Reeving a Tackle to Advantage -- What am I Missing?

[conachair]

So here's a more complex calculation...

Pretending that a Kilogram is a unit of force, ignoring friction..

and that the direction of pull is always 30deg off vertical..

what tension will be required to lift this bucket and how far off vertical will it end up..

[Seaworthy Lass]

Originally Posted by conachair

Handy site.. https://awwapp.com/
Draw up your latest force vector masterpiece. Click on menu|post then click on full size, then copy the web address and paste it into the insert image dialog box.

Not nearly as much fun if all the quirky hand drawn masterpieces are eliminated

Originally Posted by conachair

So here's a more complex calculation...

Pretending that a Kilogram is a unit of force, ignoring friction..

and that the direction of pull is always 30deg off vertical..

what tension will be required to lift this bucket and how far off vertical will it end up..

Someone else needs to tackle this. It is complicated as the bucket will shift sidewards while it is moving up. I need to walk before I can run LOL and I have been told by 2Hulls that my grasp of the basics is wrong, so I am not qualified to run yet .

The only thing I can say is that it will not be off the vertical at the end as it will shift back to the vertical if you keep pulling long enough and it will end up jammed against the block .

SWL

[conachair]

Originally Posted by Seaworthy Lass

Someone else needs to tackle this. It is complicated as the bucket will shift sidewards while it is moving up. I need to walk before I can run LOL and I have been told by 2Hulls that my grasp of the basics is wrong, so I am not qualified to run yet .

The only thing I can say is that it will not be off the vertical at the end as it will shift back to the vertical if you keep pulling long enough and it will end up jammed against the block .

SWL

Actually, looking at it it's not solve-able with a distance between the attachment point of top pulley and the point where the bucket attaches to the bottom pulley. Lets say 2m.
Got me stomped, the loads will change as there are horizontal and vertical components which end in equilibrium - but how to calculate...

[Lodesman]

Originally Posted by Seaworthy Lass

Lodesman, I think my cheer could have been heard all the way to Athens .
Very glad the penny finally dropped. So many members have put in effort to help you understand. Wish the scotch could have been passed around - great work guys .

'So many members' or too many cooks. Sorry I got lost in all the divergent thoughts. Like Dockhead, I didn't see the placement of the man pulling; embarrassed, but somewhat relieved that cleverer chaps than I also had it wrong. I've modified my diagram, which might help others who come in wondering how it works. Hope it helps.

[conachair]

Originally Posted by conachair

So here's a more complex calculation...

Pretending that a Kilogram is a unit of force, ignoring friction..

and that the direction of pull is always 30deg off vertical..

what tension will be required to lift this bucket and how far off vertical will it end up..

Think i might have figured out this one....maybe..

The horizontal force must be the same both sides, otherwise it wouldn't be stationary - Newton again.

Right hand side..
So for tension T and horizontal force H..
Sin(30)=H/T, H=T * Sin(30);

left hand side with twice the tension T (2 parts of the rope)
Sin(angle)=H/2T, H=2T*Sin(angle) but we know H so
2T*sin(angle)=T*sin(30)
sin(angle)=T/2T * sin (30), = 0.5*sin(30), = 0.25
So angle = 14.477512185929923878771034799127 (approx )

Then the ratio for the vertical load will be sin(14.4..)/sin(30) x 10Kg for the right hand side. And carry on with bits of trig for the rest of the forces.

Make sense.......

[jkishel]

You've correctly determined that the angle the blocks make with the vertical is the arcsin of 1/2 the angle at which the tension is applied. However the angle(s) depend on the tension. The weight will begin to lift with any non-zero tension as it is pulled to the right. The tension required to continue to raise the weight increases as the weight rises and swings to the right.

[conachair]

Originally Posted by jkishel

You've correctly determined that the angle the blocks make with the vertical is the arcsin of 1/2 the angle at which the tension is applied. However the angle(s) depend on the tension. The weight will begin to lift with any non-zero tension as it is pulled to the right. The tension required to continue to raise the weight increases as the weight rises and swings to the right.

In the original post to make life easier it was assumed that the angle of the hauling part would always be 30deg to the vertical, if this was the case (unlikely in the real world) the weight would swing to the left as it rises.

Need to have a think about this next one.. would the weight lift straight up when you pull the rope?

Instinctively it would - force H2 must always equal H1, the rest is ratios. I think...maybe... - care to work it out?

[David M]

Why is this so complex? Count the number of lines leading away from the running block (the block opposite of the standing block). There is nothing debatable about this.

[Paul Elliott]

Originally Posted by David M

Why is this so complex? Count the number of lines leading away from the running block (the block opposite of the standing block). There is nothing debatable about this.

If the final lead comes off the running block then the advantage multiplier will be a function of the lead angle, so "counting the lines from the running block" won't work in some cases.

As for the off-angle pull analysis, why don't we assume for the moment that the blocks are on tracks? This is exactly the case for, say, adjustable jib sheet blocks on a rail track. Once we agree on the results of the friction-free track case, then we can move on to the free running block case.

[David M]

Correct Paul, I was keeping it simple (like the original Sea Scout drawing) by not bringing up lead angles or Y configurations. I was assuming the lines make 180 degree changes when run through a block and that no angles change as the standing and running blocks get closer together.

[conachair]

Originally Posted by conachair

Instinctively it would - force H2 must always equal H1, the rest is ratios. I think...maybe...

Wrong! I think.

arcsin(2*A) is different to 2 * arcsin(A)

And I think this bit is wrong as well..

"You've correctly determined that the angle the blocks make with the vertical is the arcsin of 1/2 the angle at which the tension is applied. "

shouldn't that be.... arcsine(half of the sin of the angle at which the tension is applied) ?

[s/v Jedi]

Still at it huh?!

So here's how to understand it for some:

imagine that bucket of 10kg. If you attack a block to it, run a line through the block and holding both sides of this loop while pulling up the bucket, then you still lift 10kg but it is equally divided by half over each line: 5kg each. So now take one line in each hand and you have 5kg in each hand. Now hand one of them to a helper next to you, now you each handle 5kg. Now tie your end off a ceiling beam and you will find that the helper still has 5kg and the ceiling beam has the other 5kg and thus a 2:1 purchase was born. You yourself at that time are already at the single malts of-course as the helper can deal with it alone

[Seaworthy Lass]

Originally Posted by conachair

And I think this bit is wrong as well..

"You've correctly determined that the angle the blocks make with the vertical is the arcsin of 1/2 the angle at which the tension is applied. "

shouldn't that be.... arcsine(half of the sin of the angle at which the tension is applied) ?

Yes, I think jkishel just accidentally left out the words "sin of the" rather than misunderstanding.

Just thinking about all this in a very lazy fashion here - wine, movie & dinner happening so concentration is not there.

One comment - if we are looking at practical situation, you have raised the biggest problem: the angle of pull would change constantly as you pulled on the rope as the distance between the standing and moving blocks shortens. It makes the whole situation very complex is the only thing I can say and that is too mind boggling for me

[2Hulls]

Originally Posted by conachair

OK, I'll try once more, using conachair's very artistic drawing.

Premise: the angle of pull on the running end has no effect on the MA nor for the force to move the load.

To show that the premise may be true, let's assume it is false and test what happens if it is false, e..g., test that the angle makes a difference.

If the angle in the drawing was 0*, i.e., vertically up parallel to the rest of the rig, the load in this 3:1 system would be 1/3 the weight of the load, right?

I assume those arguing that the angle makes a difference must be using the logic that any angle from vertical must still apply the same vertical component as if the angle was vertical or 0*. So, if you postulate that the force on the rope holding the load has to increase as the angle of the force departs from vertical, i.e., increases from 0*, then to maintain the same vertical component you would have to divide the force by the cosine of the angle. This is the only trigonometry that can apply.

To keep it simple, let's neglect friction and assume a large weight that will not "swing" significantly when pulling it sideways. If this bothers you, we could modify the drawing and make the mechanical advantage 100:1 or something else very large such that the side pull was very small. But for the math we'll use the drawing MA of 3:1.

So, starting back at 0* angle, the force on the rope, F:

F = 1/3 X weight / cos 0* Cosine 0* = 1, so the force = .33 X weight. We all agreed to this a minute ago. No angle, no increase in F is needed to hold the weight. No surprise.

Now let's consider the 30* angle case of the drawing:

F = 1/3 X weight / cos 30* Cosine 30* = .866, so doing the math F = .38 X weight. A small increase. Stopping here, this appears to validate that more F is required to support the weight as you increase the angle from vertical.

But let's keep going. Let's consider a 45* angle.

F = 1/3 X weight / cos 45* Cosine 45* = .707, so doing the math F = .43 X weight, and sure enough, more F is required to hold the weight as the angle increases.

For 60*, similarly doing the math, F = .67 X weight. Still more F.

But any body getting worried yet?

Now let's consider the special case of 70.529* whose cos = .333333333

Wow, doing the math, F = 1 X weight. All of the mechanical advantage of using the block and tackle has disappeared! How can that be?

OK, consider an 89* angle. cos 89 = .01745. More wow, F = 19.1 X weight. Who believes that?

Finally, disproving that the angle makes any difference means that when it finally reaches 90*, cos 90 = 0 and division by 0 is impossible. The Limit of F as the angle approaches 0 = infinity X weight.

QED

For those arguing that the angle still makes a difference and don't agree with my proof that it can't, please pick an angle and do the math to arrive at what the force will be. Then do the math with the same logic at the extremes of the angles.

Dave

[TeddyDiver]

Originally Posted by 2Hulls

OK, I'll try once more, using conachair's very artistic drawing.

Premise: the angle of pull on the running end has no effect on the MA nor for the force to move the load.

To show that the premise may be true, let's assume it is false and test what happens if it is false, e..g., test that the angle makes a difference.

If the angle in the drawing was 0*, i.e., vertically up parallel to the rest of the rig, the load in this 3:1 system would be 1/3 the weight of the load, right?

I assume those arguing that the angle makes a difference must be using the logic that any angle from vertical must still apply the same vertical component as if the angle was vertical or 0*. So, if you postulate that the force on the rope holding the load has to increase as the angle of the force departs from vertical, i.e., increases from 0*, then to maintain the same vertical component you would have to divide the force by the cosine of the angle. This is the only trigonometry that can apply.

To keep it simple, let's neglect friction and assume a large weight that will not "swing" significantly when pulling it sideways. If this bothers you, we could modify the drawing and make the mechanical advantage 100:1 or something else very large such that the side pull was very small. But for the math we'll use the drawing MA of 3:1.

So, starting back at 0* angle, the force on the rope, F:

F = 1/3 X weight / cos 0* Cosine 0* = 1, so the force = .33 X weight. We all agreed to this a minute ago. No angle, no increase in F is needed to hold the weight. No surprise.

Now let's consider the 30* angle case of the drawing:

F = 1/3 X weight / cos 30* Cosine 30* = .866, so doing the math F = .38 X weight. A small increase. Stopping here, this appears to validate that more F is required to support the weight as you increase the angle from vertical.

But let's keep going. Let's consider a 45* angle.

F = 1/3 X weight / cos 45* Cosine 45* = .707, so doing the math F = .43 X weight, and sure enough, more F is required to hold the weight as the angle increases.

For 60*, similarly doing the math, F = .67 X weight. Still more F.

But any body getting worried yet?

Now let's consider the special case of 70.529* whose cos = .333333333

Wow, doing the math, F = 1 X weight. All of the mechanical advantage of using the block and tackle has disappeared! How can that be?

OK, consider an 89* angle. cos 89 = .01745. More wow, F = 19.1 X weight. Who believes that?

Finally, disproving that the angle makes any difference means that when it finally reaches 90*, cos 90 = 0 and division by 0 is impossible. The Limit of F as the angle approaches 0 = infinity X weight.

QED

For those arguing that the angle still makes a difference and don't agree with my proof that it can't, please pick an angle and do the math to arrive at what the force will be. Then do the math with the same logic at the extremes of the angles.

Dave

Sorry but your math is sound and angle makes a difference..