Question

In the figure, a uniform plank, with a length L of 5.19 m and a weight of 586 N, rests on the ground and against a frictionless roller at the top of a wall of height h = 1.89 m. The plank remains in equilibrium for any value of θ = 70.0° or more, but slips if θ < 70.0°. Find the coefficient of static friction between the plank and the ground.

Answer 1

lenth of ladder below the roller = 1.89/ sin 70 degree = 2.011 m

horizontal force of F =the frictional force
or F*cos 20 degree = μN

Balancing forces in vertical,
F*sin 20 degree +N - W = 0

or N = W - F*sin 20 degree

therefore μ = F*cos 20 degree / (W - F*sin 20 degree)

torque about the ground,
F*2.011 - W*L/2*cos 70 degree = 0

so F = W*L/2*cos 70 degree/2.11 = 586*5.19/2*cos 70 degree /2.11 = 246.5 N

So, μ =246.5*cos 20 degree /(586- 246.5 sin 20 degree)

= 0.462 Answer