A uniform steel beam of length 4.8 m has a mass of 510 kg. One end of the beam i

Question

A uniform steel beam of length 4.8 m has a mass of 510 kg. One end of the beam is bolted to a vertical wall. The beam is held in a horizontal position by a cable attached between the other end of the beam and a point on the wall. The cable makes an angle of 25 degrees above the horizontal. A load whose mass is 250 kg is hung from the beam at a point that is 2.9 m from the wall. A)Find the magnitude of the tension in the supporting cable? B)Find the magnitude of the force R exerted on the end of the beam by the bolt that attaches it to the wall.

Explanation / Answer

Take summation of moments about the bolt. (4.8)(T)(sin 25) - W1(2.4) - W2(2.9) = 0 where T = tension in the supporting cable W1 = weight of the uniform steel beam = 4.5 x 10^3 N W2 = weight of the load on the beam w1=4998 w2=2450 Substituting values, (4.8)(T)(0.4226) - 4998*(2.4) - 2450*(2.9) = 0 T = 9416.01 Newtons **************************************… Let Fy = vertical component of bolt force Summation of vertical forces = 0 Fy + T(sin 25) = W1 + W2 Fy = W1 + W2 - T(sin 25) Fy = 4998 + 2450 - ( 9416.01)(0.4226) Fy = 3468.79 N Let Fx = horizontal component of bolt force Summation of horizontal forces, Fx = T(cos 25) Fx = 9416.01(0.9063) Fx = 8533.72 N Fb = Resultant of force exerted by bolt = sqrt(Fy^2 + Fx^2) Fb = sqrt(3468.79^2 + 8533.72^2) Fb =9211.77N

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