A sign for the local cheese shop is hung by two strings of length 0.25 meters at

Question

A sign for the local cheese shop is hung by two strings of length 0.25 meters at either end from a 4 meter long pole with a mass of 200 kg. The sign is 2 meters long, has a height of 0.5 meters, and a mass of 700 kg. The cheese shop owners would like to go back to eating their cheese, so rather than try to screw the pole into the wall, they decide they will let the force of friction between the wall and the end of the pole do the work of keeping up the sign. They tie a string to the far right end of the pole, and find that in order to get the pole to stay in place they need to tie the top of the string at most a distance of 0.75 meters above the end of the pole. If the left end of the sign is 1 meter from the wall, what is the coefficient of static friction between the wall and the pole?

Explanation / Answer

Let tension in string = T

Angle between string and pole = atan (0.75 / 4) = 10.619 deg

Vertical component of T = T*sin10.619

Horizontal component of T = T*cos10.619

Balancing moments:

(T*sin10.619)*4 = (200*9.81)*(4/2) + (700*9.81)*(1 + 2/2)

T = 23955.7 N

Balancing forces in vertical direction,

Fv + 23955.7*sin10.619 = (200 + 700)*9.81

Fv = 4414.5 N

Balancing forces in horizontal direction,

Fh = 23955.7*cos10.619

Fh = 23545.5 N

Friction force = u*Fh

u*23545.5 = Fv = 4414.5

u = 0.187

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