Physics load / tipping point problem

[Salty]

I'm hoping someone on here like @Capt. Geoffrey Thorpe can help out with how to solve a couple problems with an engine hoist. I'm not looking for the solutions but rather the formulas / techniques to solve them.

The first one I think is easy but I want to verify that there isn't anything more complicated in the mix I'm not thinking about.

1. What is the max load a 2 ton jack can hold in the hoist below? I believe the answer is simple 4000 / 50 * 15 = 1200 lbs. Is it more complicated than that?

2. Ignoring jack limits, at what load would the hoist tip over? My head aches just trying to figure out how to calculate this.

 

[Dana]

The max load is 4000/65*15=923.

To know the tipping load, you need to know the weight of the hoist, and where its center of gravity is. If the boom was a bit shorter, or the base a little longer, it wouldn't tip at any load.

 

[Salty]

The max load is 4000/65*15=923.

To know the tipping load, you need to know the weight of the hoist, and where its center of gravity is.

I disagree with your first answer. I'm pretty sure that is incorrect.

For the second, assume the hoist has no weight and infinite strength.

 

[Dana]

Disagree all you want, but you have to sum the moments about the boom hinge point, 4000*15 = 923*65. That assumes the boom is horizontal and the jack cylinder vertical.

Since the load is beyond the base, if the hoist has no weight then ANY amount of load would tip it over.

 

[Salty]

Disagree all you want, but you have to sum the moments about the boom hinge point, 4000*15 = 923*65. That assumes the boom is horizontal and the jack cylinder vertical.

I think you are right.

 

[Salty]

Since the load is beyond the base, if the hoist has no weight then ANY amount of load would tip it over.

I see what you are saying. For the sake of finding a solution let's say it weighs 100 pounds and the CG is 20" from the left side of the base.

 

[Dana]

Nope. It's just like doing a weight and balance, sum the moments about a point.

If the jack were in the middle of the boom the working load would indeed be 2000# since relative to the pivot point, the load has a 2:1 mechanical advantage over the jack.

 

[Salty]

Nope. It's just like doing a weight and balance, sum the moments about a point.

If the jack were in the middle of the boom the working load would indeed be 2000# since relative to the pivot point, the load has a 2:1 mechanical advantage over the jack.

Sorry, I realized my mistake. See above

 

[Dana]

I see what you are saying. For the sake of finding a solution let's say it weighs 100 pounds and the CG is 20" from the left side of the base.

Then you sum moments about the end of the base. 20" from the left side is 37" from the right, which is the tipping point. That's 3700 in-lb. As drawn, the boom extends 8.4" beyond the base, so divide 3700 by 8.4 and you get 440#.

Better make the base heavier.

 

[Salty]

Then you sum moments about the end of the base. 20" from the left side is 37" from the right, which is the tipping point. That's 3700 onion. As drawn, the boom extends 8.4" beyond the base, so divide 3700 by 8.4 and you get 440#.

Better make the base heavier.

Crap. I was trying to make this way too complicated.

I'm not building a hoist, I'm just curious about the numbers printed on it that don't seem reasonable at all (they seem too high not too low).

Is onion supposed to say pound?

 

[Dana]

Damn autocorrect. In-lbs.

 

[Salty]

Damn autocorrect. In-lbs.

that makes more sense. duh

 

[Salty]

Thanks. Now I just have to weigh the hoist and find the CG. Unfortunately, it's holding up my plane right now, so that's not an option.

 

[Capt. Geoffrey Thorpe]

Dana nailed it.

If this is a commercial engine hoist, you usually can't get the boom end beyond the end of the supports so it will never tip over.
I've pushed them pretty much up to the stated load limits... (Moved a Tormach PCNC1100 CMC mill)
The jack may be rated at 2 tons, but what it delivers depends a lot on how hard you can lean on the handle.

This is the one I own https://www.tractorsupply.com/tsc/product/big-red-2-ton-engine-hoist?cm_vc=-10005 With the boom all the way out, I think it's supposed to be good for 1/2 ton. I do know it will pull a Lilac bush out of the ground at that setting.

 

[Salty]

And just based on question #1 there's no way the labels on this hoist are right.

At the numbers given above it says 2000 lbs, with a 3 ton jack that's 600 pounds over. Some of the other numbers are even more suspect.

 

[Dana]

Geoffery has a good point. It may be rated 2000 max with the boom retracted, but if your 50" dimension is with the boom extended, it would have to be be derated.

 

[Dan Thomas]

Even if the load isn't beyond the end of the jack's base you can have an unstable hoist. Trundling it forward with an engine hanging on it and stopping quick, like running up against something on the floor, can cause the engine to swing forward and pull the jack over.

 

[Sac Arrow]

I disagree with your first answer. I'm pretty sure that is incorrect.

For the second, assume the hoist has no weight and infinite strength.

If the hoist has no weight, it will simply tip over, at any load, as configured.

 

[Lachlan]

Even if the load isn't beyond the end of the jack's base you can have an unstable hoist. Trundling it forward with an engine hanging on it and stopping quick, like running up against something on the floor, can cause the engine to swing forward and pull the jack over.

But it’s unlikely you’re going to roll it that fast. Besides, you only need to worry about what you jack up against, what you jack above, or tipping the jack over, if it’s operated by a jack off.

 

[Salty]

I put 100 lbs of counter weight on the other side of the hoist just for extra measure. It ain’t going anywhere now.

 

[Dan Thomas]

I put 100 lbs of counter weight on the other side of the hoist just for extra measure. It ain’t going anywhere now.

That's the stuff. Peace of mind.

 

[Craig]

But it’s unlikely you’re going to roll it that fast.

It doesn't take much movement to get a hanging load going in a pendulum motion, and exceed the balance limitation. Done it before with as little movement as a couple of feet. When it happened to me, I was fortunate to have the load only a foot or so above the ground and was able to dump the jack before it went over.