A community for students.
Here's the question you clicked on:
 0 viewing
DLS
 4 years ago
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)
DLS
 4 years ago
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)

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Position of Cm = m1x1 + m2x2 + m3x3 / (m1+m2+m3)

DLS
 4 years ago
Best ResponseYou've already chosen the best response.0I didn't understand the coordinates,is it 1,0 2,0 & 3,0?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There are three Coordinates....For the 1 St one..CM is @ (1,2,3)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First Find the coordinates of m1 m2 and m3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.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...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait .. i know.. on call with gf :P.. 2 mins

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok.. 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
Ask your own question
Sign UpFind more explanations on OpenStudy
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
 Engagement 19 Mad Hatter
 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.