[QUOTE=Defective Detective]
Stranger, you’re usually fabulous with physics (especially theoretical physics), so I regret to say that in this instance, you are mistaken.
If you and I lift a piano of moderate weight, say 300 pounds, and it is level, we will each carry 1/2 of the load. However, if I lift my end higher than yours, I will certainly be carrying less than 150 lbs, and it has nothing to do with ergonomics. If I put each end of the piano on a scale, one higher than the other, they would register very different weights. What I want to figure out is, how does that change relate to the angle at which the mass is supported? Surely someone remembers their statics class better than I…
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When I say that each of the workers will be carrying a proportionate share of the load, I’m assuming that “a horse is a sphere to make the math easier” (which is an old engineering joke and Q.E.D. is rolling his eyes about now). If we’re talking about a rectangular box in which the center of mass is at the geometric centroid, then taking it by opposite corners–that is, the uphill guys grab the bottom forward corners, and the downhill guys grab the upper rear corners–will keep the load centered between them, whereas you are assuming that the workers each hold the bottom corners regardless. Now, of course it usually doesn’t work like that; in order to physically keep hold of the box and make sure it doesn’t fall backward the guys on the downhill usually have to get underneath it, getting closer to the centroid, and the guys at the top may have to rotate it up, and thus taking an even smaller portion of the load. But in an ideal world, where there are handles on all of the corners and the stairs don’t get in the way, each worker would be carrying only his percentage of the load. The only reason the piano movers would end up taking a higher portion of the load is because they don’t stay at the outer extents of the piano; the guys in back get gypped into moving closer to the c.g.
Regarding your latter question–what is the effect on static load when you rotate a object “hinged” at one end–it depends on how high the c.g. is when it is in the initial orientation. The higher the c.g. the faster load will be transfered to the hinged end, and the easier it is to cause it to tilt over center and fall; conversely, the lower it is, the further you have to push it to make it to over center, and thus it is more stable. (This is, all things being equal, why passenger cars have much lower incidence of rollovers than high-centered SUVs.) You can visualize this graphically by drawing a box and following the path of the c.g. as it rotates about a corner, and you can calculate this as a function of the horizontal component of the vector from the origin (hinged corner) to the c.g.
For instance, if the box is 8 feet wide and the c.g. is located 3 feet up from the bottom, the distance to the hinged corner is r= 5cos(arctan(3/4)+θ ). where θ is the angle at which you’ve rotated the box (so initially θ=0° ), while the horizontal distance to the lower non-hinge corner of the box where you are lifting is R=8cos(θ ). The ratio between these two (r/R) will give you how much of the load you’re carrying. (Technically you need to take the moment about the hinge point and set it equal to zero, then do the algebra, but it works out to the same thing.) Once you tilt the box more than 53.1° (for this particular example, I’ll leave it to you to back through the trigonometry), r≤0, the c.g. will be over the hinge, and now you’re going to have to pull back in order to keep it from falling over away from you.
I’ve looked online for an image to show this but didn’t find anything that satisfied me as being clear (although there are a lot of good statics tutorials online) but you can find this in any basic physics or statics book.
Stranger