A uniform 6.0m long ladder of mass 17.0 kg leans against a smooth wall (so the force F_w exerted by the wall, , is perpendicular to the wall). The ladder makes an angle of 22.0 with the vertical wall, and the ground is rough.Determine the coefficient of static friction at the base of the ladder if the ladder is not to slip when a 75.0kg person stands three-fourths of the way up the ladder.
N=normal force from ground
The Attempt at a Solution
I thought that if I take the sum of torques and the sum of the forces I could determine the static friction but I'm stuck.
Assuming that my axis of rotation is the point where the ladder is leaning against the wall
τ= [1.5 x (75)(9.8)(cos(22))] + [3 x (17)(9.8)(cos(22))] + [Nsin(22)]=0
Y-direction: sum of forces
X-direction: sum of forces
I'm stuck because the coefficient of static friction can't be zero. Can somebody please help?