How Is the Tension in the Right Support Wire Calculated?

[pkossak]

Two wires support a beam of length L=20 m and mass 330
kg as shown in the figure above. A box of mass 220 kg hangs
from a wire which hangs from the beam a distance x=15 m
away from the left edge of the beam. What is the tension in
the RIGHT support wire? (in N)

I just did an (at least I thought...) similar problem to this, and got the answer correct. For the life of me, I don't know why this would be different. This is what I've been doing..

center of beam = 0 m, so

10 m*(330/2 kg * 9.81) + 5 m*(220 kg * 9.81) = 10 m* x N

the only difference is that the weight of the beam wasn't given in the question I got correct, and I was solving for the weight of box. why would this be different?

 

[Astronuc]

You are correct that the beam contributes a tension of (330/2 kg * 9.81 m/s²) on each wire suspending the beam.

Check your moments. The sum of the moments is zero (statics). Look at the moments about the left side (left pivot).

Also think about superposition. The beam is evenly distributed by the wires, the 220 kg box is not.

 

[thepatientmental]

First of all, great job on solving the previous problem correctly! It's always good to review and understand the concepts before moving on to more complex problems like this one.

In this case, the weight of the beam is given as 330 kg, so we need to take that into account when calculating the tension in the right support wire. The key concept here is that the beam itself is being supported by two wires, so the weight of the beam is being distributed between the two wires.

To solve this problem, we need to use the principle of moments, which states that the sum of the moments on an object must equal zero for it to be in equilibrium. In this case, we can say that the sum of the moments on the beam must equal zero, since it is not rotating or moving.

So, using the principle of moments, we can set up the equation:

(10 m * 330 kg * 9.81 m/s^2) + (5 m * 220 kg * 9.81 m/s^2) - (15 m * T) = 0

Where T is the tension in the right support wire. Solving for T, we get:

T = (10 m * 330 kg * 9.81 m/s^2) + (5 m * 220 kg * 9.81 m/s^2) / 15 m

T = 2589 N

So, the tension in the right support wire is 2589 N. It may seem counterintuitive that the weight of the beam affects the tension in the right support wire, but this is due to the distribution of weight between the two wires. I hope this helps clarify the concept for you!