Question

Two bodies of masses 3kg and and 12 kg are placed at a distance 12 m. A third body of mass 0.5kg is to be placed at such a point that the force acting on this body is zero. Find the position of that point

Answer 1

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/r2

=Gm1m3/x2 ....(1)

Force of attraction on m3 by m2(F2)=GMm/r2

=Gm2m3/(12-x)2 ...(2)

Siince the net force on the body is zero

Therefore F1=F2

Gm1m3/x2=Gm2m3/(12-x)2

m1/x2=m2/(12-x)2

(12-x)2=x2 * m2/m1

144+x2-24x=x2 * 12/4

144+x2-24x=4x2

4x2-x2+24x-144=0

3x2+24-144=0

3x2+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.

Answer 2

Answer:

here is ur answer.

Explanation:

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.