mw+mb)*g*cosθ] / (mb*g)

Finally our answer is... us = [(mw+mb)*cosθ] / mb

i hope thats right? loll :P thanks

7. Oct 30, 2009

### rl.bhat

Fsmax = (mw+mb)*g*cosθ... since fsmax is equal to us*Fn we can sub it in...
It should be
Fsmax = (mw+mb)*g*sinθ*cosθ... since fsmax is equal to us*Fn we can sub it in...
Since wedge is moving down ward, fn = mb*g - (mw+mb)*g*sinθ*sinθ.

Last edited: Oct 30, 2009