A uniform steel beam of length 5.00m has a weight of4.50*10 3 N. One end of the

Question

A uniform steel beam of length 5.00m has a weight of4.50*103 N. One end of the beam is bolted to avertical wall. The beam is held in a horizontal position by acable attached between the other end of the beam and a point on thewall. The cable makes an angle of 25.0o above thehorizontal. A load whose weight is 12.0*103 N ishung from the beam at a point that is 3.50m from the wall. Find a) the magnitude of the tension in the supporting cable and b)the magnitude of the force exerted on the end of the beam by thebolt that attaches the beam to the wall.

Explanation / Answer

This is exactly like the crane problem... same process: . horizontal forces:   Rx   - T cos25 = 0 .    vertical forces:     Ry   + T sin25   - weight of beam - weight of load = 0 . Torques... use wall end of beam for reference: .        torquescounterclockwise = torques clockwise .             T* L * sin25     =    weight of beam * (L/2)   +   weight of load *3.5 .           T *5.00 * sin25   =    4500 * 5.00 /2      +    1200 *3.5 .     T sin25  =    2250 +   840 .    T = 3090 / sin25 =   7312 Newtons     isthe tension .    Rx = T cos25 = 7312 cos25 =    6627 .    Ry = 4500 + 1200 - 7312 sin25 =   2610 . Magnitude = (66272 + 26102 )1/2   =    7122 Newtons    is the force from the bolt .

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