A uniform beam 3.65m long and weighing 2700Ncarries a 3050N weight 1.50m from th

Question

A uniform beam 3.65m long and weighing 2700Ncarries a 3050N 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.

How strong does the wire have to be? That is, what is the minimum tension it must be able to support without breaking? (N)

What are the horizontal component of the force that the hinge exerts on the beam?(N)

What are the vertical component of the force that the hinge exerts on the beam? (N)

Explanation / Answer

I am guessing the 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(-2700n), therefore;

Sum of torque at the hinge;

0=-2700X2-3050(1.50)+(tension from the cable)X3.65meters

solve for the tension in the cable and you get
t= .... N upward
This is the minimum amount the cable has to support.
The sum of the forces in the y direction is zero.


Fy=..........

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