Let's talk pulley physics.

Aioua

So, my roommates and I decided to rig a pulley and basket from our deck to our patio underneath. (Beers go down, burgers come up. Aw yeah.)

So, we have a beam catilevered out to hold the pulley, like so:

Now, having all the weight on the end of the cantilever makes it kinda weak, so we were thinking of adding another pulley to move some of it inward, like this:

Now, they're sure this will reduce the force on the left pulley. I'm thinking that's right, but, I'm not sure why. Looking around the intertrons, there's lessons on pulleys about the forces on the weight and the rope, but I didn't find anything about the forces on the pulley/ceiling interface.
So how would one go about measuring the force on the pulley itself?

Boutros

lets assume all of the ropes are purely vertical.

Then, the vertical force on the beam from the single pulley is 2W, where W is the weight of the object being suspended. The moment at the supported end of the cantilever beam is 2WL, where L is the distance from the support to the pulley.

In the two pulley arrangement the force on the beam from each pulley is W. The moment is now equal to WL + Wl, where l is the distance from the support to the near pulley. Since this distance is much smaller, the total moment is reduced. if the near pulley is at the support there is no moment and you have cut the moment in the beam due to the applied load in half, to WL.

Boutros

I didn't explain why the force on the beam from the pulley is W or 2W. It's because when the pulley is static, all of the vertical and horizontal forces add to zero.

In case 1, (assuming the ropes both pull straight down) the tension in the rope is W, so the force down is 2W. Since the total force in the vertical direction has to be 0, the force in the member connecting the pulley to the beam must also be 2W, in the opposite direction (also tension, but up in this case).

In case 2, the tension in the rope is still W, but there is only W pulling down for each pulley. There is now a horizontal force to be reconciled, but it is largely irrelevant to the beam since this horizontal force will be axial to the beam, and cause no moment (for beams moment is all we care about 99% of the time).

FunkyWaltDogg

All of the above is correct, but I drew a free-body diagram and I'm not about to let it go to waste!



For setup #1, you have these equations:

Force (vertical):
[INDENT]F[SIZE="1"]p[/SIZE] - W - F[SIZE="1"]diag[/SIZE]*sin(θ) = 0[/INDENT]
Torque:
[INDENT]W*r - F[SIZE="1"]diag[/SIZE]*r = 0 =>
W = F[SIZE="1"]diag[/SIZE][/INDENT]
Thus
[INDENT]F[SIZE="1"]p[/SIZE] - W - W*sin(θ) = 0 =>
[COLOR="Lime"]F[SIZE="1"]p[/SIZE] = W*(1 + sin(θ))[/COLOR][/INDENT]

For setup #2, it's even simpler:

Torque (same as before):
[indent]W = F[SIZE="1"]horiz[/SIZE] = F[SIZE="1"]down[/SIZE][/INDENT]
Force (vertical):
[COLOR="Lime"][indent]F[SIZE="1"]p1[/SIZE] = W
F[SIZE="1"]p2[/SIZE] = W[/indent][/COLOR]

Really, #2 here and the case Boutros discusses (where the pulling rope is vertical) are both special cases of #1. Any way you slice it, Fp = W*(1 + sin(θ)), which means the closer to horizontal the rope is, the less weight the pulley is bearing. The existence of the second pulley is basically irrelevant.

A careful observer might notice I left out the shear force from the pulley's support in diagram #1, but it doesn't really come into play in calculating the downward force on the pulley.

Aioua

Okay, cool guys!
I had kinda figured it out... With thought-experimenting, I had concluded that with a single pulley the downward force was 2W, but since that's kinda unintuitive at first glance (why would it weigh twice as much?), I thought some double checking was in order.

Ruckus

It will not reduce the force necessary to lift your load. To do that the load would need a pully as well, and your system would look something like this:

F
| /o

---

>
| |
| |
|o|
-|-
L

Where F is a fixed point, L is the Load, o are pulleys and -> is the end of the rope that needs pulling. This system would reduce the force to lift the Load by half (at the end of the rope), but in exchange 2x as much rope needs to be pulled (eg if you need to move the load 10 feet vertically, your going to be pulling 20 feet of rope horizontally).

Metalbourne

The short answer is, you're right. half the downward force is on the first pulley and half the downward force is on the second pulley.

Aioua

Haha, just realized. We can do it like this:

Same weight distribution, plus mechanical advantage!

EDIT: Haha, just realized. Ruckus already posted it. :oops:

Metalbourne

moving everything closer to where the beam is secured to the wall will reduce bending stress

Aioua

Metalbourne wrote: »

moving everything closer to where the beam is secured to the wall will reduce bending stress

Yeah, but it needed to be that far out to clear the deck. Otherwise we'd just stick it directly on the roof beam.

Dunadan019

you can figure out if the beam will fail with a couple of simple calculations but you need to have actual dimensions on the beam and locations of the pulleys. knowing what wood its made out of would help but I suspect that you can easily make the beam more than beefy enough to pick up 50 lbs.

the trick is how you mount the beam which will experience the highest shear and moment.

zilo

Why not just put the grill on the deck?

Rderdall

zilo wrote: »

Why not just put the grill on the deck?

That doesn't solve the beer getting down.

Aioua

zilo wrote: »

Why not just put the grill on the deck?

Not as manly as rigging up pulleys. :P

zilo

Laziness is manly.

L Ron Howard

Real men eat their meat raw and bloody, right from the dead carcass itself.

Metalbourne

L Ron Howard wrote: »

Real men eat their meat raw and bloody, right from the dead carcass itself.

Right after they use a pully to hoist it up to their apartment

Lemming

Aioua wrote: »

Haha, just realized. We can do it like this:

Same weight distribution, plus mechanical advantage!

EDIT: Haha, just realized. Ruckus already posted it. :oops:

You prooobably shouldn't actually do that, because as soon as you let go of that box it's going to immediately fly to the right until the horizontal forces cancel out.

Willeth

I'm sure there's a way to involve a second pulley that would magnify the effort you put into it, so that instead of hauling the rope in equal measure to the distance down to the ground, you'd pull and let out a much shorter length of rope to let the load move the full distance. I'm not a physics guy, but I know that this should be possible (and probably what you're trying to do with the image in post 8).

EDIT: More like this (terrible online image editor yay):



This should reduce the amount of rope you have to pass through without making the load shift horizontally (and maybe reduce the amount of effort required for heavy loads? I seem to remember something about that from high-school physics, increasing the magnification with each loop).

Eat it You Nasty Pig.

there is no problem that cannot be created, and subsequently solved, with a system of pulleys

Lemming

Willeth wrote: »

I'm sure there's a way to involve a second pulley that would magnify the effort you put into it, so that instead of hauling the rope in equal measure to the distance down to the ground, you'd pull and let out a much shorter length of rope to let the load move the full distance. I'm not a physics guy, but I know that this should be possible (and probably what you're trying to do with the image in post 8).

EDIT: More like this (terrible online image editor yay):



This should reduce the amount of rope you have to pass through without making the load shift horizontally (and maybe reduce the amount of effort required for heavy loads? I seem to remember something about that from high-school physics, increasing the magnification with each loop).

You have that backwards; it'll magnify the force, but you have to pull twice as much rope, and the object will move half as far. You can also set it up so that you pull less rope, but the force required to pull it will be twice as much, and the thing you're pulling will move twice the distance of the rope you've pulled.

Aioua

Lemming wrote: »

You have that backwards; it'll magnify the force, but you have to pull twice as much rope, and the object will move half as far. You can also set it up so that you pull less rope, but the force required to pull it will be twice as much, and the thing you're pulling will move twice the distance of the rope you've pulled.

Nitpick, that should be "or" not "and".

Also, for the life of me, I cannot envision how to have the advantage going the other way (pull a little rope, move a lot of distance), using just a rope and pulleys. Wouldn't you need gears, or a belt/pulley system?

Boutros

You could do it, it's the same principle in reverse. If you pull on the basket with your latest system you pull 2 feet of rope on the other end for every foot you move the basket. So if you have a linkage on the pulling end with more pulleys than on the lifting end you have to pull extra hard but not as far. It's kind of a bad idea though, magnifying the tension + adding the friction from lots of pulleys would make lifting anything of a decent weight quite hard.

Improvolone

Pulleys can easily add a ton of friction to he point of "I don't believe this!"

Aioua

Boutros wrote: »

You could do it, it's the same principle in reverse. If you pull on the basket with your latest system you pull 2 feet of rope on the other end for every foot you move the basket. So if you have a linkage on the pulling end with more pulleys than on the lifting end you have to pull extra hard but not as far. It's kind of a bad idea though, magnifying the tension + adding the friction from lots of pulleys would make lifting anything of a decent weight quite hard.

Oh, durr!

I AM THE KING OF PHYSICS! ¬_¬

Kruite

Why don't you add a line supporting the weight of the beam from the building.

Aioua

Kruite wrote: »

Why don't you add a line supporting the weight of the beam from the building.

We can't reach the non-deck side of the roof beam without a two story ladder, which we don't have. Making a cantilever strong enough to support the weight is easier.