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