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anonymous

  • 5 years ago

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?

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  1. anonymous
    • 5 years ago
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    is that 4 g mass distributed all on the x-axis at those points or are there other y -coordinates, or 3d coords?.

  2. anonymous
    • 5 years ago
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    thry sre sll distributed on the x axis

  3. anonymous
    • 5 years ago
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    they are all*

  4. anonymous
    • 5 years ago
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    i did the problem and got 4x-8/16 the answer is 2 but i do not know what i did wrong

  5. anonymous
    • 5 years ago
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    give me a little, I'm working it out

  6. anonymous
    • 5 years ago
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    k

  7. anonymous
    • 5 years ago
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    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

  8. anonymous
    • 5 years ago
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    shouldn't the denominator be 4x4x4x4

  9. anonymous
    • 5 years ago
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    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

  10. anonymous
    • 5 years ago
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    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

  11. anonymous
    • 5 years ago
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    yes thats what i meant , its suppose to be 16

  12. anonymous
    • 5 years ago
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    so shouldnt it be 4(x)+4(-6)+4(1)+4(3)/16=0

  13. anonymous
    • 5 years ago
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    so then it would be 8

  14. anonymous
    • 5 years ago
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    right

  15. anonymous
    • 5 years ago
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    when i solved for it it become 4x-8/16=0

  16. anonymous
    • 5 years ago
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    the 16 automatically falls out in the beginning being multiplied by 0

  17. anonymous
    • 5 years ago
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    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

  18. anonymous
    • 5 years ago
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    oh thats so funny

  19. anonymous
    • 5 years ago
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    i didnt even realize that

  20. anonymous
    • 5 years ago
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    It happens all of the time

  21. anonymous
    • 5 years ago
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    ty u so much im gonna become a fan do u think u can help me with another problem of center and mass

  22. anonymous
    • 5 years ago
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    sure, what's the question no problem by the way

  23. anonymous
    • 5 years ago
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    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

  24. anonymous
    • 5 years ago
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    i plugged into the formula but im stuck on the integration part, i think

  25. anonymous
    • 5 years ago
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    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

  26. anonymous
    • 5 years ago
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    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

  27. anonymous
    • 5 years ago
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    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

  28. anonymous
    • 5 years ago
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    do u have a cramster account

  29. anonymous
    • 5 years ago
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    I do.

  30. anonymous
    • 5 years ago
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    if u can u look at the problem and help me

  31. anonymous
    • 5 years ago
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    hughes hallet 5th ed number 23

  32. anonymous
    • 5 years ago
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    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.

  33. anonymous
    • 5 years ago
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    what chapter?

  34. anonymous
    • 5 years ago
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    8.4

  35. anonymous
    • 5 years ago
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    ohh sry i meant to say 25

  36. anonymous
    • 5 years ago
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    number 25

  37. anonymous
    • 5 years ago
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    no drry 23

  38. anonymous
    • 5 years ago
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    23 is right

  39. anonymous
    • 5 years ago
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    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.

  40. anonymous
    • 5 years ago
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    k

  41. anonymous
    • 5 years ago
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    does that make sense at all?

  42. anonymous
    • 5 years ago
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    yah ill figure it out ty

  43. anonymous
    • 5 years ago
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    i have to go but ill be back later ty for ur help

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