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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.