The question: A plank, of length L = 3.4 m and mass M = 22.0 kg, rests on the ground and on a frictionless roller at the top of a wall of height h = 1.60 m (see the figure). The center of gravity of the plank is at its center. The plank remains in equilibrium for any value of θ >= 59° but slips if θ < 59°. a.Calculate the magnitude in newtons of the force exerted by the roller on the plank when θ = 59°. b.Calculate the magnitude in newtons of the normal force exerted by ground on the plank when θ = 59°. c.Calculate the magnitude in newtons of the friction force between the ground and the plank when θ = 59°. image: http://smg.photobucket.com/albums/v231/er1smesp00n/?action=view¤t=physics-1.gif so I was just solving it as if it was a normal plank against wall style question Sum of Forces in X direction=Force of roller - frictional force=0 Sum of Forces in Y Direction=Normal(upward force from ground)-mass(gravity)=0 Sum of Torques = Force of roller(height)-mass(gravity)cos(59°)(1.7) I know these formulas are wrong since the Normal force does not equal mass*gravity like it should according to the equations above. How do you solve this problem.