1. The problem statement, all variables and given/known data In Fig. 6-37, blocks A and B have weights of 51 N and 24 N, respectively. (a) Determine the minimum weight of block C to keep A from sliding if μ_s between A and the table is 0.20. (b) Block C suddenly is lifted off A. What is the acceleration of block A if μk between A and the table is 0.10? http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c06/fig06_35.gif 2. Relevant equations 3. The attempt at a solution I started to draw some free body diagrams, but it got rather carried away. There's only a handful of forces involved, so instead, I will analyze block B then block A, giving the forces that are affecting them each. On block B, there are exactly 2 forces. The first force is the force of gravity, call it F_gB, in a downward, negative y-axis, direction. Opposing this force, is the force of tension, call it T, which has equal magnitude but opposite direction. Then we have F_gB = T for both magnitudes. For block A, we have 4 forces. The first force is the T mentioned earlier, caused by the string. It moves in the positive x-direction. We also have the force of gravity, caused by both block A and block C. For that, we'll denote F_g = m_A * g + m_C * g = m_A * g + W_C With W_C being the weight of block C. The normal force F_N is just the opposite of the force of F_g, but point up. The force of friction is moving toward the left, and given the coefficient of friction, we have F_f = mu_k * F_N = mu_k * (m_A*g + W_C) Since block A refuses to move, we have F_net,x = 0 = T - F_f Calculating T and F_f, we have T = m_B * g = 24*9.8 = 235.2 F_f = 0.2*(51*9.8 + W_C) = 99.96 + 0.2*W_C T - F_f = 235.2 - (99.96 + 0.2*W_C) W_C = 135.24 / 0.2 = 676.2 N But this is not correct. I've come up with the same answer twice. My feeling is that messed up on the force of gravity on block A, or the normal force on A.. I'm not completely sure.