Question

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?

Answer 1

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

Answer 2

thry sre sll distributed on the x axis

Answer 3

they are all*

Answer 4

i did the problem and got 4x-8/16 the answer is 2 but i do not know what i did wrong

Answer 5

give me a little, I'm working it out

Answer 6

k

Answer 7

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

Answer 8

shouldn't the denominator be 4x4x4x4

Answer 9

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

Answer 10

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

Answer 11

yes thats what i meant , its suppose to be 16

Answer 12

so shouldnt it be 4(x)+4(-6)+4(1)+4(3)/16=0

Answer 13

so then it would be 8

Answer 14

right

Answer 15

when i solved for it it become 4x-8/16=0

Answer 16

the 16 automatically falls out in the beginning being multiplied by 0

Answer 17

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

Answer 18

oh thats so funny

Answer 19

i didnt even realize that

Answer 20

It happens all of the time

Answer 21

ty u so much im gonna become a fan do u think u can help me with another problem of center and mass

Answer 22

sure, what's the question
no problem by the way

Answer 23

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

Answer 24

i plugged into the formula but im stuck on the integration part, i think

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

do u have a cramster account

Answer 29

I do.

Answer 30

if u can u look at the problem and help me

Answer 31

hughes hallet 5th ed number 23

Answer 32

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.

Answer 33

what chapter?

Answer 34

8.4

Answer 35

ohh sry i meant to say 25

Answer 36

number 25

Answer 37

no drry 23

Answer 38

23 is right

Answer 39

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.

Answer 40

k

Answer 41

does that make sense at all?

Answer 42

yah ill figure it out ty

Answer 43

i have to go but ill be back later ty for ur help