A uniform beam 3.55 long and weighing 2550 carries a 3850 weight 1.50 from the f

Question

A uniform beam 3.55 long and weighing 2550 carries a 3850 weight 1.50 from the far end, as shown in the figure below (Figure 1) . It is supported horizontally by a hinge at the wall and a metal wire at the far end. A) How strong does the wire have to be? That is, what is the minimum tension it must be able to support without breaking? B) What are the horizontal component of the force that the hinge exerts on the beam? C) What are the vertical component of the force that the hinge exerts on the beam?

Explanation / Answer

1.Take moments about the hinge and get T= 5812.5 Newtons. 2. Once you get that you can sum forces in the y direction to get the y force at the hinge. 3487.5 - 2300 = 1187.5 Newtons upward Then do the x direction. That would be (4/5)T= 4,650 Newtons away from the wall for the x force at the hinge. 2.. Suppose the vertical wall is on left hand side and a uniform beam AD is hinged at 'A' to vertical wall and held horizontally by a 5m cable BE attached at point B on the beam and to the wall at point E, 3.0m above the hinge. The force that the hinge exerts on the beam = F Tension in the cableBE = T The beam AD also carries 800N weight at the far end D. Weight of beam= W=1500 N, Weight of beam acts at mid point C of beam Point E on the wall is 3 m above point A Point B on the beam is 4 m from point A Point C is mid point of beam AD and C is 4.5 m from point A ABE is a right angle triangle, 3 m long side AE on wall and cable BE is hypotenuse of length 5m Suppose cable BE makes angle O with beam AD From point A drop perpendicular AF on cable BE Perpendicular distance of cable BE from point A = AF Angle ABE is angle O sinO = 3/5 = 0.6 cosO = 4/5 = 0.8 AF /AB=sinO=0.6 AF =AB*0.6=4*0.6 =2.4 m AF = 2.4 m As beam is rotational equilibrium,torque about any point must be zero. Taking torque about point A on wall through which force F exerted by the hing passes,the torque of F is zero Torque of tension = T*AF=2.4 T..........anticlockwise Torque of weight of beam = 4.5*1500=6750 Nm ...clockwise Torque of weight at far end = 9*800=7200 N clockwise As beam is rotational equilibrium; anticlockwise torque = clockwise torque 2.4T = 6750+7200 =13950 Tension = T =13950 /2.4 =5812.5 N 1 The minimum tension in the cable to support without breaking should be 5812.5 N

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