# Inclined plane and friction

## 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:

## 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:
$$F=\mu (\frac{M}{2}+m)g$$
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:

$$N=\frac{Mg}{2}cos(\theta)$$

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:

$$\mu (\frac{M}{2}+m)g=\frac{Mg}{2}cos(\theta)sin(\theta)$$

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

tiny-tim
Homework Helper
Assume there is no friction between wedge and block,

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:

$$N=\frac{Mg}{2}cos(\theta)$$
Hi learning_phys!

No … consider the forces on the block

there is no friction, so the only forces are Mg/2 and N …

the horizontal components balance anyway …

so resolve vertically, and that gives you … ?

are you sayin N=Mg/2? but the normal force should be perpendicular to the blocks surface right? (that's why I have the cosine term)

tiny-tim
Homework Helper
but the normal force should be perpendicular to the blocks surface right?
Yes (because there is no friction).
are you sayin N=Mg/2?
No … resolve vertically.

tiny-tim
Homework Helper
ok, i get N=Mg/(2cos(theta))
??
That's it!

awesome

so I would get:
$$\mu (\frac{M}{2}+m)g=\frac{Mg}{2cos(\theta)}sin(\theta )$$

tiny-tim
$$\mu (\frac{M}{2}+m)g=\frac{Mg}{2cos(\theta)}sin(\theta )$$