Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

DLS Group Title

If the centre of mass of three particles of masses 10,20 and 30g is at the point(1,-2,3)then where should a fourth particle of mass 40g be placed,so that the combined centre of mass of the system is (1,1,1)

  • one year ago
  • one year ago

  • This Question is Closed
  1. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @mayankdevnani

    • one year ago
  2. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @Yahoo!

    • one year ago
  3. Yahoo! Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Position of Cm = m1x1 + m2x2 + m3x3 / (m1+m2+m3)

    • one year ago
  4. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I didn't understand the co-ordinates,is it 1,0 -2,0 & 3,0?

    • one year ago
  5. Yahoo! Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    There are three Co-ordinates....For the 1 St one..CM is @ (1,-2,3)

    • one year ago
  6. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    10-40+90/60 =1?

    • one year ago
  7. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i didnt get it what to do

    • one year ago
  8. Yahoo! Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    First Find the co-ordinates of m1 m2 and m3

    • one year ago
  9. Yahoo! Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    (1,-2,3 ) x , y , z x axis of Cm = m1x1 + m2x2 + m3x3 / (m1+m2+m3) y axis of Cm = m1y1 + m2y2 + m3y3 / (m1+m2+m3) etc...

    • one year ago
  10. Yahoo! Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    u got it..)

    • one year ago
  11. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks :D then?

    • one year ago
  12. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @Mashy

    • one year ago
  13. Mashy Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    wait .. i know.. on call with gf :P.. 2 mins

    • one year ago
  14. Mashy Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    ok.. so let xcm be the centre of mass of the three particle system then \[Mxcm = m1x1+ m2x2 + m3x3 \] now .. when the fourth particle is added.. the centre of mass shifts hence \[M'xcm' = m1x1+ m2x2 + m3x3+ m4x4\] or \[M'xcm' = Mxcm + m4x4\] the only thing you have to find now is x4.. so calculate.. do the same in other 2 axes

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.