"The coefficients of friction are u_s = 0.40 (static) and u_k = 0.30 (kinetic) between all surfaces of contact. Determine the force P for which motion of the 60-lb block is impending if cable AB (A) is attached as shown, (B) is removed." (sorry the image is blurry, i took it on my camera phone) Force P acts to the left, the top block (we will call block A) weighs 40 lb and the bottom (block B) weighs 60 lb. Friction = Coefficient of Friction * Normal Fnet = 0 _____ I decided to do part B first, since it seems easier. My free body diagram of block B looks like this: Leftward force, P Downward weight, W_B = 60lb Downward force of A on B, F_A = 40lb Upward normal, N_(A+B) = 100lb Rightward force of static friction = (0.40) * 100 = 40lb Rightward force of A on B, due to the force of friction acting on A = (0.40) * 40 = 16lb If this is all correct, then a simple sum of the forces shows that P = 40 + 16 = 56lb. *NOW* Part A has got me a bit confused. We add the rope and pulley system, and now we appear to have another force acting on block B... the question is, what is this force, and where does it come from? Unless i'm mistaken, the other forces acting on block B (the ones already listed above) do not change. If we look at the top block on it's own, then in order for it to stay motionless, the force of the rope must be equal to the force of friction which acts to the left on block A, and equals 16 lb. Therefore, the tension in the rope acting to the right on block B must also equal 16lb... but this seems strange to me, because that means that there are TWO 16-lb forces acting to the right on block B, in addition to the rightward 40-lb force due to friction. If this is all correct, then force P in case (A) would equal 16+16+40 = 72lb to the left. But i'm just not sure if that's right. Any help would be appreciated, and if my explanation is at all confusing, i'll be happy to attempt to clarify... Also, if anyone knows how to directly put the image into the post, i'd appreciate it.