anonymous
  • anonymous
3 point mass of 4g placed at x=1,3,-6. where should a point fourth mass of 4 gm be placed to make the center of the mass at the orgin?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
is that 4 g mass distributed all on the x-axis at those points or are there other y -coordinates, or 3d coords?.
anonymous
  • anonymous
thry sre sll distributed on the x axis
anonymous
  • anonymous
they are all*

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anonymous
  • anonymous
i did the problem and got 4x-8/16 the answer is 2 but i do not know what i did wrong
anonymous
  • anonymous
give me a little, I'm working it out
anonymous
  • anonymous
k
anonymous
  • anonymous
it looks like the sum of all of the moments of mass (1g) * distance from the origin divided by the sum of the masses has to equal zero. so\[[(1g)(-6)+(1g)(1)+(1g)(2)+(1g)x]/(4g)=0\], x solves to be 2, which makes intuitive sense because on the -x side of the system you have a moment of -6g and if you were to add the moments on the +x side you would have 6g and a center of mass at 0
anonymous
  • anonymous
shouldn't the denominator be 4x4x4x4
anonymous
  • anonymous
no because the total amount of the mass is 4g, distributed over 4 points right?. Now if there are four points with 4g a piece, you would still use the same equation but instead of one gram *x you would use 4g*x and the denominator would be 4*4
anonymous
  • anonymous
center of mass = sum of the moments of mass / the sum of all of the masses. Whereas a moment is the mass *a considered distance
anonymous
  • anonymous
yes thats what i meant , its suppose to be 16
anonymous
  • anonymous
so shouldnt it be 4(x)+4(-6)+4(1)+4(3)/16=0
anonymous
  • anonymous
so then it would be 8
anonymous
  • anonymous
right
anonymous
  • anonymous
when i solved for it it become 4x-8/16=0
anonymous
  • anonymous
the 16 automatically falls out in the beginning being multiplied by 0
anonymous
  • anonymous
essentially it does not matter what the denominator is because the center of mass has to fall at zero in which case the numerator is the deciding factor
anonymous
  • anonymous
oh thats so funny
anonymous
  • anonymous
i didnt even realize that
anonymous
  • anonymous
It happens all of the time
anonymous
  • anonymous
ty u so much im gonna become a fan do u think u can help me with another problem of center and mass
anonymous
  • anonymous
sure, what's the question no problem by the way
anonymous
  • anonymous
a rod of length 2 meter and denisty =3-e^-x kg per meter placed on the x-axis with its ends at x=+1 and -1. find the coordinate of the center mass
anonymous
  • anonymous
i plugged into the formula but im stuck on the integration part, i think
anonymous
  • anonymous
so if the density of the rod is 3-e^(-x) the moment should be \[\rho \int\limits_{-1}^{1}xf(x)dx\]. But that formula is for uniform density. This question is saying that the density is changing in y for values of x. Unless I'm interpretting it wrong. Is the shape of the rod 3-e^-x from -1 to 1? If so, then we need avalue for the density coefficient. I would think
anonymous
  • anonymous
otherwise it is the density value of that function at zero, which 2, times the integral mentioned before,, where the f(x) = the line y=0, and therefore the center = 0
anonymous
  • anonymous
That has to be the case because, the question would not have mentioned that the rod lies on the x-axis. It would have said something like it follows the shape of the curve of a function above or below the x-axis on that interval
anonymous
  • anonymous
do u have a cramster account
anonymous
  • anonymous
I do.
anonymous
  • anonymous
if u can u look at the problem and help me
anonymous
  • anonymous
hughes hallet 5th ed number 23
anonymous
  • anonymous
I'm trying to pull it up, In the mean time I was thinking thatmaybe we could sue that density eq, and calculate the mass, and the moments at -1 and 1, then use the same formula as before.
anonymous
  • anonymous
what chapter?
anonymous
  • anonymous
8.4
anonymous
  • anonymous
ohh sry i meant to say 25
anonymous
  • anonymous
number 25
anonymous
  • anonymous
no drry 23
anonymous
  • anonymous
23 is right
anonymous
  • anonymous
it's not letting me get to either one for whatever reason, might be a prob with the site. Truly though, if it explicity says that 3-e^(-x) is the density of the rod. then they might be just asking for you to intuit that you can pull out the mass, (because the rod is 2m) and then calculate what the center would be, via the last problem.
anonymous
  • anonymous
k
anonymous
  • anonymous
does that make sense at all?
anonymous
  • anonymous
yah ill figure it out ty
anonymous
  • anonymous
i have to go but ill be back later ty for ur help

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