Two people carrying a beam of 50 kilogram , both are lifting 25 KG. Now they going up a staircase ,. Why is the bottom guy lifting more of the weight then the top guy. Is centre of gravity shifting , but why ???
Cheers John
Hi John:
The center of gravity does stay in the center of the beam. When the beam is level there is only force in the vertical direction due to gravity, so both people experience the same weight downward. It's actually a special case of a more complicated problem.
You have to take "moment" into account. Think of the weight of the beam as being concentrated at its center. Moment is like torque. If you think of one man as a pivot, the weight is causing a torque equal to the weight times the distance from the man. So if the weight is 50 kgf and it is two meters away (the beam being 4 m long) it would cause a moment of 50 x 2 = 100 m-kgf. In other words, if you didn't have help from the guy on the other end, you would have to twist your end with a torque of 50 m-kgf to hold the other end up.
But the guy on the other end is helping you: he is lifting upward and equalizing the moment in the other direction. If he is 4 meters away and equalizing 100 m-kgf of moment that takes 100 / 4 = 25 kgf. This may seem like the hard way to arrive at the same number, but it is universal for any tilt of the beam.
Let's look at the extreme case of the beam tilted entirely vertical. Now the 50 kg is entirely on one guy and there is NO moment. That's because the weight is acting downward, but at no distance, so the beam balances in the air and you experience no torque. So the guy at the other end doesn't have to equalize anything and he sees no force at all. For different angles of tilt the men will experience forces somewhere in between these extremes. In fact, the exact force that each man experiences (even if you place the wieght off center) can be calculated using the science of "statics".
Hope this helps.
Don
Dear John:
As eromlignod said, the center of gravity is not shifting. The nature of lifting is comprised of cupping one's hand and laying the beam in it (plus whatever gripping strength is added). The man up the stairs then is providing one support force in this fashion. And this net force is not normal to the direction gravity, but normal to the incline of the beam.
The lower man is providing an additional force. This force is required to keep the beam from sliding into him/her.
If the beam were fitted with pins at each end, and graduated spring scales, and each person held their scale in line with the action of gravity, and held the needles at 25 kg, the beam would go upstairs uniformly or stay stationary, no matter the inclination of the beam.
David A. Smith
The previous answers are incorrect.
If the beam is thin, the C.G is still midway between the two men, and they both sustain half of the total weight.
However, if the beam is thick, like a piece of furniture, and the men are lifting it from the bottom, as they normally would, then the C.G. is higher than the line between the men's hands. The C.G. and the two pairs of hands form a triangle. When they are on the stairs this triangle is tilted, moving the C.G. away from the upper man and toward the lower one. This puts more than half of the load on the lower man.
If that's not clear, draw a careful picture of two men carrying a chest of drawers up some stairs. Draw the C.G. and measure the vertical displacement of it from each of the two lifting points.
Mitchell Timin, former lisenced M.E.
The guy on the bottom takes a horizontal component of the gravity loads that the guy on the top does not have to carry.
Sincerely,
Donald L. Phillips, Jr., P.E. Worthington Engineering, Inc.
145 Greenglade Avenue Worthington, OH 43085-2264snipped-for-privacy@worthingtonNSengineering.com (remove NS to use the address)
614.937.0463 voice 208.975.1011 fax
So if the beam is at an 85 degree incline you're saying that the top guy will support the same load?
Nonsense.
Don
So if the beam is at an 85 degree incline you're saying that the top guy will support the same load? Nonsense.
Apparently you've never balanced a baseball bat on your hand. Who's holding up half the weight *then*?
Don
So if the beam is at an 85 degree incline you're saying that the top guy will support the same load? Nonsense.
Apparently you've never balanced a baseball bat on your hand. Who's holding up half the weight *then*?
Don
There is no horizontal component.
Don
Dear Don A. Gilmore:
Get tired of the anagram?
David A. Smith
Actually after thinking about this some more, this problem is not as clear-cut as it appears at first glance. Actually, when the beam is inclined either guy could carry more or less of the burden than the other. So, in a way it is statically indeterminate unless you know more about how each man is contributing.
It's ambiguous because there are two basic ways that a free body diagram could be constructed.
So I guess it all boils down to which guy is more macho and which one is being a pussy.
Don
Don Gilmore has hit on the answer......it is a question of support compliance (stiffness). From a practical POV the guy in the downstairs position typically is in a better position to apply more force & thus carries more of the load.
Bob
It is true that if the beam is vertical then the problem is statically indeterminate, and either man might carry any portion of the load.
But if we assume we are well away from vertical (as any stairs would be) and if we assume that both men support only vertical forces (which they should, otherwise they would be pushing against each other) and if we assume a very thin beam, then the C.G. is midway between the two men. The moment arm of their force vectors about the C.G. is the same, hence the forces must be the same.
I suppose there may be a tendency, in practice, for the two men to push against each other to some extent, especially if the upper man does not have an actual handle of some sort. The upper man may push backwards in order to prevent slippage, and of course the lower man must balance this force. Now if the two are not applying vertical force only, but are also pushing against each other, this will incline both force vectors toward the C.G. A sketch of the inclined beam, the C.G., and the force vectors shows that the moment arm becomes longer for the upper man's force and shorter for the other. This will require the lower man's force to be larger than the upper man's.
m (not licensed in spelling)
For a 10 inch deep beam held at its bottom corners, at a 45 degree slope the CofG shifts 5 inches back. For a 10 foot long beam, that's a shift of 5 ft 5 in / 4 ft 7 in or 18%
But at increasing angles, it's easy for the upper bearer to reduce uplift, but impossible for the lower bearer, without dropping the beam.
Brian W
Mitchell, your spelling may not be the best :-) but I'm with you!
Brian W
Actually, there *would* be a horizontal component of force (do a simple free body diagram). If there were not, you would be able to lean a beam, with rollers at both ends, against a wall without it slipping to the ground. This is why ladders need rubber feet at the bottom.
As I mentioned, this problem has multiple solutions depending on how you model it. Here are just two of many scenarios:
Situation 1: A man hangs the beam out of an upper-story window of a building. He now supports all the weight. Then a man below starts pulling the bottom end outward. Assuming he only pulls perpendicularly to the end of the beam, he will both introduce a horizontal force and increasingly experience a vertical force as the angle increases. When the beam is at 45 degrees, the horizontal force (tension) will be greatest. When the beam is level, the horizontal force has gone away and he now supports half the weight of the beam.
Situation 2: A man balances a beam straight up vertically in the air, supporting all the weight. Then a man above tips the beam slowly over. Assuming he only pushes perpendicularly to the beam there will again be a horizontal component, only this time it will be compressive with respect to the beam. Again it will be maximum at 45 degrees and again the beam will be equally loaded and with no remaining horizontal component when level.
There are many other scenarios as well. What if the beam is pivoted at its left end and leans against a wall with a roller at its right end? What would the compressive force be when the beam is just five degrees above horizontal? A lot! In fact it theoretically approaches infinity as we become level.
This really turns into a physiological problem for which no exact solution exists. We have to consider how each man holds the beam (are there handles? What stance do his surroundings allow him?).
Don
There may be a horizontal component of force, but there does not have to be. A man is not a wall. He has hands. If there are handles, or a decent handhold, the force may be vertical. But if there is not a handle of good handhold, then the upper man may be forced to push backwards as well as vertically. If this happens then the load on the lower man is increased as I explain in my previous post.
Then want makes it fall slide down the stairs when you set it down?
Sincerely,
Donald L. Phillips, Jr., P.E. Worthington Engineering, Inc.
145 Greenglade Avenue Worthington, OH 43085-2264snipped-for-privacy@worthingtonNSengineering.com (remove NS to use the address)
614.937.0463 voice 208.975.1011 fax
The normal reaction when a large flat surface is sitting across the edges of several steps will be at 90 degrees to the angle it makes... Then it has a horizontal component...
But for the guys carrying it, standing on the flat part of the steps.. there should be no horizontal component... (if they are carrying it that way...)
FWIW, whenever I help someone move a large item up stairs, I get the top person to push out and control the horizontal position, The guy on the bottom takes all the weight (from the outside edge), and pushes against the horizontal component that the top guy applies to keep it level... (this only works if the item is not too heavy for one person to lift, but it keeps a lot of dents out of walls due to the control that is achieved)
Al...
Unless the top guy is lifting straight up (in a plumb direction), and he's probably not, there will be a horizontal component.
I think what it all boils down to is that both guys will try to make their job as easy as possible. The bottom guy is pretty much screwed and will support most of the weight. The top guy will probably push up and outward on the top of the beam, approximately perpendicularly to the beam. This is his easiest option. So the model of a pivot at the bottom and a tangential force at the top is probably the most accurate.
Don
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