Overhead crane and pulleys problem

[Marinel]

The problem bellow is a real problem and I would like to get some ideas in order to overcome it.

As you can see in the attached doc, we have two winches ( drum 1 and 2 ) on the same shaft of motor M1, two moving pulleys ( 3 and 4 ) and a fixed pulley (5).
The load is 130 Tons, and we can measure that with a load cell at pulley 5 ( 32,5 Tons) after the lifting of the load.
But during the time as the load is rising we measure only 30 tons! Note that we have a cable of steel with 30mm diam. and the rising speed is about 1m/min.

When we lower the load, the load cell measures 36,2 Tons. and as soon it stops we get again 32m5 Tons.

When the load gets the floor than we read 0 Tons (of course!)

Can someone explain this? This is a real case and my problem is to adjust the safety relay to 110% of nominal load, which is impossible with this beahviour of the system!

 

[A.T.]

The load is 130 Tons, and we can measure that with a load cell at pulley 5 ( 32,5 Tons) after the lifting of the load.

Intuitively I would think that Ft = 2 x F1 here (not 4 x F1 as denoted in your PDF)

But during the time as the load is rising we measure only 30 tons! Note that we have a cable of steel with 30mm diam. and the rising speed is about 1m/min.

When we lower the load, the load cell measures 36,2 Tons. and as soon it stops we get again 32m5 Tons.

Could it be that the deformation of the steel cable at pulley 3 & 4 changes the force balance between the two cable segments going from/to the pulley? It seems that the side to which the cable goes is taking more load.

 

[Marinel]

You are right! The force at pulley is 2XF1 and not 4X.

In fact I have a display showing the real load and that's whya mmy mistake.

I thought that the problem is concerned with the deformation of the rope but I cannot explain how it works. The length of each rope ( distance from 3 to 5 ) is about 10 meters.

Could someone explain it??

 

[A.T.]

Could someone explain it??

Isn't it a general phenomenon of lossy pulley dynamics, that the outgoing rope has a higher force, because it includes the force needed to overcome friction? The bearings at the axis of pulley 3 & 4 have some friction. To overcome that friction you need some torque. So the outgoing rope must have a greater force than the incoming rope, to create that torque.

 

[Marinel]

OK, I can understand what you say, but that is just a feeling. Can you or someone describe it mathematically?
The feeling I have already but that's all!I would like to have the evidence of the phenomenon.

 

[Marinel]

May be we can simplify the problem using the arrangement below.
What will be the force in the rope when it goes up with no acceleration? And what happens if we stop the lifting?

 

[A.T.]

May be we can simplify the problem using the arrangement below.
What will be the force in the rope when it goes up with no acceleration? And what happens if we stop the lifting?

For your simplified arrangement:
https://www.physicsforums.com/attachment.php?attachmentid=37850&d=1312849616

At constant speed the force in the right rope will be always the weight of the green mass (Ft). Due to friction in the upper pulley the force in the left rope will be more than Ft when lifting the weight, and less than Ft when lowering it. So the force at the pulley axis F1, which is the sum of the rope forces will vary accordingly.

Can you or someone describe it mathematically?

At constant speed all the forces acting at each pulley must balance each other and all the torques acting at each pulley must balance each other. If you introduce a friction torque into the equation, the force vectors must still sum to zero, but to balance the friction torque they have to be redistributed.