Finding the Centre of Mass of a Cubical Container

AI Thread Summary
The centre of mass of a cubical container with uniform density and negligible thickness can be determined using symmetry principles. Given the edge length of 40 cm, the x and z coordinates of the centre of mass are both 20 cm. The y coordinate is considered to be 0 due to the absence of a top on the container. To show work, one can illustrate the symmetry of the cube and explain how the coordinates are derived from its uniform dimensions. Thus, the centre of mass is located at (20, 0, 20).
suspenc3
Messages
400
Reaction score
0
A cubical container has been constructed from metal plate of uniform density and negligible thickness. The box has an edge length of 40 cm.

Find the x, y, & z coordinates of the centre of mass of the box.

Im guessing that the x and y coordinates are found by the symmetry..But how do i find the y coordinate?

Any help would be appreciated
 
Physics news on Phys.org
Now that I think about it, Wouldnt the y coordinate 0?

If this is right, is their any method to show my work, or is it all just found by symmetry?
 
suspenc3 said:
A cubical container has been constructed from metal plate of uniform density and negligible thickness. The box has an edge length of 40 cm.
Find the x, y, & z coordinates of the centre of mass of the box.
Im guessing that the x and y coordinates are found by the symmetry..But how do i find the y coordinate?
Any help would be appreciated

I'm guessing you mean how do you find the z coordinate. Since a box has uniform length in every dimension it is easy to find the COM in every component through symmetry.
 
what I meant was : "remember that the box has no top
I can find the centre of mass in the x & z directions through symmetry

How do i find the y component? then i thought about it and realized that it would be "0"...therefore the centre of mass is (20, 0, 20)?

How do I show work for this?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top