1. The problem statement, all variables and given/known data Two triangular wedges are placed next to each other on a flat floor. Both triangles have two 45 degree angles where they touch the floor and a 90 degree angle up in the air. The Triangular wedges have a mass of m. A cube of mass M is balanced on the wedges so it's weight is even distributed over both of them. The coefficient of static friction is less than one. What is the largest M that can be balanced without the wedges moving? If this isn't making sense then it's problem #6 on this http://www.physics.ohio-state.edu/undergrad/greStuff/exam_GR9277.pdf 2. Relevant equations Frictional force = Normal force * Coefficient of friction Normal Force = mgcos(theta) *cos(45) = sqrt(2)/2 3. The attempt at a solution So my idea was to make the force of the big block onto the little block ((M/2)*g*(sqrt(2)/2)) equal the static friction force. The normal force acting up from the bottom of the small block i believe would be ((M/2)g + mg). So i set ((M/2)*g*(sqrt(2)/2)) = ((M/2)g + mg)μ. The answer i'm looking for is ((2μm) / (1-μ)) and i must have set it up wrong because I don't see that happening.