There are 3 blocks on a table. One of the blocks, block C, is in the middle of the table with mass 2M. There are two pulleys on the edges of the tables. The pulley on the left is attached to block C and is attached to block A (which has a mass of M) such that it is hanging off of the pulley and is along the side of the table. The pulley on the right is also attached to block C and is attached to block B (Which has a mass of 2M) such that it is hanging off of the pulley and is along the right side of the table. They are released from rest and then accelerate with a mag of 0.500 m/s^2. What is the coefficient of kinetic friction (Uk) between the sliding block (block C, I am assuming - since it is attached on both sides by the rope holding the other two blocks that are hanging from the pulley) and the table? I had 3 free body diagrams two were vertical (blocks A and B) and one was horizontal (block C) Ta | * | Mg Tb | * | 2Mg Ta--*--Tb (with forces of friction going towards the left since the right side has a block with mass 2M) ** also from the block there is a N force up and the Fg force down the block Then I had: A) Ta - Mg = ma = 0 Tb = ma + mg B) Tb - 2Mg = 2Ma = 0 Tb = 2Ma + 2Mg C) Tb - Ta - f = 2Ma Tb - Ta - UkN = 2Ma 2Ma + 2Mg - Ma - Mg - UkMg = 2Ma Uk = a/g + 1 Uk = 0.5/9.8 + 1 = 1.05102 Is my process correct?