three particles of masses 1kg, 2kgand 3kg are subjected to forces of (3i-2j+2k)N, (-i+2j-k)N and (i+j+k)N respectively. What will be the magnitude of acceleration of centre of mass of the system?
2006-09-03 19:15:41 · 6 answers · asked by sweepie 2 in Science & Mathematics ➔ Physics
ans is underoot 22/6 but how!!?
2006-09-03 20:55:30 · update #1
a(cm) = ( m1a1 + m2a2 + m3a3)/(m1 + m2 + m3)
= {(3i-2j+2k)+(-2i+4j-2k)+(3i+3j+3k)}/6
a(cm) = 2/3 i + 5/6 j + 1/2 k
|a(cm)| = ((16+25+9)/36)^1/2
= 1.178 m/s2
2006-09-06 04:31:57 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Assumption: System of particles moves as rigid body.
Total force=3i+j+2k N
Total mass=6 kg;
magnitude of force=sqrt(3^2+1^2+2^2)
=sqrt(14)
magnitude of Acceleration=sqrt(14)/6 m/s^2
2006-09-05 02:18:36 · answer #2 · answered by arun 1 · 0⤊ 0⤋
as accelaration of centre of mass is= m1a1+m2a2+m3a3/m1+m2+m3 (let us take accelaration as a vector not magnitude) now m1a1 is the forces applied on mass m1 and similarly for others, so a=3i-2j+2k-i+2j-k+i+j+k/3+2+1=
3i+j+2k/6, now magnitude of accelaration will be square root of 9+1+4/36=square root of 14/36
2006-09-06 09:52:16 · answer #3 · answered by mahaveersoganiappu 2 · 0⤊ 0⤋
3/1i - 2/1j + 2/1k = 3i - 2j + 2k ms-2
Magnitude of acceleration = Root of (3^2 + (-2)^2 + 2^2)
same goes to the next questin..
2006-09-04 03:49:47 · answer #4 · answered by Mr. Logic 3 · 0⤊ 0⤋
F1+F2+F3+... =( ma+m2+m3+.... ) * a (cneter)
we shoul calculte the total F.
Fi= 3-1+1=3
Fj=-2+2+1 = 1
Fk = 2-1+1=2
F^2 = Fi^2 + Fj^2 + Fk^2 = 9+1+4=14
F= 14^0.5
a= F/m1+m2+m3
= 14^0.5 / 6
2006-09-04 05:34:30 · answer #5 · answered by M M 1 · 0⤊ 0⤋
dear sweety-pie;
if nothing else works-
try this:
mug-up all of them;
re-produce them in-toto;
(put faint crossing/cutting lines - here and there)
prof. would pick the "ONE",
that one of them ought to be your "saviour";
best of luck
many-que!!!
2006-09-08 02:12:51 · answer #6 · answered by many-ques!!! 2 · 0⤊ 0⤋