1. The problem statement, all variables and given/known data Consider the following system: A mass 4m is suspended with a massless string from a massless pulley wheel. On the other end of the string is another massless pulley wheel with a mass 3m suspended from a massless string on it, and a mass m suspended on the other side. Find the acceleration of the mass 4m. 2. Relevant equations Use the Lagrangian formulation. L = T - U d/dt (dL/dq*) = dL/dq q* is the time derivative of q. dL/dq* is the generalized momentum of q. dL/dq is the generalized force of q. Let X be the distance from the the 1st pulley wheel to the mass 4m measured down. Let X' be the distance from the 1st pulley wheel to the 2nd pulley wheel measured down. Let Y be the distance from the 2nd pulley wheel to the mass 3m measured down. Let Y' be the distance from the 2nd pulley wheel to the mass m measured down. 3. The attempt at a solution I already know the answer is not zero. However I keep getting zero. Constraints: length of the 1st rope is a constant K = X+X'. length of 2nd rope is a constant C = Y+Y' X = K - X' X* = -X'* Y = C - Y' Y* = Y'* T = 1/2 (4m) (X*)^2 + 1/2(3m)(Y*)^2 + 1/2(m)(Y*)^2 Kinetic energy is the sum of the kinetic energy of the 3 masses. U = -4mgX - 3mg(X'+Y) - mg(X'+Y') X is the distance from 1st pulley wheel to the mass 4m measured down. As X increases, potential energy decreases as expected. Same for the mass 3m. X'+Y is the total distance of 3m from the 1st pulley wheel and as this increases the potential decreases. Same for mass m. U = -4mgX - 3mg(K-X+Y) - mg (K-X+(C-Y)) = -4mgX - 3mgK + 3mgX - 3mgY - Kmg + mgX - Cmg + Ymg U = 0 mgX - 4mgK - Cmg -2 mgY. L = T + U dL/dx = 0 d/dt(dL/dx*) = m d2/dt2(x) = 0. This says there's no acceleration of the mass 4m which is associated with the x direction. That's incorrect. Why?