Equilibrium of a Uniform Rod with Attached Particle Question

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The discussion revolves around the equilibrium of a uniform rod AB, measuring 3 meters in length and weighing 120N, supported at points C and D. The reactions at C and D are defined as Rc and Rd, respectively, with the condition that Rc equals twice Rd. The participant is tasked with demonstrating that the weight W of a particle attached at point E, where AE equals x meters, can be expressed as W = 60/(1-x). The participant has derived the reactions before the weight was applied but struggles to integrate these results with the new conditions introduced by the weight W.

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I'm stuck! This is the question:
A uniform rod AB has length 3m and weight 120N. The rod rests in equilibrium in a horizontal position, smoothly supported at points C and D, where AC=0.5m and AD=2m. A particle of weight W Newtons is attached tto the rod at a point E where AE=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).

I've worked out the reactions at C and D before W is applied, (40N and 80N I think) and by substituting 2Rc=Rd into 180+Wx=0.5Rc+2Rd (where Rc and Rd are the reactions at C and D) I got 180+Wx=4.5Rc, but now I'm stuck. I can't figure out how to bring the 'pre-W' results and the 'post-W' results together, and I'm not even really sure if that's what I'm supposed to be doing! Could someone point me in the right direction please?
 
Last edited:
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icvotria said:
I'm stuck! This is the question:
A uniform rod AB has length 3m and weight 120N. The rod rests in equilibrium in a horizontal position, smoothly supported at points C and D, where AC=0.5m and AD=2m. A particle of weight W Newtons is attached tto the rod at a point E where AE=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).

I've worked out the reactions at C and D before W is applied, (40N and 80N I think) and by substituting 2Rc=Rd into 180+Wx=0.5Rc+Rd (where Rc and Rd are the reactions at C and D) I got 180+Wx=4.5Rc, but now I'm stuck. I can't figure out how to bring the 'pre-W' results and the 'post-W' results together, and I'm not even really sure if that's what I'm supposed to be doing! Could someone point me in the right direction please?

I think you are missing a 2 in this equation

180+Wx=0.5Rc+Rd

and it looks like you used the 2 to get the equation that followed

To bring your results together you need one more piece of information. What is the final Rc + Rd in terms of the weight of the rod and the added weight, W?
 
Last edited:
OlderDan said:
I think you are missing a 2 in this equation

180+Wx=0.5Rc+Rd
That missing 2 was a typo, sorry. I've corrected it in the original question. I can't try to do the question now cos I'm piiiiiissed but I'll try again in the morning. Or maybe the afternoon. Thanks for the advice though.
 

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