Question

pls give me the step by step ans for this question guys​

Answer 1

Let a coordinate system be introduced in the system where,

• O is taken as the origin.

• x axis should be along PQ such that +x is from O to Q, and -x is from O to P.

• y axis should be along RO such that +y is from O to R.

Given that the mass of each particle is 1 kg.

So the gravitational force of attraction experienced by the mass placed at R due that at P is,

$\displaystyle\longrightarrow F_P=\dfrac{G}{2L^2}\ \sf{along\ RP}$

since the length of RP is $L\sqrt2.$

And the gravitational force of attraction experienced by the mass placed at R due that at Q is also,

$\displaystyle\longrightarrow F_Q=\dfrac{G}{2L^2}\ \sf{along\ RQ}$

since the length of RQ is $L\sqrt2.$

Since $RP=RQ=L\sqrt2$ and $PQ=2L,$ we see that $\triangle PQR$ is a right triangle, right angled at R. So the net force acts along $RO.$

Hence the net gravitational force experienced by the mass placed at R due the two masses at P and Q is,

$\displaystyle\longrightarrow\Sigma F=\sqrt{\left(F_P\right)^2+\left(F_Q\right)^2+2\cdot F_P\cdot F_Q\cos90^o}$

$\displaystyle\longrightarrow \Sigma F=\sqrt{\left(\dfrac{G}{2L^2}\right)^2+\left(\dfrac{G}{2L^2}\right)^2}$

$\longrightarrow\Sigma F=\sqrt{\dfrac{G^2}{4L^4}+\dfrac{G^2}{4L^4}}$

$\longrightarrow\underline{\underline{\Sigma F=\dfrac{G}{\sqrt2\,L^2}\ \sf{along\ RO}}}$

Answer 2

Question:

Two particles each of mass 1kg are placed at P and Q such that PO=OQ=L.

The gravitational force experienced by another 1kg mass placed at R, where OR = L is

Theory :

⇒Newtons law of Gravitation

${\purple{\boxed{\large{\bold{F=\frac{Gm_{1}m_{2}}{r{}^{2}}}}}}}$

Where , G = Universal gravitation constant

and r is distance between two Masses.

Solution :

Given : Two particles each of mass 1kg are placed at P and Q such that

PO=OQ=L and OR = L

we have to find The gravitational force experienced by another 1kg mass placed at R .

__________________________

PO=OQ=L and OR = L

⇒PR = RQ = √2L

$\bf F_{rp} = \bf \frac{Gm {}^{2} }{( \sqrt{2}l) {}^{2} } =\bf \frac{Gm {}^{2} }{2l {}^{2} } = F$

$\bf F_{rq} = \bf \frac{Gm {}^{2} }{ ( \sqrt{2}l) {}^{2} } = F$

Now resultant of these forces

$\bf F_{net} = \sqrt{F{}^{2} + F{}^{2} }$

$= \sqrt{2} F$

$= \sqrt{2} \times \bf \frac{Gm {}^{2} }{2l {}^{2} }$

$= \bf \dfrac{Gm {}^{2} }{ \sqrt{2}l {}^{2} }$

put m = 1 kg

$= \bf \dfrac{G}{2l {}^{2} } \: along \: ro$

correct option 1)