A block rests on a wedge inclined at angle B. The coefficient of friction between the block and plane is mu (but lets just call this M). The wedge is given horizontal acceleration A. Assuming tan B > M (essentially that the block will slide down), find the minimum and maximum acceleration for the block to remain on the wedge without sliding. I said the horizontal projection on to the surface would be -mA(cos(B)) and it's normal would be mA(sin(B)), thus friction would be MmA(sin(B)). For the block, friction is -Mmg(cos(B)) and force due to gravity would be -mg(sin(B)). I said the minimum would be where the friction of the block due to gravity equals the other forces. So: -Mmg(cos(B))= -mA(cos(B)) + MmA(sin(B)) - mg(sin(B)), which simplifies to: (divide by cos(B) to obtain tan(B)) -Mg = -A + MA*tan(B) - g(tan(B)) (g(tan(B)) - Mg) = A(M*tan(B) - 1) or.. A = g(tan(B) - M)/(M*tan(B) - 1) the book says if I plug in 45 degrees, I should get: A= g(1-M)/(1+M)... which is similar, but I am thinking I may have just gotten lucky or put in a wrong negative sign somewhere.