- #1
icvotria
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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?
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?
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