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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?
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is that 4 g mass distributed all on the xaxis at those points or are there other y coordinates, or 3d coords?.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thry sre sll distributed on the x axis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i did the problem and got 4x8/16 the answer is 2 but i do not know what i did wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0give me a little, I'm working it out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it 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
 5 years ago
Best ResponseYou've already chosen the best response.0shouldn't the denominator be 4x4x4x4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no 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
 5 years ago
Best ResponseYou've already chosen the best response.0center 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
 5 years ago
Best ResponseYou've already chosen the best response.0yes thats what i meant , its suppose to be 16

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so shouldnt it be 4(x)+4(6)+4(1)+4(3)/16=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so then it would be 8

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when i solved for it it become 4x8/16=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 16 automatically falls out in the beginning being multiplied by 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0essentially 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
 5 years ago
Best ResponseYou've already chosen the best response.0i didnt even realize that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It happens all of the time

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ty u so much im gonna become a fan do u think u can help me with another problem of center and mass

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sure, what's the question no problem by the way

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a rod of length 2 meter and denisty =3e^x kg per meter placed on the xaxis with its ends at x=+1 and 1. find the coordinate of the center mass

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i plugged into the formula but im stuck on the integration part, i think

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if the density of the rod is 3e^(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 3e^x from 1 to 1? If so, then we need avalue for the density coefficient. I would think

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0otherwise 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
 5 years ago
Best ResponseYou've already chosen the best response.0That has to be the case because, the question would not have mentioned that the rod lies on the xaxis. It would have said something like it follows the shape of the curve of a function above or below the xaxis on that interval

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do u have a cramster account

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if u can u look at the problem and help me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hughes hallet 5th ed number 23

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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
 5 years ago
Best ResponseYou've already chosen the best response.0ohh sry i meant to say 25

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it'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 3e^(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
 5 years ago
Best ResponseYou've already chosen the best response.0does that make sense at all?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yah ill figure it out ty

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have to go but ill be back later ty for ur help
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