I could use any help you can give with this physics problem.

Question

A uniform ladder 10.0 m long and weighing 60 N rests against a frictionless vertical wall as shown in the figure. If the ladder is just on the verge of slipping when it makes 50 degree angle with ground. Set the conditions of static equilibrium for the ladder.

a.) set the conditions of static equilibrium for the ladder.

b.) What is the force from the wall?

c.) What is the coefficient of static frictional force of the ground?

Answer 1

Consider the weight 60N of the ladder to be centered at its midpoint. In the following I found it easier to decide between sine and cosine by looking at limits as the angle increases or decreases, rather than by geometry. Let F_w be the normal force of the wall on the top end of the ladder, and let pivot point be at the bottom.

a) Sum of torques Στ = F_w • 10 • sin 50° - 60 • 5 • cos 50° = 0.

Sum horizontal forces at bottom: ΣF_x = 60 • cos 50° - f = 0.

Sum vertical forces at bottom: ΣF_y = F_N - 60 • sin 50° = 0.

b) F_w • 10 • sin 50° = 300 • cos 50°, F_w = 30 cot 50° = 25.2 N.

c) F_N = 60 sin 50° = 46.0 N. Friction force f = 60 cos 50° = 38.6 N = μ_s F_N. Therefore μ_s = f / F_N = 0.84.