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

  • #1
tomlib
13
2
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
 

Attachments

  • úloha na nakloněné břemeno.png
    úloha na nakloněné břemeno.png
    15.7 KB · Views: 13
Physics news on Phys.org
  • #2
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:

1712856040348.png


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)##

##\ ##
 
Last edited:
  • Like
Likes berkeman

Similar threads

Replies
2
Views
1K
Replies
7
Views
901
  • Mechanical Engineering
Replies
8
Views
862
  • Mechanics
Replies
16
Views
4K
Replies
11
Views
1K
Replies
23
Views
4K
Replies
4
Views
3K
Replies
10
Views
1K
Back
Top