How Do You Calculate Tension in Each Wire for a Hanging Painting?

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Homework Help Overview

The problem involves calculating the tension in two wires supporting a painting of mass 3.20 kg. The wires connect the painting to hooks on a wall, and the setup includes specific measurements for the length of the wires and the distance between the hooks.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of tension, considering the symmetry of the setup and whether to divide the total tension by two at different stages of the problem. Questions arise about the role of angles and the implications of symmetry in the tension calculations.

Discussion Status

Participants are actively engaging with the problem, clarifying the relationship between combined tension and individual wire tension. Some guidance has been offered regarding the mathematical approach, particularly in how to handle the symmetry of the situation.

Contextual Notes

There is mention of a separate problem involving a non-symmetrical arrangement, indicating that participants are considering different scenarios in their reasoning.

kristen151027
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I'm working on the following question:

"A painting of mass 3.20 kg hangs on a wall. Two thin pieces of wire, each 0.250 m long, connect the painting's center to two hooks in the wall. The hooks are at the same height and are 0.330 m apart. When the painting hangs straight on the wall, how much tension is in each piece of wire? (It is assumed that the wire is massless)"

I found the angle between one of the angles and the horizontal line connected the two hooks (48.7 degrees). Then, I used the weight (mg = 31.36) and the angle to calculate the tension in the rope (41.7 N).

I know that the answer is half of that, but at what point should I divide by two in order to find the tension in each piece? My guess is that I divide the final answer by two. I guess I'm unsure of whether tension is the same throughout the whole wire.
 
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Since they specifically refer to two pieces of wire, that's how you should treat it. You know (by symmetry) that each piece of wire has the same tension. Since two wires pull up on the painting, that force that you calculated is really the combined tension of both wires.
 
So I can solve for the combined tension and then divide that answer by 2 (since they're symmetrical), or should I divide earlier in the problem? (I have another problem that involves a non-symmetrical arrangement, so I'm trying to figure out when to make the distinction.)
 
Since you've solved for the combined tension, just divide by 2. Mathematically, what you are doing is this:
[tex]T\sin\theta + T\sin\theta = mg[/tex]

or:
[tex]2T\sin\theta = mg[/tex]
 
Okay thanks, that makes sense. I was just getting confused about why T is multiplied by [tex]sin\theta[/tex]. I was solving it a different way and didn't see that. Thanks :smile:
 

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