# Two people carrying an object up stairs

1. May 21, 2011

### jschwalbe

Hi all--

First time poster, long time physics lover. ;)

My dad and I got into a discussion the other day since we are about to start moving out of one house and into another. When carrying a uniform object (such as a couch, or large slab of granite), is it advantageous to be at the top vs. the bottom, or does each carrier support the same amount of weight?

My approach to the problem was to envision a uniform wooden board supported by rope at the ends.. tilting the object would have no effect on the tension in either rope.. they would be equal. Drawing out a force diagram with the CoM at the middle and at various angles, each rope would indeed seem to have the same amount of tension. Thus my argument is that it doesn't matter realistically which side of the (uniform) couch you're on, it's still heavy. ;)

My dad's approach was to go to the extremes and say, what if it was straight up and down (instead of the 30-45 deg incline of most stairs); in which case the person on the bottom would be carrying 100% of the burden. He then theorized that it would be better to be on the top because you would be supporting less. I do not think that this question can be answered by looking at the extremes, and disagree with his reasoning.

Thoughts, physics lovers?

2. May 22, 2011

### rcgldr

If getting a good grip on the object wasn't an issue, then orientation wouldn't be much of an issue. In the vertical case, the person above could be supporting all the weight, or the person below, or some combination. In the real world, it's probably difficult to get a good pulling grip on most objects, so the person on the bottom is usually lifting more of the weight.

3. May 22, 2011

### Quinzio

Instinctively someone carrying a weight is trying to make the less effort.
If a weight is carried by two, none really is able to feel the effort of the other, so each one is honestly trying to make the less effort as possible.
If two people are carrying a long object (say a ladder) up on a stair (which is usually at 45°), easily the one on the top will finally push 1/4 of the weight, and the one on the bottom the 3/4 of the weight.

This is because the one on the top is just required to counterbalance the torque given by mg in the center of mass, so he pushes:
$$mg/2\sqrt2$$
He's not really cheating, but he has no exact means to tell how much he should push. The ladder doesn't fall, he did his job.
As it is clear in the drawing what happens is that the man on the bottom can carry up to 3 time that weight carried by the man on the top.
That is why if you're a man and a woman, let always the woman be on top :).

If you and you dad want to be sure you carry each half of the weight, you should support the weight with a short piece of rope, so the tension of the rope is always upward. But like this is more uncomfortable.

Last edited: May 22, 2011
4. May 22, 2011

### jschwalbe

Thanks for the replies.. however they seem to differ! Would anyone be willing to help me understand with a force diagram? (Esp with the 2nd post from Quinzio.)

Thanks!

5. May 22, 2011

### nrqed

Your dad is completely right.

Yes, looking at the extremes is the best way to understand it. At 0 degrees, the two support the same weight (mg/2) whereas at 90 degrees, the person at the bottom supports all the weight whereas the person at the top supports nothing.

To find the two forces at an arbitrary angle, one must use torque in addition to force.
At any angle, we have F1 + F2 = mg (where F1 and F2 are the upward forces exerted by the two persons). For an arbitarty angle theta, we can use that the net torque must be zero to get a relation between F1 and F2. That will show that the force at the bottom is always larger than the force at the top, for theta different than zero.

Last edited: May 22, 2011
6. May 22, 2011

### Quinzio

This is an example, I believe it's correct:

7. Jan 4, 2014

### vega32

I'm wondering the same thing. Can anyone else please confirm this solution? Thank you very much!

8. Jan 4, 2014

### SteamKing

Staff Emeritus
It's always better to be at the top. If the top guy drops his load, the bottom guy gets run over by it. Same result if the bottom guy drops the load. Either way, the bottom guy gets screwed.

9. Jan 7, 2014

### Jano L.

Back to physics, in the rare case the thing has center of mass beneath the plane of grip, i.e. if you move a bathtub and hold it on the top, it is easier for the lower guy. If it is ladder, both exert the same force, and the most usual case, if it is wardrobe hold from below, it is easier for the guy on the top.

10. Jan 8, 2014

### sophiecentaur

That picture is ok if you are saying that the top guy is supporting the object on a smooth tray - or on the flat of his hand.
In practice, the top guy will (or should) be pulling up the slope. He could actually cancel the force acting down the slope completely and leave the bottom guy with no more than (his share of ) the normal force to deal with. The force along the slope can be shared in any way the two guys arrange it. If the top corner is smooth and you can't get a grip on it. a loop of rope around the object will allow the top guy to do his fair share.

11. Dec 5, 2015

### RiverM

I'm not a physics guy, but like questions like this. The center of gravity of the object would be one factor, and the other factor would be the angle of the object. If you were to hold either end of a rectangular wooden block in your hands, and move one hand up and the other down, you'd see that the weight would shift to the lower hand. So the guy at the bottom is bearing more weight.

Though I always like to be at the bottom because I can get a better grip on it.

12. Dec 5, 2015

### CWatters

I recommended reading post #2 carefully. Its not possible to know who is carrying the most weight even if the stairs are vertical.

For example consider what happens if the person at the bottom picked up a chair on his own and passed it up to the man at the top. At various times they both carry the weight of the chair on their own or in some variable proportion that ranges from 0 to 100% of the load.

13. Dec 8, 2015

### RiverM

Well yeah, but the real question here is, how is the weight distributed between the two carriers. Now how much force each one is going to apply.

14. Dec 8, 2015

### CWatters

Yeah ok if neither is allowed to apply torque. One man can carry a rod (eg a fishing rod) without needing anyone at the other end. All of the weight is at his end whichever end he chooses to be at.

15. Dec 8, 2015

### jbriggs444

If neither is allowed to apply a torque, you still have a concern that the force that each applies may not be purely vertical. Without resolving that, no definite answer is possible. If both apply purely vertical forces then the principles in #5 and #9 apply.

16. Dec 8, 2015

### sophiecentaur

I think you must mean a special (the simplest) case, in which they are just applying a force vertically. The share of the weight that each one gets will depend upon the position of the Centre of mass of the load. If you take moments about the CM, the clockwise moment for the guy at one end will be equal and opposite to the anticlockwise moment from the other guy. That will give you one equation.
Force1X(horizontal distance from CM to man1) = Force2X(horizontal distance from CM to man1)
You will need another equation and that is that the sum of the Forces is the total weight.
Two simultaneous equations will give you the two force values.
If the CM is half way along then the forces will be equal. (most trivial case)
But this is not a realistic situation and could be very hard for the two carriers to achieve unless they actually were holding the load with force meters and could make sure they were supporting half the weight and only lifting vertically. In practice, one of them will be applying more force by pulling / pushing in the horizontal plane and also applying some torque by multiple contact with the load.

17. Dec 9, 2015

### RiverM

If you lift one end of a barbell off the ground, and stand it vertically, it gets very easy when it's almost vertical.

If the barbell weighs 50 pounds, and you lift one end, you'd initially be lifting about 25 pounds right? And the ground would be holding the other 25 pounds. But when the bar is almost upright, you'd barely be holding any weight, just a slight push is all that's needed and the ground would be holding all 50 pounds when the bar is upright.

For this example, I think you can think of the lower person as being like the ground. Both ends always stay equal distance from the center of gravity, but the closer the higher end gets to being verticle, the more the weight is transferred to the lower end.

18. Dec 10, 2015

### CWatters

.

That's correct if you allow all the weight to be transferred to the ground. If you were to hang the top end from a metal stand with the bottom end touching the ground you wouldn't be able to calculate what percentage is carried by the stand or the ground.

19. Dec 10, 2015

### sophiecentaur

I would disagree with this. IMO, you are ignoring the horizontal forces involved in your situation of lifting the top of the barbell. If the barbell were resting on ice then there could be no horizontal forces, until the CM is directly below the upper lifting force, there will be equal shares (principle of moments). If you actually try the experiment, you will be pulling slightly sideways if you want to make the force you are applying greater than half the weight (assuming the CM is halfway along)
The situation of two guys carrying an object up stairs is far more complicated than the simple model that is being assumed on this thread.

20. Dec 10, 2015

### RiverM

Why couldn't you measure it? You could put a little scale between the bar and the stand, that would tell you the weight that is being bore by each end.

I think the question the original poster was driving at is, when you have an unlevel, symetrical object (like a cylinder or recangle), would it naturally place more load on the lower end?

If you lay one end of a bar on a scale, and lift the other end up till it's horizontal, you'll see that the weight on the scale will increase. It would start out roughly half and half at the bottom, then gradually increase till there was nothing in your hand and all the weight would be on the scale. So in this example, the scale could be a person's hands, holding the couch. The force they actually apply doesn't chance the amount of weight that the object is applying.

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