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[Summary: If two forces are such that the component of one of these forces forms a couple with the other, than which of the two cases (obtaining resultant via adding forces or obtaining moment and a resultant force via the method shown below) are correct? Read further for clear and detailed explanation]

Let two forces, F1 and F2, act on a rigid body and have a common point of application at Point A. Then their resultant force, R, can be calculated by the normal use of parallelogram as shown in the image below:

Adding Forces F1 and F2 to get Resultant Force R

Now, since forces can be moved along their action lines in rigid bodies (statics), let the same forces be moved along their action lines as shown in the image below (Img.2):

Adding forces F1 and F2 to get moment along with resultant force

Assume that the y-component of F2 is equal but opposite to F1. This means that the y-component of F2 and the force F1 act as a couple, and cause Moment in the rigid body, which is equal to F1 multiplied by the distance between them. In addition to this moment, we also have the x-component of F2 left over, which is the resultant force on the rigid body.

So, does this mean that, given the following conditions:

1) If any two forces Fa and Fb have intersecting action lines,

2) A line perpendicular to the action line of Fa at the point of intersection divides the plane (or rigid body) in two parts,

3) And Fb points towards a different plane (or different part of the rigid body) than Fa,

Then we have a resultant force (found by the parallelogram law) and moment at the same time?

But then what is the point of application, O, of the resultant force found in the second Image (Img. 5)?

But if this is so, then shouldn't the parallelogram be only defined in a way where the two forces are directed towards the same plane?

Or is this entire thing true? Can the forces actually be moved and resolved in a way they are here? Why or why not?

[Sorry if the post is messy and incomprehensible. I tried, as much as I can, to explain the question in the most organized fashion...]

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There is only one possible resultant for two forces acting through a single point, and it is not possible for those forces to produce a moment about said point.

It is valid to move a force along it’s line of action.

It is valid to break a force into x and y components.

In the situation you describe, if you moved the force along its line of action and then broke it into x and y components, you would still solve the three equilibrium equations. Sum of forces in the y direction. Sum of forces in the x direction. And sum of the moments about a point. The moments about the point produced by the x and y components would cancel. (In your writeup, you seemed to be ignoring the moment produced by the y component, which would be incorrect)

Free Body Diagram

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  • $\begingroup$ You can actually demonstrate that $\sum M_P = 0$ by defining $r = s\dfrac{F_{2y}}{F_{2x}}$, which makes it $\sum M_P = F_{2x}\left(s\dfrac{F_{2y}}{F_{2x}}\right) - F_{2y}s = s(F_{2y}-F_{2y}) = 0$. $\endgroup$ – Wasabi Feb 1 '19 at 1:49
  • $\begingroup$ @Wasabi, you and Cable Stay are right. My answer just skipped to assuming zero sum of moments, by working off the intersection of the two forces, which of course eliminates moments. $\endgroup$ – kamran Feb 1 '19 at 2:26
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Any two force regardless of their graphic representation if they have a point mass destination have no torque unless the mass is at a fixed distance from a center fixed as a rotation arm. Or if they act on two separate points on a real object.

In your case it is ok to break the forces into their components, however you have to start from the intersection point, A, add all x components together, y components together and then apply the resultants which will give you the exact outcome of resolving the forces by parallelogram method.

You have not yet added the components in your second sketch, if you do the F2y will be cancelled by F1.

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  • $\begingroup$ As described by @cablestay's answer, you can actually break the forces into components wherever you like. The math will still work out such that $\sum M = 0$. $\endgroup$ – Wasabi Feb 1 '19 at 1:50

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