Centre of gravity for multiple objects

[aab234]

Hi. This is my first post so apologies if I do anything wrong.

I have a homework question.

*calculate the distance of the centre of gravity from support A for a framework containing 4 crates at 52kg per crate.

I'll describe the diagram:

A is on the far left. 3m from A are 2 crates stacked on each other (so 104kg), 6m from A there is another crate, then another crate at 9m from A.
I'm disregarding support B at 12m from A.

My answer is that as the crates ate the same mass, we'll call them m.
I'll use x as the unknown distance I'm looking for. So:

3 (2m) + 6 (m) + 9 (m)

19m = 5mx
x = 3.8m

So the centre of gravity is 3.8m from point A.
Am I correct?

As I've seen pop up a few times on here I'm studying with ICS and they're...a little lacking in quality educational material so I've had to work it out on my own so if I'm shockingly incorrect a point in the right direction would be great.

Thanks.

 

[Simon Bridge]

Welcome to PF;
Does the answer make sense?

All masses are in a line? ... imagine they are sitting on a beam and don't worry about the supports for now.
The center of gravity is the position x where you would put a pivot and the beam would balance.

x=3.8m is 0.8m to the right of the 2 stacked crates ... a single crate 1.6m on the other side would balance it, but you have two crates much further away than that so the whole thing tips over clockwise.

If you sketch it out to scale and draw in a pivot at your value for x, you intuition should tell you the same. The same intuition should give you an idea of where the pivot should go, which will tell you if your maths comes out right.

 

[Chestermiller]

You analyzed it pretty OK, but you have to do the math correctly. The left hand side of your equation should be 21m, and the right hand side should be 4mx (you only had 4 crates, not 5).

Incidentally, Welcome to Physics Forums!

Chet

 

[aab234]

Doh! Maybe I should learn to add up first. Sigh...

So let's try that again.
3 (2m) + 6 ( m ) + 9 ( m )
21m = 4mx
x = 5.25

I've drawn a diagram and 5.25 metres from A as a fulcrum looks like it world balance so I think this is right.

Thanks for the help guys. I'll return the favour one day, when I've learned to add up obviously. How embarrassing.

 

[aab234]

*Would balance.

Seems I can't spell either. Not my finest hour I have to say.

 

[Simon Bridge]

2am effect? - have another coffee ;)

I've drawn a diagram and 5.25 metres from A as a fulcrum looks like it world balance so I think this is right.

... diagrams are the backbone of physics :)

If you rely too much on remembering equations, you cna get tied up in simple mistakes like this.
If you remembered about the balance-point rule, that wouldn't happen. Also you would not need to memorize the equation. Just remember that the sum of the moments must be zero.