A uniform beam 3.55m long and weighing 2550N carries a 3850N weight 1.50m from t

Question

A uniform beam 3.55m long and weighing 2550N carries a 3850N weight 1.50m 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

In this casemy assumption is cable goes straight up.
If this is true there is no tension in the x direction (horizontal direction).
From the hinge, the sum of the torque equals zero.Because the center of gravity is at 2 meters, the torque from the beam is ;
2mX(-2500n),
Sum of torque at the hinge;
0=-2500X2-3500(1.50)+(T)X4 meters
Here T is the tension in cable in vertical direction
T= 25625 N upward
This is the minimum amount the cable has to support.
The sum of the wt is 4000N down. The sum of the forces in the y direction is zero.
0= 21875-3500-2500+force at the hinge in the y direction(fy)
Fy=3437.5 N in upward direction
Answer

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