B How Does Inclined Load Affect Weight Distribution When Moving a Cabinet?

[tomlib]

TL;DR Summary

How the weight changes on the sides of the body/cabinet if it is tilted

Hello. I want to ask for advice. I know I probably won't understand it anyway, but maybe some information will help me. I did not find a solution to this problem on the Internet, which in itself I do not understand. We moved the cabinet and I can't calculate how much the weight on the opposite side will change if the cabinet is lifted on one side.
Likewise if it is held on the bottom or top edge. Is it necessary to take into account that the cabinet has mass in the sides and not in the center of the body? I was able to determine that if the cabinet was hung at one end only, the suspension point would bear the full weight, if it were raised at the other end, this weight would gradually decrease, but how and by how much? Perhaps the crank relation of cosine on a circle applies. Similarly, I would be interested, if two movers carry a cabinet down the stairs, how much weight will the lower mover carry. The inclined plane is applied here, but also the force that must lift the cabinet upwards. Again, there is probably a paradox in view of the above, that it seems that the top end at the top of the stairs will be lighter. Can someone give me some information? I would also welcome a logical reasoning.
So I'm actually interested in cases where the cabinet is attached at points A and B and then also at points C and B. The slope is 27°, the total weight of the cabinet can be 42 kg.

(I used Google Translator)

tomlib

 

[BvU]

Hi,

Key concept here is torque : the case is not moving and not rotating.
If the mass of the case is not distributed asymmetrically, the center of mass is in the middle as drawn here:

and equilibrium means two requirements are satisfied:

1. Force balance : $F_a + F_b + mg = 0$. Since $mg = -420$ N you get $F_a + F_b = 420$ N

2. Torque balance: take the center of mass as axis of rotation, then no rotation takes place if
$x_a\times F_a - x_b\times F_b = 0$

With $AC = p$ and $AB=q$ you have $AD = \sqrt{p^2+q^2}$ and with an angle of $\theta = 27^\circ$ you have $x_a+x_b=p\cos\theta$ and $x_a = {1\over 2} (p\cos\theta - q \sin\theta)$ and of course $x_b = {1\over 2} (p\cos\theta + q \sin\theta)$

$\$