Find Center of Mass: Attached Problem

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jacy
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hi,
I am finding the center of mass in this problem. I have it as an attachment. Please take a look, thanks.
 

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Your attachment shows three masses, but I can't tell their positions from what's written. (It's just not clear to me.)
 
Doc Al said:
Your attachment shows three masses, but I can't tell their positions from what's written. (It's just not clear to me.)

Here is that file again. Thanks for your help.
 

Attachments

OK, now it's a bit clearer. The first thing to realize is that the masses are not point masses, but have length. Assuming that they are uniform, each mass has its own center of mass, right at its center. When calculating the center of mass of the system, you need to measure the distance of the center of each mass from your reference point.

Give it another shot.
 
Doc Al said:
OK, now it's a bit clearer. The first thing to realize is that the masses are not point masses, but have length. Assuming that they are uniform, each mass has its own center of mass, right at its center. When calculating the center of mass of the system, you need to measure the distance of the center of each mass from your reference point.
Give it another shot.


Thanks again, so the distance for mass 1 will be 1m from my reference point, for mass 2 it will be 6m, for mass 3 it will be 8m. Am i correct.

center of mass = (20(1) + 30(6) + 40(8))/ 90
= 5.78 m
 
Recheck the distance to the cm of the third mass.
 
jacy said:
so the distance for mass 1 will be 1m from my reference point, for mass 2 it will be 6m, for mass 3 it will be 8m. Am i correct.
Measured from the left edge of mass 1, I'd say that your distances are correct for masses 1 and 2, but not for mass 3.
 
Doc Al said:
Measured from the left edge of mass 1, I'd say that your distances are correct for masses 1 and 2, but not for mass 3.


Thanks, the distance for mass 3 will be 10m from the left edge of mass 1, correct.

center of mass = (20(1) + 30(6) + 40(10))/90
= 6.67 m
This will be the answer, thanks for ur help.