This is a problem from Mechanics by Kleppner and Kolenkow 1. The problem statement, all variables and given/known data Two identical masses M are pivoted at each end of a massless pole of length L. The pole is held leaning against frictionless surfaces at angle θ, as shown, and then released. Find the initial acceleration of each mass. Code (Text): Note: | o | \ | \ | \ | \ |_____o____ 2. Relevant equations y^2+x^2=L^2 yy''+(y')^2+xx''+(x')^2=0, if the masses are at initially at rest then yy''+xx''=0, x'' = (-y/x)y'' = -tan(θ) y''. The forces that act on the upper mass is the weight (Mg), normal force (Ny) , and force of the pole (F) so the equations are: My'' = Mg-Fsin(θ) and Ny = Fcos(θ) (No use, can be omitted) The forces that act on the lower mass is the weight (Mg), normal force (Nx), force of the pole (F) and friction (f) so the equations are: Mx'' = Fcos(θ)-f and Nx-Fsin(θ)-Mg=0 3. The attempt at a solution By manipulation, I ended up getting y'' = (-g)/(1+µtan(θ)+(tan(θ))^2) and so x'' can easily be derived. My question here is that K&K didn't include Fsin(θ) in the second equation of the lower mass. This will result to Nx=Mg which then leads to a different answer. Why did K&K exclude Fsin(θ) knowing that it is a force from the pole that contributes to the downward force for the lower mass?