Painting hung by a wire suported by two nails

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A 10 kg painting is hung using a wire supported by two nails, with a wire length of 1.3 m and a horizontal distance of 1.2 m between the attachment points. The discussion focuses on calculating the tension in the wire and the force exerted on each nail, emphasizing the importance of understanding force diagrams and vector addition. It is noted that the lengths of the wires are less significant than the angles formed, which dictate the tension distribution. The weight of the painting acts through its center, and the vertical components of tension are used to balance the gravitational force. The problem is resolved by recognizing that the horizontal components of tension cancel each other out.
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1. A 10 Kg painting is hung with a wire supported by two nails. The length of the wire is 1.3m. A) what is the tension in the wire? B) what is the magnitude of the force exerted on each nail by the wire (assume that the tension is the same in each part of the wire).

In the diagram you are given the distance between the far left and far right points on the painting to which the wire is attached: which is 1.2 m.

the wire is symetric and there is a distance of 0.4m on the x-axis between each point. I suppose a picture of the wire kind of represents a 2 dimensional pyramid with its top cut off and no bottom; if that helps.

2. Lots of vector manipulation...
3. This should be a pretty easy problem but I was having difficulty conceptualizing the force diagrams and the individual forces on each point, namely, how does the length of the wire relate to the tension as well as how do I distribute the total weight of the painting as a gravity force at each point. Once I know this it should just be a simple matter of vector addition.
 
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SteveM- said:
1. A 10 Kg painting is hung with a wire supported by two nails. The length of the wire is 1.3m. A) what is the tension in the wire? B) what is the magnitude of the force exerted on each nail by the wire (assume that the tension is the same in each part of the wire).

In the diagram you are given the distance between the far left and far right points on the painting to which the wire is attached: which is 1.2 m.

the wire is symetric and there is a distance of 0.4m on the x-axis between each point. I suppose a picture of the wire kind of represents a 2 dimensional pyramid with its top cut off and no bottom; if that helps.

This should be a pretty easy problem but I was having difficulty conceptualizing the force diagrams and the individual forces on each point, namely, how does the length of the wire relate to the tension as well as how do I distribute the total weight of the painting as a gravity force at each point. Once I know this it should just be a simple matter of vector addition.


Hi SteveM-! Welcome to PF! :smile:

As you've probably worked out, the lengths of the wires don't matter, only the angles do.

Just work them out using geometry.

The weight of the painting is through its centre. Just use vertical components of the tension in the usual way. :wink:
 
Ok, thanks that makes sense. The answer I got basically said that all the tension in the wires is in the negative y direction because the x direction components are pointing opposite and cancel out. This problem is done.
 
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