As shown in the figure below, a uniform beam is supported by a cable at one end

Question

As shown in the figure below, a uniform beam is supported by a cable at one end and the force of friction at the other end. The cable makes an angle of = 30°, the length of the beam is L = 1.75 m, the coefficient of static friction between the wall and the beam is s = 0.480, and the weight of the beam is represented by w. Determine the minimum distance x from point A at which an additional weight 2w (twice the weight of the rod) can be hung without causing the rod to slip at point A. x = m

Please show all work. Thank you.

Explanation / Answer

Horizontally, for tension T and normal force on beam Fn, we've got

Fn = Tcos30º = 0.866T

We know that at the threshold, Ff = µ*Fn = 0.480 * 0.866T = 0.4157T

where Ff is the friction force at the beam

Vertically, we've got Ff + Tsin30º =0. 4157T + 0.5T = 0.9157T = w + 2w = 3w

so T = 3.2762w

Finally, consider the moment about the left end of the beam.

It must be zero, or the beam would rotate.

M = 0 = Tsin30º * L - w*L/2 - 2w*x = 3.2762w * 0.5 * L - w*L/2 - 2w*x

w cancels from all terms:

3.2762* 0.5 * L - L/2 - 2x

1.1381L = 2x

x = 0.56905L

Since L = 1.75m,

x = 0.9958 m

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