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Moments of despair: exam tomorrow

It's driving me crazy...

I thought you could take moments from any point you wished.

Take this problem:

A uniform rod AB has length 3m and weight 120N. The rod rests in equilibrium in a horizontal position, smoothly suppoirted at points C and D, where AC = 0.5m and AD=2m. A particle of weight W newtons is attached to the rod at a point E where EA= x metres. The rod remains in equilibrium and the magnitude of the reaction at C is now twice the magnitude of the reaction at D.

Show that W = 60/(1-x)

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Anyway this query has to do with the taking of the moments:

I did like this, taking them from B

Wx + (120 * 1.5) = (2R * 2.5) + (R * 1)

Wx + 180 = 6R

Now I have the answer and they took moments from A

Wx + (120 * 1.5) = (2R + 0.5) + (R + 2)

Wx + 180 = 3R

What did I do wrong?. Surely they both should yeild the same result

Exam tomorrow. Help...

Reply 1

RObTRIP

u did it wrong in the first part where u resolve from B. the distance from the weight W to B is (3-x). I think that is where you get an error from.

as you're probably in a rush because of exam here is the solution:

moments about c:
(x-0.5)W + 120 = 1.5R
2W(x-0.5) + 240 = 3R (1)

moments about D:
60+(2-x)W=3R (2)

setting 1 and 2 equal to each other gives:

60+2W-xW=2Wx-W+240
3W-3Wx=180
W-Wx=60
W=60/(1-x)

:smile: