In the figure, a uniform plank, with a length L of 4.63 m and a weight of 631 N, rests on the ground and against a frictionless roller at the top of a wall of height h equilibrium for any value of ? = 70.0° or more, but slips if ? 70.00. Find the coefficient of static friction between the plank and the ground. 1.96 m. The plank remains in Roller Number Units the tolerance is +/-296 Solution Given, L = 4.63 m ; W = 631 N ; h = 1.96 m ; theta = 70 deg We know that F(fric) = us N In X direction F(fric) - N1 sin(theta) = 0 In Y direction N2 + N1 cos(theta) - mg = 0 Considering the sum of torques N1 h/sin(theta) - mg L cos(theta)/2 = 0 Eliminiating putting N2 in F = us N, solving for us we get us = L sin^2(theta) cos(theta)/(2h - L sin(theta)cos^2(theta)) us = 4.63 sin^2(70) cos70/(2 x 1.96 - 4.63 sin70 cos^2(70)) = 0.41 Hence, us = 0.41 .

1. In the figure, a uniform plank, with a length L of 4.63 m and a weight of 631 N, rests on the
ground and against a frictionless roller at the top of a wall of height h equilibrium for any value
of ? = 70.0° or more, but slips if ? 70.00. Find the coefficient of static friction between the
plank and the ground. 1.96 m. The plank remains in Roller Number Units the tolerance is +/-296
Solution
Given,
L = 4.63 m ; W = 631 N ; h = 1.96 m ; theta = 70 deg
We know that
F(fric) = us N
In X direction
F(fric) - N1 sin(theta) = 0
In Y direction
N2 + N1 cos(theta) - mg = 0
Considering the sum of torques
N1 h/sin(theta) - mg L cos(theta)/2 = 0
Eliminiating putting N2 in F = us N, solving for us we get
us = L sin^2(theta) cos(theta)/(2h - L sin(theta)cos^2(theta))
us = 4.63 sin^2(70) cos70/(2 x 1.96 - 4.63 sin70 cos^2(70)) = 0.41
Hence, us = 0.41