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The weight of a body acts through the center of mass of the body. If every particle of the body is attracted by earth, then why do we assume that the weight acts through the center of mass? I know that this is true but I can't understand it. Does it mean that the earth does not attract other the other particles of the body ? Wouldn't it mean that girders would not need any support at the periphery if we erect a pillar at the center?

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  • $\begingroup$ I'm confused about your second question. Do you mean to say that you want to erect a pillar at the center of the earth? $\endgroup$ – probably_someone Dec 30 '16 at 9:56
  • $\begingroup$ I think u misunderstood that. So gravity acts through center of mass of a body but then why do girders bend at the periphery even if the gravity acts at the center? $\endgroup$ – AScientist Dec 30 '16 at 12:14
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    $\begingroup$ Gravity acts at all points of a body. In classical mechanics, you can invoke the concept of the center of mass in order to view the situation more compactly, as it could rule out the unnecessary details about the size or shape of the rigid body under consideration. Instead of a bulky body motion, you can simply illustrate it as a simple particle motion, where the point particle represents the center of mass (or the entire) system, without failing any associated dynamics related to it. $\endgroup$ – UKH Dec 30 '16 at 13:08
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    $\begingroup$ @unnikrishnan I think you ought to expand the comment into an answer. $\endgroup$ – AScientist Dec 30 '16 at 14:41
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    $\begingroup$ Weight acts at the center of mass of a rigid body because "center of mass" is defined as the point where the weight action is centered. $\endgroup$ – Hot Licks Dec 31 '16 at 0:53
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The other answers here, which show that gravity does not exert a torque on an object, are correct. However, they rely on the following implicit step of logic to get to the answer the OP wants:

An object that has a force acting on it, but no torque acting on it appears as if it is being pulled from its center of mass.

This is true in the case of ideal rigid bodies only. In the case of elastic objects, OP is absolutely correct, in that gravity does indeed act on each individual particle in the object. This is why girders bend under their own weight, among other things.

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    $\begingroup$ The curvature of elastic objects under gravity is quite complicated, and likely beyond my powers of explanation this early in the morning (greetings from UTC -5). Maybe this will help? google.com/url?sa=t&source=web&rct=j&url=http://… $\endgroup$ – probably_someone Dec 30 '16 at 14:34
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    $\begingroup$ I think this answer is not correct. Generally the centre of gravity does not coincide with the centre of mass unless the gravitational field is uniform. When that is the case, CG and CM coincide regardless of the rigidity of the body. For bodies much smaller than the Earth, the gravitational field is uniform to a good approximation. The rigidity of the body has nothing to do with the matter. It is a Red Herring. Gravity acts on all particles of the body whether it is rigid or elastic. $\endgroup$ – sammy gerbil Dec 30 '16 at 21:40
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    $\begingroup$ Considering his question was about girders, which typically aren't planet-sized, there is only one relevant approximation here, which is the one I specified. You are technically correct, though. $\endgroup$ – probably_someone Dec 31 '16 at 7:52
  • $\begingroup$ Girders would bend even if pulled from the center of mass. A better example would be a two by four held up on your shoulder bending even when you're supporting the center of mass. $\endgroup$ – candied_orange Jan 1 '17 at 16:00
  • $\begingroup$ Hence why I said, "bending under their own weight". That eliminates any doubt as to where it's being pulled. $\endgroup$ – probably_someone Jan 1 '17 at 17:43
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As a point of clarification which perhaps has not been made as clearly in the other answers: No, the weight of a body does not act through the center of its mass, and no such assumption is necessary. However, one can show (see the answer by @tomph) that the sum of all gravitational forces (which indeed do act on any small part of the body) can be replaced equivalently by a single force through the object's center of mass, if that object can be thought of as rigid. The "equivalently" in this statement refers to the fact that, when we calculate e.g. the forces required to hold such an object in place (by "pillars", say), the result will be exactly the same whether we use the single weight force acting through its center of mass or the actual weight distribution of the object.

In short, the model of the single force acting through the center of mass is a very convenient simplification, but it's neither a necessary assumption nor does it reflect reality. As others have said, once we want to describe the behavior of deformable bodies or discuss interior load distributions in a body, this model is no longer adequate or useful, or even correct, as pointed out by @probably_someone.

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    $\begingroup$ One cannot actually show this in general because it's wrong. It's true (a) at the surface of a planet, where the gravitational force is uniform throughout space and (b) for spherically symmetric bodies on scales where gravity is not uniform. These account for the vast majority of cases that come up in real life. $\endgroup$ – Peter Shor Jan 1 '17 at 14:47
  • $\begingroup$ Two answers: First, I think you would have to consider my answer in the context of the question it was addressing, which was clearly a question by someone without a high level of formal training (and I will note that this is not meant as a slight). For the scenario this person had in mind, I am willing to bet that my answer was appropriate without going to great lengths to establish the conditions under which it hold (and can is "true"). $\endgroup$ – Pirx Jan 1 '17 at 17:43
  • $\begingroup$ As a second answer, taking a more formal perspective, if we consider the $L_2$-space of all square-integrable force density distributions, then my answer is "generically true", in the sense that it is true for all possible force density fields in the above set, with the exception of a subset of measure zero (the subset comprising the distributions that produce a finite moment with zero net force). $\endgroup$ – Pirx Jan 1 '17 at 17:48
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The fact that a weight of a body seems to act on its center of mass is a direct consequence of the weight being a parallel force field (obviously assuming that the body is small enough for the field $\textbf{g}$ to be constant over it).

Consider for simplicity a discrete system formed by N particles, each of mass $m_i$. The total external force acting on this system is \begin{equation} \textbf{F}_{tot} = \sum_{all \,particles} \,m_i \textbf{g} = M \textbf{g} \end{equation} where $M$ is the total mass of the system.

In order to prove that the body is not rotated around its center of mass by its weight alone we consider the torque exterted by weight with respect to the center of mass: $$ \tau _{cm} = \left(\sum_{all\,particles} m_i \textbf{r}'_i\,\right)\times\,\textbf{g} = \left(\sum_{all\,particles} m_i (\textbf{r}_{i} - \textbf{r}_{cm})\,\right)\times\,\textbf{g} = (M\textbf{r}_{cm} - M\textbf{r}_{cm})\times\textbf{g} = \textbf{0} $$ where the primed coordinate are with respect to the center of mass.

Thus, as you can see, the weight of a body exerts no torque on the body. From these result we can infer that, of course, every particle constituting the body is subject to weight, but the total effect is exactly the behaviour the body would have if the weight acted only on a particle having the mass of the total system and located in the center of mass.

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  • $\begingroup$ I'm pretty sure it's also true for non-constant gravitational fields as long as they are conservative (curl = 0), not sure how to show that though. $\endgroup$ – Jason S Dec 30 '16 at 19:15
  • $\begingroup$ @JasonS Mmh. Are you saying that any conservative field has the same kind of effect? $\endgroup$ – tomph Dec 30 '16 at 19:49
  • $\begingroup$ @JasonS and what about the electric equivalent? For example, an electrostatic field ($\nabla \times \textbf{E} = 0$) has a non-zero torque on an electric dipole, with respect to the center of the dipole itself, $\textbf{p} \times \textbf{E}$. $\endgroup$ – tomph Dec 31 '16 at 7:48
  • $\begingroup$ Maybe not then. I guess this means that tidal forces can exert a torque on a rigid body? $\endgroup$ – Jason S Jan 1 '17 at 17:37
  • $\begingroup$ @tomph Could you just tell me how you get $\textbf{(r prime)}_{cm}=(\textbf{r}_{cm} - \textbf{r}_i)$ ? $\endgroup$ – The Cryptic Cat Feb 15 '17 at 1:27
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First of all, lets look at the difference between center of mass and center of gravity in a general case.

Center of mass of a rigid body is a hypothetical point where the whole mass of the body is "assumed" to be concentrated. It's not actually true that the mass is concentrated only at a single point of he body. The distribution of mass is balanced around the center of mass and the average of the weighted position coordinates of the distributed mass defines its coordinates. If the body has a uniform density throughout, then the center of mass lies at the centroid of the body.

Now, center of gravity is the average location of the "weight" of the body, whereas the center of mass is the average location of the "mass" of the body. Hence, in general, the center of mas and center of gravity of a body are not equal. However, in a uniform gravitational field (a point on earth is a very good approximation where your body lies), these two points coincide due to the fact that the weight, $W=mg$, where $m$ is the mass of the body and $g$ is the acceleration due to gravity. This means that if we set the constant $g$ as one, or express weight in terms of $g$, then weight is numerically equal to mass. So, center of mass coincides with the center of gravity here.

Why do we assume weight acts through the center of mass?

As I have quoted in the above paragraph, the weight of a body, in general acts through the center of gravity, not the center of mass, However, a body on the surface of earth feels a uniform gravitational field and so the center of gravity coincides with the center of mass.

The center of mass is a hypothetical point where the whole mass of the body is assumed to be concentrated. Its not actually the point where the whole mass is accumulated. The mass (the amount of matter) spreads continuously throughout the body. But, when it comes to study the motion of a body, which we do actually by studying the variation in the position coordinates associated with the body with respect to time, the COM is really helpful. But, how do we associate coordinates to a body? If the body is bulky, then you can't actually specify the coordinates, instead you specify the volume of the body it occupies in the space. But, you may know that this "volume" is completely irrelevant and unnecessary for a rigid body to explain its dynamics.

To avoid such difficulty, we make use of the center of mass. You can plot the trajectory of the body in the space by tracing down the motion of the center of mass of the body as a function of time. This approach does not fail any dynamics of the body under consideration. So, we could attach the coordinates (frame of the body) on the center of mass of the body.

Gravity acts at all points of the body. The concept of center of mass allows one to study the body or a system of bodies (which I think as the most useful purpose of the concept of center of mass) in a more compact way by simplifying the problem (or by deleting unwanted details). You can assume (on Earth) that the weight of the body acts through the center of mass. An important aspect of this consideration is that the center of mass of a rigid body does not change during its motion. Also, if you consider the complex cases, for example, a two body problem, which can be resolved to a one body problem (that's really a great relief) by invoking the center of mass.

If you are not convinced in the ease of doing this approximation, consider a body moving through a gravitational field under the influence of some external force. try to resolve the Force components to get the resultant force. Without the concept of center of mass, you need to resolve it for the entire points (or particles) constituting the body.

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Force that acts on the center of mass does not exert any torque on an extended body. So, gravity 'acting on the center of mass' means a force that accelerates, but does not rotate, its target.

There is tidal torque on the Earth by the Moon, but this is because the Earth is not a rigid body, and changes shape (and weight distribution) with tides. That means that the Earth is effectively polarized by a gravity field, and that polarization (tidal lobes) slows the Earth while adding to the Moon's orbital angular momentum. This is not due to gravity directly, but due to the time-dependence of Earth's shape change (it isn't a time-reversible effect, though a conservative force field, like gravity, is).

There is clearly torque, too, in a Cavendish apparatus Cavendish torsion balance where the two objects are designed to act against a torsion spring by gravity force. So, the claim of action on the center of mass is sometimes false.

One can argue, of course, that a point mass exerts no torque on a rigid object, because no equal-and-opposite torque can meaningfully be exerted on the point object by gravity. It is difficult, though, to generalize that argument.

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By definition of the center of mass is the point whose acceleration is found by $$\vec{a}_{cm} = \frac{1}{m} \vec{F}_{net}$$

In fact, gravity acts on all the bodies and as far is the linear motion of the center of mass the location of where the force is acting doesn't matter. The location of where a force is applied only matters for the rotational motion.

Take a free floating bar in space and apply a load on one end of it. The body is going to translate and rotate. If you work out the equations you will see that the translation of the center of mass does not depend on the location of the force.

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protected by Qmechanic Dec 30 '16 at 12:29

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