- #1

gnits

- 137

- 46

- 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.

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.

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 = wNot 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.

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