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Two people carrying an object up stairs

by jschwalbe
Tags: carrying, force, lifting, object, people, stairs
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jschwalbe
#1
May21-11, 10:59 PM
P: 2
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?
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rcgldr
#2
May22-11, 02:25 AM
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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.
Quinzio
#3
May22-11, 03:40 AM
P: 558
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:
[tex]mg/2\sqrt2[/tex]
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.

jschwalbe
#4
May22-11, 01:07 PM
P: 2
Two people carrying an object up stairs

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!
nrqed
#5
May22-11, 01:16 PM
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Quote Quote by jschwalbe View Post
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!
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.
Quinzio
#6
May22-11, 03:10 PM
P: 558
This is an example, I believe it's correct:
vega32
#7
Jan4-14, 04:45 AM
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I'm wondering the same thing. Can anyone else please confirm this solution? Thank you very much!
SteamKing
#8
Jan4-14, 08:54 AM
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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.
Jano L.
#9
Jan7-14, 01:38 PM
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Quote Quote by SteamKing View Post
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.


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.
sophiecentaur
#10
Jan8-14, 04:11 AM
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Quote Quote by Quinzio View Post
This is an example, I believe it's correct:
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.


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