Recent content by urbanupstart

Ok, I appreciate your reponses Tinytim, but I still need more help. Please go on...

Hmmm, centre of mass of the bar from the pivot? I don't know, please help:wink:

ok, then w*4.5 + w* 1.5 ?

ok, then the moment of the bar is w*6m?

Ok, that would be 1500Nm(for the first), I don't know how to figure out the bar moment (as distance is zero and zero* anything is zero), but on the other hand, if it is the sum of the other two moments, then it would be 1950Nm? and 450Nm(for the right-hand weight)

I'm so sorry, but I mis-read the answer. It should be 700N :redface: Can anyone still help me with this please?

Ah, that was my attempt to figure out the bar weight lol! Thanks Tim, My level of understanding of this is pretty low though. How would I calculate the bar moment? Would you be kind enough to show me the calcs, so I can take a look please. Sorry to be so dense:blushing:

1. A heavy uniform bar is in equilibrium in the arrrangement shown (I cannot upload the picture, but the bar is resting on a pivot with two weights at either end. 1.5m to the left of the pivot with 1000N weight at the end. 4.5m to the right of the pivot with a 100N weight at the end) Question...