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In free body diagrams, such as a beam attached horizontally to a wall, $F_g$ is always shown acting on the center of gravity of an object.

My question - is this the case in real life, where gravity only acts on this point of the object? Or is gravity acting on all parts of the object, but that point is at the exact center of all the force?

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    $\begingroup$ "gravity only acts on this point of the object" FYI, the center of gravity of an object isn't always a part of the object. $\endgroup$
    – H Walters
    Commented Sep 10, 2017 at 17:29
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    $\begingroup$ Many engineers would be out of business if there were no inner forces in buildings :-). $\endgroup$
    – Peter - Reinstate Monica
    Commented Sep 10, 2017 at 22:24
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    $\begingroup$ It is important to understand that interaction in nature is very immediate. Change of any kind is only transferred between adjacent space/time volumes. It's really like Conway's game of life, just with more parameters. All abstraction we see, all elegance and efficiency and symmetry is just emergent. The arrow of the force acting on the center has been drawn by a human, Nature doesn't know where the center is; she is blind. $\endgroup$
    – Peter - Reinstate Monica
    Commented Sep 10, 2017 at 22:29

3 Answers 3

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Gravity (treated as homogenous) is acting the same on all parts of the object, but if the object is rigid, internal forces allow the simplification that the centre of mass is where all the force acts.

Torque: There is equal mass on both sides of the centre of mass so there is no net torque about it. If the pivot is at the centre of mass, the object will not turn, it will balance.

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    $\begingroup$ When you say simplification, do you mean in our calculations? If so, would you mind explaining how this works? $\endgroup$
    – Inertial Ignorance
    Commented Sep 9, 2017 at 20:31
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    $\begingroup$ It works because lots of little masses all connected together do behave just like one big mass in a certain location (the c or m). $\endgroup$
    – JMLCarter
    Commented Sep 9, 2017 at 20:52
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    $\begingroup$ It is the average position of all element masses that make up the body. It's valid to sum forces to that point because the element masses are rigidly connected. $\endgroup$
    – JMLCarter
    Commented Sep 9, 2017 at 21:14
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    $\begingroup$ that all seems correct. Except the process of using center of mass means once you have calcualted it you then dont' have to do the summation of all the gravitational forces acting on each small masses. $\endgroup$
    – JMLCarter
    Commented Sep 9, 2017 at 21:26
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    $\begingroup$ The mathematical reason this works in everyday objects is linearity, i.e., the fact that the gravitational field can be considered homogenous. $\endgroup$
    – Hagen von Eitzen
    Commented Sep 10, 2017 at 15:31
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Gravity pulls in every single particle in the object.

This means that gravity causes a torque at every single particle around some other point.

  • If you look at a point to the left, then all those torques sum up to a net torque which pulls clockwise.
  • If you look at a point to the right, then they sum up to a net torque pulling counter-clockwise.
  • If you look at a point in between, then some torques pull clockwise and some counter-clockwise.

At some special point, the clock-wise and counter-clockwise torques cancel exactly out.

Since we can see that things do not start to rotate when they fall, we must assume gravity when averaged out to be pulling in this point. If we averaged it out to pull in any other point, there would be a net torque and the object would spin - and we know/see that this does not happen.

So, let's call this point the centre of gravity, while keeping in mind that we can average out gravity to pulling in this point as a simpler model to work with.

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    $\begingroup$ I think you left out a step. Your "special point" actually is a special line, and if the gravitational force pulled on any point on that line, the object would not rotate. The final step, is to explicitly rotate the object. Then, a different line is defined. Each unique rotation of the object gives a different line, and the place where all of those different lines intersect is the center of mass. $\endgroup$
    – Solomon Slow
    Commented Sep 25, 2017 at 21:14
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Simplifications to mechanics problems are often made with respect to the COG because it is the average location of all the mass in an object. For example, picture a half-filled plastic water bottle on a 45-degree angle. The weight of this object at one end is very different from the other end, and so if you were to translate this object across, let's say the xy-plane while keeping the angle intact, it would therefore only make sense to calculate where the average location of where the mass in the water bottle is located, and just use that point in your calculation because it is one, simple calculation as opposed to a much large calculation. Here is a helpful link about COG if you are interested in it.

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