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- Thread starter Mr Davis 97
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Well, not always, as your example illustrates. You can certainly draw anI know that the objects that we work with in introductory physics are point particles.

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So it's not possible to represent the situation I am talking about with a point-like free-body diagram?Well, not always, as your example illustrates. You can certainly draw anextendedfree body diagram where the points of application of the forces are indicated and the distances between them are shown.

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It would be somewhat odd to represent the picture as a point object. Why would you want to? (Depending upon your purpose, the location of the forces may not matter.)So it's not possible to represent the situation I am talking about with a point-like free-body diagram?

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Well I am trying to find at which angle the wires approaches makes with the horizontal makes the tension approach a maximum. It would seem I would have to use Newton's 2nd law on a point particle to find out.It would be somewhat odd to represent the picture as a point object. Why would you want to? (Depending upon your purpose, the location of the forces may not matter.)

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Don't get hung up on the diagram. Newton's 2nd law applies to extended objects just as it does to point particles.Well I am trying to find at which angle the wires approaches makes with the horizontal makes the tension approach a maximum. It would seem I would have to use Newton's 2nd law on a point particle to find out.

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Don't get hung up on the diagram. Newton's 2nd law applies to extended objects just as it does to point particles.

So if I were to apply F = ma to the system I have described, it would be something like ##\displaystyle \sum \vec{F} = \vec{T} + \vec{T} + \vec{W} = \vec{0}##, which would mean ##\displaystyle T = \frac{mg}{2\sin \theta}##?

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In general, if an extended object is acted upon by forces applied at different points of the object, and is in equilibrium, the particle model fails, and you have to satisfy conditions for translational equilibrium (sum of forces = 90), and rotational equilibrium (sum of torques = 0).So if I were to apply F = ma to the system I have described, it would be something like ##\displaystyle \sum \vec{F} = \vec{T} + \vec{T} + \vec{W} = \vec{0}##, which would mean ##\displaystyle T = \frac{mg}{2\sin \theta}##?

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Sure. (Assuming symmetry, of course.)So if I were to apply F = ma to the system I have described, it would be something like ##\displaystyle \sum \vec{F} = \vec{T} + \vec{T} + \vec{W} = \vec{0}##, which would mean ##\displaystyle T = \frac{mg}{2\sin \theta}##?

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