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Let's say I have the following diagram:

enter image description here

Since it's a fixed beam, I know that there will be three components acting on point A:
1) Force acting in x direction
2) Force acting in y direction
3) Moment

However, I don't understand why If I take moment about point "A", I would need to consider the moment Ma but not the x and y forces acting on that point.

I always thought that taking moment about specific point cancels all the forces acting on that point.

I would really appreciate if someone could clarify that for me.

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  • $\begingroup$ What od you mean by "cancels ... forces?" Do you mean they go to zero magnitude, or they don't contribute to torque about that point? $\endgroup$ – Bill N Jun 4 '17 at 17:43
  • $\begingroup$ I meant that they don't contribute to torque about that point but I'm still confused how the moment at that point does contribute. $\endgroup$ – Student Jun 4 '17 at 18:02
  • $\begingroup$ There isn't a separate moment, other than what arises from forces not acting co-linearly through that point. Sometimes a problem may specify a moment, but rest assured that it arises from some force that has a line of action that doesn't intersect that point. In other words, you can't have a moment (in mechanics) without having an associated force. In this problem, the forces are C, D, E, and B produce moments, along with the weight of the beam. $\endgroup$ – Bill N Jun 4 '17 at 18:56
  • $\begingroup$ Is the $M_{max}=-0.8 kN\cdot m$ a specified moment? About A? Is it the maximum torque due to everything, or a separate torque? If it's separate, it arises from some unknown force acting at an unknown location in an unknown direction, resulting in that particular torque. $\endgroup$ – Bill N Jun 4 '17 at 19:07
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We can apply the concept of torque only about a fixed axis(I.e. an inertial frame) and in this specific question calculating torque about the pivot will also help you as torque due to hinge forces will be zero(as their arm length is zero)! But in the case of non-fixed systems we calculate torque about COM which is attributed to the fact that torque of pseudo forces about COM in COM from will be zero(hence behaving as a fixed axis).

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  • $\begingroup$ Are you saying that if this structure was mounted to the wall of an elevator which accelerated upward relative to the ground at $\vec{A}$, then we couldn't apply the concept of torque? $\endgroup$ – Bill N Jun 4 '17 at 19:08
  • $\begingroup$ @BillN You can, as long as you choose a pivot that is inertial relative to the particles. Then the torques are exactly the same because the situation is equivalent to an inertial frame. A nice choice is the CoM. Elevators, carousel, orbiting spacecraft are all not inertial but can be modelled as such $\endgroup$ – lucky-guess Jun 5 '17 at 0:08
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Because A is a pivot. The moments about A can only be evaluated because it is inertial, (or not accelerating relative to C,D,E and B). When we define the moments about a point is such and such, we mean a pivot that is not accelerating itself. It is an easy misconception to think that moments are taken about particles, they are not.

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  • $\begingroup$ That's not true. You can calculate moments (aka, torques) about any point at any time. How you use those moments, and whether they are constant is another matter, but you can always do the calculation. According to your statements, one shouldn't calculate the moments about the hinges of a door while someone is opening or closing it because the hinge is accelerating relative to the door handle. $\endgroup$ – Bill N Jun 4 '17 at 17:49
  • $\begingroup$ the hinge is not accelerating though $\endgroup$ – lucky-guess Jun 4 '17 at 17:50
  • $\begingroup$ Relative to the moving handle it is. One can also calculate the moments about the handle or any other point. It may be more difficult to solve the system of equations you get, but you can certain get valid equations. $\endgroup$ – Bill N Jun 4 '17 at 17:51
  • $\begingroup$ yeah ok. but i said inertial point $\endgroup$ – lucky-guess Jun 4 '17 at 17:52
  • $\begingroup$ unless you wish to find instantaneous torque. in any case, the logic still holds $\endgroup$ – lucky-guess Jun 4 '17 at 17:55
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I am afraid you are having a huge misconception. Force balancing and moment balancing are two independent things. If you ensure force balance on a body, that doesn't imply that moment will be automatically balanced. A body which is in static equilibrium (like in your example) is balanced in forces and moments both. If you assume that there is some moment about point A and use moment balance to find that "assumed moment", forces will still be unbalanced.

If you draw forces correctly on a correct positions, where they are actually applied then you will get complete consistent picture of both moments and forces. Force balance doesn't require any information of positions of application of forces, it just requires only the forces. But moment balance (as it is $r\times F$) , on the other hand, requires the vector $r$ i.e. distribution of forces on the body i.e. the point or region of application of forces.

What happens at point A is: There is a distribution of forces at point A because Point A is not actually a point, it is a small region where forces are distributed, generally non uniformly. Now we don't know how the distribution of forces is. So what we do is just a simple trick (This trick can be justified using simple mathematics of cross products) : We say that, ok we don't know the distribution but we do know that the forces will produce some NET FORCE and a NET MOMENT about point A. And so we replace that unknown distribution of forces by an unknown Net Force and an unknown Net Moment. So you can solve for them by Force balance and Moment balance together.

But after doing this all, the distribution of forces on A still remains unknown and luckily no one asks that in problems/assignments/exams.

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