To find the reaction in a system at a ring

[gnits]

Homework Statement

To find the reaction in a system at a ring

Relevant Equations

moments

Could I ask for a hint as to where to go next with this question please?

I've done this first part, to find the reaction on the wall. Here's my diagram:

I've labelled the internal forces at B in red.

In green I've shown the reaction at the ring.

So I need to find sqrt(Rx^2 + Ry^2) = R.

So I need to find Rx and Ry in order to calculate R.

For the whole system:

Vertically: R_C + Ry = 3w
Horizontally: R_A + Rx = F

For BA only:

Vertically Y = w
Horizontally: R_A = X
Moments about B gives: R_A * a sin(30) = w * (a/2) * cos(30) which yields R_A = sqrt(3)*W/2 as required.

Now that we know R_A, for whole system:

Taking moments about C leads to: sqrt(3)*Rx + Ry = w

Not sure how to proceed.

If I take moments about C for BC only, this leads to the same equation sqrt(3)*Rx + Ry = w

If I take moments about D for BC only this yields: R_C - sqrt(3)*F = w/2

All my answers above are consistent with the book answer, but I can't see how to proceed.

Book answers is: R = sqrt(Rx^2 + Ry^2) = w/2

So, I am trying to find Rx and Ry in order to calculate R.

Thanks for any pointers.

 

[TSny]

The ring is smooth. Can you deduce from this anything about the direction of the reaction force $\vec R$?

 

[gnits]

Yes, of course, thank you. The reaction force R must be perpendicular to CB, and so using my already derived formula sqrt(3)*Rx + Ry = W and replacing Rx with R cos(30) and Ry with R sin(30) I find that R = w/2 as required. Thank you very much indeed.