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Taking moments about a point

by Shaybay92
Tags: moments, point
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Feb29-12, 01:30 AM
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Conceptually, what does it actually mean to take the 'moments about a point' on a body, even if that point is not the center of rotation of the body (center of mass say). For example, we could take the moments about a point not even 'in' the body, so what does this value represent?

I am asking this as part of an Engineering Mechanics/Statics course. As such, we have to ensure a body is in equilibrium by forcing the sum of moments about any one point to be equal to zero.
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Feb29-12, 04:20 AM
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tiny-tim's Avatar
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Hi Shaybay92!
Quote Quote by Shaybay92 View Post
Conceptually, what does it actually mean to take the 'moments about a point' on a body, even if that point is not 'in' the body
Conceptually, what do moments mean anyway?

Remember, the sum of moments about any point will be zero (in equilibrium)

the reason we take care to choose a specific point is usually that there's an unknown force (usually a reaction force), and the only way we can avoid including it is to choose a point about which its moment is zero!
Feb29-12, 04:57 AM
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You are looking for a deeper meaning in the operation of taking moments. I suggest that you might be better off just starting to go through the process of actually doing it, following the 'rules' scrupulously. A lot of your worries will be sorted out by just getting familiar with moments.
You don't say at what level you are operating here and there are a lot of different levels of answer that could be given.
The principle of moments tells you that the sum of moments about ANY point will be zero. The moment of a force about a point (or turning effect) is defined as force times perpendicular distance. To keep a rigid body in equilibrium you first need the sum of the forces to be zero (Newton's First Law) so it won't accelerate off in some direction. This is often ignored when they first introduce you to Moments but it's important.

Now, unless all the forces act through just one point, there will have to be a minimum of three forces in play for rotational equilibrium (again, no always pointed out).

Skip to the end of this if you want to but the next few lines set the scene
Two kids on a (massless but totally rigid) seesaw is a simple example with two weight forces and one supporting force. To get a balance (equilibrium), the heavier kid needs to sit closer to the middle until their two moments are equal and opposite. If you took the pivot and moved it away from the middle but also introduced the same upwards force (the sum of both their weights) to the middle point then the seesaw is in exactly the same situation as before. It doesn't 'know' that the pivot has been moved and it still has exactly the same forces acting at the same points so it still won't rotate and nor will it move upwards or downwards, cos it's in equilibrium in BOTH senses. If you now take moments about the new point, there will be two contributions (moments) in one direction, balancing the one contribution in the other. It's still in equilibrium so the sum of the newly calculated moments must also be zero.
You could extend the seesaw in one direction by 100m and take moments about the extended end point and the moments would still sum to zero.

Final bit: There will be no force on this remote point, because the three forces and moments already have canceled each other out SO the seesaw structure doesn't even need to be there (rigid) because no force or torque is being transmitted through it from this virtual pivot point.

The point that we choose for taking moments is always at our convenience. We usually choose our point to be one through which one of the forces acts so that particular Moment is zero - leaving us with a simpler equation to solve.

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