Representing Free body diagrams

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SUMMARY

This discussion focuses on representing tension forces in free-body diagrams for extended objects, specifically a picture frame suspended by a wire. Participants emphasize that while introductory physics often treats objects as point particles, it is possible to create extended free-body diagrams that indicate the points of force application and their distances. The application of Newton's 2nd law is confirmed to be valid for both point particles and extended objects, with the equation for tension derived as T = mg/(2sin θ) under equilibrium conditions.

PREREQUISITES
  • Understanding of Newton's 2nd Law
  • Familiarity with free-body diagrams
  • Knowledge of equilibrium conditions for extended objects
  • Basic trigonometry for analyzing angles and forces
NEXT STEPS
  • Study the application of Newton's 2nd Law to extended objects in physics.
  • Learn how to construct extended free-body diagrams for complex systems.
  • Explore the concepts of translational and rotational equilibrium in detail.
  • Investigate the role of symmetry in simplifying physics problems involving forces.
USEFUL FOR

Physics students, educators, and anyone interested in understanding the complexities of force representation in mechanics, particularly in scenarios involving extended objects and tension forces.

Mr Davis 97
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I have a simple question. I know that the objects that we work with in introductory physics are point particles. Thus, say we have a picture frame that is put on a wall. There is a wire holding it up, where the wire is attached to the top right and the top left corners of the frame. The passes through a nail so that the whole picture frame stays up. the wire makes a triangular shape with the picture frame. My question is how do we represent the tension forces on the frame in a free-body diagram if the picture frame is not a point particle, since the tension forces are acting on the corners of the frame.
 
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You can reduce the picture frame to a beam between the two points where the wire is attached. Symmetry allows to ignore rotations of the beam, so you can simplify the problem sufficiently to get back to point-like objects.
 
Mr Davis 97 said:
I know that the objects that we work with in introductory physics are point particles.
Well, not always, as your example illustrates. You can certainly draw an extended free body diagram where the points of application of the forces are indicated and the distances between them are shown.
 
Doc Al said:
Well, not always, as your example illustrates. You can certainly draw an extended free body diagram where the points of application of the forces are indicated and the distances between them are shown.
So it's not possible to represent the situation I am talking about with a point-like free-body diagram?
 
Mr Davis 97 said:
So it's not possible to represent the situation I am talking about with a point-like free-body diagram?
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.)
 
Doc Al said:
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.)
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.
 
Mr Davis 97 said:
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.
Don't get hung up on the diagram. Newton's 2nd law applies to extended objects just as it does to point particles.
 
Doc Al said:
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}##?
 
Mr Davis 97 said:
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}##?
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).
 
  • #10
Mr Davis 97 said:
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}##?
Sure. (Assuming symmetry, of course.)
 

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