One end of a uniform meter stick is placed against a vertical wall. The other en

Question

One end of a uniform meter stick is placed against a vertical wall. The other end is held by a lightweight cord that makes an angle theta with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.35. Let the angle between the cord and the stick is theta = 11(degree). A block of the same weight as the meter stick is suspended from the stick, as shown, at a distance x from the wall. What is the minimum value of x for which the stick will remain in equilibrium?

Explanation / Answer

Let angle between wall and stick be and tension in cord be T.

Taking moments about contact point with wall,

W*x + W*(L/2)*Sin = TCos( + )*(LSin) - TSin( + )*(LCos)

W*x + W*(L/2)*Sin = -T*L* Sin............(1)

Balancing horizontal direction forces, R = TSin( + )

Friction force F = *R = 0.35*TSin( + )

Balancing vertical direction forces,

W + W = F + TCos( + )

2W = 0.35*TSin( + ) + TCos( + )

T = 2W / [0.35*Sin( + ) + Cos( + )]

Putting it in (1), W*x + W*(L/2)*Sin = -2W / [0.35*Sin( + ) + Cos( + )] *L* Sin

Putting = 11 deg and L = 1 m we get,

W*x + W/2*Sin = -0.381618*W / [0.35*Sin( + 11) + Cos( + 11)]

x + 1/2*Sin = -0.381618 / [0.35*Sin( + 11) + Cos( + 11)]

Xmin = 0.0956 m

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