Maths problem with pulleys

Maths problem with pulleys

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Ebo100

Original Poster:

490 posts

211 months

Thursday 29th September 2016
quotequote all
Firstly please excuse the photo; it was the closest to my problem I could find on google. smile



How can I calculate the vertical deflection of the centre weight?

In real life at the left side I have a furnace tube the expands vertically up, in normal operation the counterweight at the right side falls very nearly to the floor. However in the rare occasion where the plant is shut down the expansion is greater by approx 70mm and so there is a sag in the cable. We want to add a small weight to the cable to keep the cable in tension so the cable cannot jump off the pully.

If the distance between pully centres is 1700mm and we know that the sum of the two side of the triangle is 1770mm. The tension weight will be approx 400mm from the RHS pully how can I calculate the vertical deflection if I don't know any of the angles?

Any help would be appreciated.








Flooble

5,571 posts

107 months

Friday 30th September 2016
quotequote all
Not sure I follow if this is a simple trig question (doubt it, you sound like you know what you're doing) or a more complicated physics question (but again moments of a force about a fulcrum is easy to calculate and you sound like you'd have no trouble doing that either).

Hence I think I've missed something obvious in the question.

Would it not be easier to just experiment - add a small weight and see if that solves the problem. I know it's not very pure, but sometimes a bit of practical engineering is the best way. Especially as the expansion sounds to be a bit variable anyway.

Ebo100

Original Poster:

490 posts

211 months

Friday 30th September 2016
quotequote all
Thanks for the reply, this was a case of trying to over-think a problem.

The forces in this case are not that important as our new tension weight is heavy enough to take up more than the weight of the anticipated sag of cable but acting against a 2 tonne counterweight.

I was getting hung up (no pun intended) on the position of the new tension weight being off centre, giving awkward angles. However the vertical deflection will be the same where-ever it will be applied across the 1700mm span. So applying it at the centre line gives two easily solvable triangles and a total vertical deflection of 260mm downward.

Thanks for the interest. thumbup