Proving Equilibrium in a System of Forces Using Moment Equations

[kirab]

Diagram is attached.

P = forces
M = free-moving moment
C = looks like a pin connection
L = distance between A and C, and C and B

With the system in equilibrium.

Question: Write the moment equations about point A, B and C and prove that they are all the same.

so basically I end up with:

(Counter-clockwise moment is positive.)

Summation of moments about point A: P(sin(90-theta))(2L) - M - Cx(sin(90-theta))(L) + Cy(sin(theta))(L) = 0

point B: P(sin(theta))(2L) + Cx(sin(90-theta))(L) - Cy(sin(theta))(L) - M = 0

point C: P(sin(90-theta))(L) - M + P(sin(theta))(L) = 0

which means that

P(cos(theta))(2L) - Cx(cos(theta))(L) + Cy(sin(theta))(L) = P(sin(theta))(2L) + Cx(cos(theta))(L) - Cy(sin(theta))(L) = P(cos(theta))(L) + P(sin(theta))(L) = M.

I don't see how these can be equal. Any mistakes seen here in the approach to solving this problem? Maybe pin supports Cx, Cy don't create moments?

Thanks for the help!

 

[Redbelly98]

Please post homework questions in the homework section of PF.

 

[PhanthomJay]

Your attempt is excellent. Cx and Cy (at the support point C) do create moments about B or A. A and B are not pin suppports, they are just the ends of the member that rotates about C under the applied forces and couple, until equilibrium is reached. Note that by applying the equilibrium equations sum of Fx = 0 and sum of Fy = 0, you can solve for Cx and Cy in terms of P. That will greatly simplify the equations in a manner that will clarify their equality.

 

[kirab]

Sorry for posting in the wrong sub-forum :(

Thanks for the help!

My issue is: when I do summation of forces in x and y, I get

P = (Cy(cos(theta)) + Cx(sin(theta)))/(cos(theta)+sin(theta)) from sum of Fx and

P = -(Cx(cos(theta))+Cy(sin(theta)))/(sin(theta)-cos(theta)).

Which I'm not sure if they're equal?

Either way, I plugged the 1st P into
P(cos(theta))(2L) - Cx(cos(theta))(L) + Cy(sin(theta))(L) and
P(sin(theta))(2L) + Cx(cos(theta))(L) - Cy(sin(theta))(L) and
P(cos(theta))(L) + P(sin(theta))(L)

(all equal to M, this is the last line of my initial post)

When I plugged P into those, I got a huge mess that doesn't clarify the equality. Any tips?

Thanks

 

[PhanthomJay]

Sorry for posting in the wrong sub-forum :(

Thanks for the help!

My issue is: when I do summation of forces in x and y, I get

P = (Cy(cos(theta)) + Cx(sin(theta)))/(cos(theta)+sin(theta)) from sum of Fx and

P = -(Cx(cos(theta))+Cy(sin(theta)))/(sin(theta)-cos(theta)).

Which I'm not sure if they're equal?

Either way, I plugged the 1st P into
P(cos(theta))(2L) - Cx(cos(theta))(L) + Cy(sin(theta))(L) and
P(sin(theta))(2L) + Cx(cos(theta))(L) - Cy(sin(theta))(L) and
P(cos(theta))(L) + P(sin(theta))(L)

(all equal to M, this is the last line of my initial post)

When I plugged P into those, I got a huge mess that doesn't clarify the equality. Any tips?

Thanks

I am not sure from where you are getting your sum of Fx and Fy equations. Cx and the upper P are the only 2 forces in the x direction, and Cy and the lower P are the only 2 forces in the y direction. So in the x direction, you have Cx -P =0, and in the y direction, you have Cy -P = 0. Solve for Cx and Cy.

 

[kirab]

I am not sure from where you are getting your sum of Fx and Fy equations. Cx and the upper P are the only 2 forces in the x direction, and Cy and the lower P are the only 2 forces in the y direction. So in the x direction, you have Cx -P =0, and in the y direction, you have Cy -P = 0. Solve for Cx and Cy.

OH! Yeah My sum of Fx and Fy came from looking at my altered diagram... which had the member lying horizontally, so I stupidly assumed that the x-coordinate system was horizontal and that the y-coordinate system was perpendicular to that, simply from looking at my diagram which has been rotated :P

Yes, with that P = Cx = Cy equation, it can be easily shown that sum of Ma = sum of Mb = sum of Mc. Thanks for the help.