A 10.0 m long massless beam is resting perpendicular to a vertical wall while be

Question

A 10.0 m long massless beam is resting perpendicular to a vertical wall while being held in place at the far end by a cable making an angle of 37 degrees with the beam (see below). Also attached to the beam at a distance x = 5.0 m from the wall is a weight of F, 5500 N. While the wall exerts a normal force F on the beam, it also exerts a static friction force (IFJ s u, Fx) on the beam. Find the minimum coefficient of static friction u, so that the beam does not slip. (a) 0.75 (b) 0.48 (c) 0.97 (d) 0.21 (e) 0.33 37.0

Explanation / Answer

Normal force FN = T*cos(37)

where T is the tension in the string

let the axis of rotation is at left end of the beam

then torque acting on the beam about that axis is zero

(T*sin(37)*10) - (5500*5) = 0

T = 4570 N

Net force acing on the beam is zero

T*sin(37) - Fg + fs = 0

(4570*sin(37)) - 5500 + fs = 0

fs = 2750 N

mu_s*FN = 2750

mu_s*T*cos(37) = 2750

mu_s = 2750/(T*cos(37)) = 2750/(4570*cos(37))

mu_s = 0.75

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