Question about "moments"
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Question about "moments"

[From: ] [author: ] [Date: 11-10-04] [Hit: ]
instead of moment.For our purposes today, they are the same thing.Moment of a force is the same thing as torque.There is a subtle distinction, but I am not going to mention it.......
i have a question asking me to "find the reactionary forces at a point" (on a cantilever beam, if that makes a difference). there are only forces in the y direction, so i can find the Y-component from the point; obviously the x-component is 0.

however, is the "moment" about that point a reaction force? ive been under the impression that the moments at any point is 0. however it seems theres a different definition for moments?

im a little confused if someone can clarify this- if it helps you visualize, my diagram shows a beam connected to a wall (via point A) with a force acting down on it at the opposite end. under my previous understanding of moments, the "moment about A" should be 0; however that is not the case obviously with just one other force acting on the beam. so i guess theres a different "moment" i must include in the "moment" equation? would this be a reaction force?

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You are probably more familiar with the word TORQUE, instead of moment. For our purposes today, they are the same thing. "Moment of a force" is the same thing as torque. There is a subtle distinction, but I am not going to mention it.

What a torque/moment is, is a rotational equivalent of a force. Forces that act directly at the center of a body cause it to translate (move and retain orientation). Forces that act obliquely however, have the ability to cause a rotation. Torques/moments are the way to quantify this rotational twisting/bending "force".


The full definition is a cross product of a vector from the reference point to the point of force application with the force vector itself.

M = r cross F

r is the radius vector from reference point to the point where the force is applied
F is the force as expressed as a vector

In MOST simple examples, the forces are perpendicular to the member, and often it is easy to see that r and F are perpendicular. This makes your cross products easy to do, because they are a simple multiplication.
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