Question

two point masses A and B have masses in the ratio 1:4 the point masses are at distance from point mass B , be a point mass m placed so that at that point the resultant force on it is zero?​

Answer 1

Answer:

Explanation:

Let a point mass C is placed at a distance of x m from the point mass A as shown in the figure

Here, MAMB=43

Force between A and C is

FAC=GMMAx2..........(i)

Force between B and C is

FBC=GMMB(1−x)2..........(ii)

According to given problem

FAC=13FBC

∴GMAMx2=13(GMMB(1−x)2) (using (i) and (ii))

MAx2=MB3(1−x)2orMAMB=x23(1−x)2

43=x23(1−x)2or4=x2(1−x)2

or 2=x1−xor2−2x=x

3x=2orx=23m

Answer 2

Given:

Ratio of two point masses $=1:4$

Separation between the two point masses $=1m$

To find:

The distance from the point B, where a point mass m should be placed so that the resultant force on m is zero.

Solution:

We have been given the ratio of two point masses of mass $m_{A}$ and $m_{B}$ as $1:4$ and the point masses are separated by a distance $R=1m$.

Lets assume the point mass $m$ is located at $P,$ which is at a distance of $x$ meters from $A$.

Hence, the distance of point $P$ from $B$ $=(1-x)$ meter

Now,

According to the question, the force on point mass $m$ is zero, that is,

$F_{AP}+ F_{BP}=0$

$F_{AP}=- F_{BP}$

$G\frac{m_{A} *m }{x^{2} } =-G\frac{m_{B}*m }{(1-x)^{2} }$

$\frac{m_{A} }{x^{2} }= \frac{m_{B} }{(1-x)^{2} }$

$\frac{m_{A} }{m_{B} } =\frac{x^{2} }{(1-x)^{2} }$

We know,  $\frac{m_{A} }{m_{B} } =\frac{1}{4}$

$\frac{x^{2} }{(1-x)^{2} }=\frac{1}{4}$

$4x^{2} =(1-2x+x^{2} )$

$3x^{2} +2x-1=0$

$3x^{2} +3x-x-1$ $=0$

$3x(x+1)-1(x+1)=0$

$(3x-1)(x+1)=0$

$x=-1 , \frac{1}{3}$

Distance can never be negative hence, the distance of point mass $m$ from $A$ is $\frac{1}{3}$ meters and that from point $B$ is $\frac{2}{3}$ meters.

Final answer:

Hence, the distance of point mass m from B so that the resultant force on P is zero is $\frac{2}{3}$ meters.

Although your question is incomplete, you might be referring to the question below.

Two point masses A and B having masses in the ratio 1 : 4 are separated by a distance 1 m. At what distance from point mass B, a point mass m should be placed so that at that point the resultant force is zero.