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## Homework Statement

There are two wedges, each with mass m, placed on a flat floor. A cube of mass M is balanced on the wedges. Assume there is no friction between wedge and block, and the coefficient of static friction is less than 1 between the floor and wedges. What is the largest M that can be balanced without moving?

diagram here:

http://sites.google.com/site/learningphys/

## Homework Equations

## The Attempt at a Solution

If we split the block in half due to symmetry, the question will get easier.

force of friction is given by:

[tex]F=\mu (\frac{M}{2}+m)g[/tex]

the M/2 comes from taking only half of the block

i want to equate the force that the block has on one of the wedge.

Still considering half of the diagram, i want the normal force of (M/2)g to the wedge's surface. The normal force is give by the usual inclined plane equation, mgcos(theta)

So normal force of (M/2) is:

[tex]N=\frac{Mg}{2}cos(\theta)[/tex]

this normal force pushes in the opposite direction of the force of friction... but we have to consider the horizontal component, which tags on a sin(theta) term

I finally get:

[tex]\mu (\frac{M}{2}+m)g=\frac{Mg}{2}cos(\theta)sin(\theta)[/tex]

And then i solve for mu. How close am I to being correct?