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# FORCE USING A STATIONARY SHEAVE

## FORCE USING A STATIONARY SHEAVE

(OP)

Can someone please try and explain this in as simple terms as possible? It is extremely unintuitive to me and I really want to make sense of it in my mind.

I understand how it works in theory but not in reality.

If you look at my little sketch.......Figure (A) shows 1000Kg pulling on a sheave which is attached to the ground.
I can just wrap my head around how the Force that is required to hold the Sheave is 20KN (2,000Kg) I just picture one person pulling on one side with the pull of 1,000Kg equivalent and the same on the other side. These two added together will give the 2,000Kg reading on the load gauge shown.

I then decided to take a very similar example but it was this one that I just couldn't rationalise in my mind.
Image B shows a little person weighing 100Kg. He ties his rope to a little Eye Bolt and straps himself to the rope.
It is clear that if there was a gauge measuring his mass/weight (sorry for using interchangeably) it would read 100Kg.

If however in Figure (C) the stick person fed the rope through the Eye Bolt and secured the end of the rope to the ground, the reading on the Gauge would now read 200Kg.

This stretches my mind too far......Can someone please make some sense of this? Again, from the first example, I can appreciate how it works in theory, but why is it so un obvious in the other example?

Thanks!!

### RE: FORCE USING A STATIONARY SHEAVE

Your mistake is neglecting friction and motion through the use of the word stationary. Pullies and ropes are not stationary. They reach a stable equilibrium after motion has occurred. Motion is a requirement.

### RE: FORCE USING A STATIONARY SHEAVE

(OP)
Thanks for your reply. Sorry, I shouldn’t use the term sheave if considering motion. I am considering a static equilibrium reached.I’ve been thinking about it more and realise that the Reading On B and C would both be 100Kg. I imagined myself being attached to the rope and instead of attaching it to the ground, just holding onto the other end of the rope. Of course this would still read 100Kg. I assume that one side of the rope would have a load of 50Kg in it and I would have to hold onto the other end with a load of 50Kg.

Is this correct?

### RE: FORCE USING A STATIONARY SHEAVE

#### Quote (Munro85)

Reading On B and C would both be 100Kg

You were right the first time.

Neglecting friction, your three examples are drawn correctly.

Again, neglecting friction, any tension in the rope will be the same everywhere in that rope. If you apply tension on one and and anchor the other end of a rope that's wrapped over a pulley, you do in fact get twice the rope tension trying to move the pulley.

This is exactly why we use pulleys in the first place, and stack them together to create block and tackle - because they multiply rope forces and create mechanical advantage.

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