two bodies of masses 3 kg and 12 kg are placed at a distance 12 m a third body of mass 0.5 kg is to be placed at such a point that the force acting body is zero find the position of that point
plzzzzz help me ..............
Answers
Answered by
0
Answer:
4m from 3kg and 9m from 12 kg body
Explanation:
Fnet=0
where Enet=0
GM2/x^2=GM1/(12-x)^2
M2=12 kg
M1=3 kg
here x is from heavy mass
Answered by
1
Answer:
m1=3kg ,m2=12kg , r=12m , m3=0.5kg
Let its distance from m1 be x m and from m2 be (12-x)m
Force of attraction on m3 by m1(F1)=GMm/r^2
=Gm1m3/x^2 ....(1)
Force of attraction on m3 by m2(F2)=GMm/r^2
=Gm2m3/(12-x)^2 ...(2)
Since the net force on the body is zero
Therefore F1=F2
Gm1m3/x^2=Gm2m3/(12-x)2
m1/x^2=m2/(12-x)2
(12-x)^2=x^2 * m2/m1
144+x^2-24x=x^2 * 12/4
144+x^2-24x=4x^2
4x^2-x^2+24x-144=0
3x^2+24-144=0
3x^2+36x-12x-144=0
3x(x+12)-12(x+12)=0
(x+12)(3x-12)=0
x=12/3
x=4 (neglecting x= -12)
So, the object is at a distance of 4m from m1 and 8m from m2.
Explanation:
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