Consider two blocks connected by an ideal rope that passes through an ideal pulley fixed on the corner of the wedge as shown in the figure. The blocks have masses m1 and m2. The block of mass m1 is on the incline surface (which is frictionless) while the block of mass m2 hangs up in the air. Find the angle θ at which the two blocks are in equilibrium (i.e., not moving).. 1. The problem statement, all variables and given/known data N = Normal =m1g T = m2g 2. Relevant equations f = ma 3. The attempt at a solution x - direction N sin θ = T cos θ m1g sin θ = m2g cos θ cos θ = (m1/m2)sin θ y - direction N cos θ + T sin θ = m1g m1g cos θ + m2g sin θ = m1g substitute cos θ = (m1/m2)sin θ (m21 / m2) sin θ + m2 sin θ = m1 sin θ = m21 m2 / m1g + m2 θ = sin-1(m21 m2)/ m1g + m2 But the angle is actually something else according to the answer key.