Question

answer....................​

Answer 1

QUESTION:

If four identical particles of mass m are kept at the four vertices of a sqaure of side length a the gravitational force of attraction on anyone of the particle is.

ANSWER:

$\sf F_{net} = \dfrac{Gm^2}{a^2} \left({ \sqrt{2} }+ \dfrac{1}{2} \right)$

GIVEN:

• Four identical particles of mass m are kept at the four vertices of a sqaure.

• Side length of the square is a.

TO FIND:

• The gravitational force of attraction on anyone of the particle.

EXPLANATION:

DIAGRAM:

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0) \linethickness{0.2mm}\qbezier(0,0)(0,0)(2,0)\qbezier(2,0)(2,0)(2,2)\qbezier(2,2)(2,2)(0,2)\qbezier(0,2)(0,2)(0,0)\put(0,0){\vector(1,0){2}} \put(0,0){\vector(0,1){2}}\put(0,0){\vector(1,1){1}}\put(0,0){\vector(1,1){1}}\put(0,0){\vector(1,1){1}}\put(0,0){\vector(1,1){1}}\put(0,0){\vector(1,1){1}}\put(0,0){\vector(1,1){0.6}}\put( - 0.2, - 0.3){ \bf C }\put( - 0.2, 2.1){ \bf B }\put( 2, 2.1){ \bf A }\put( 2, - 0.3){ \bf D }\put(1.1,0.8){ \bf F_{AC} }\put(0.58,0.3){ \bf F_R }\put(- 0.75,0.9){ \bf F_{CB} }\put(0.58, - 0.3){ \bf F_{CD} }\end{picture}$

Let we calculate the force on the particle C.

$\boxed{ \bold{ \large{ \gray{F =\dfrac{Gmm}{r^2}}}}}$

$\sf F_{CD} = F_{CB} = \dfrac{Gm^2}{a^2} = F$

$\sf We \ want \ to \ \ find \ the \ resultant \ of \ F_{CD} \ and \ F_{CB}$

$\sf F_{R} = \sqrt{(F_{CD})^2 + (F_{CB})^2 + 2F_{CD}F_{CB}Cos\ \theta}$

$\sf F_{R} = \sqrt{F^2 + F^2 + 2F(F)Cos\ 90^{ \circ} }$

As the each angles of the square is 90°.

$\sf F_{R} = \sqrt{2F^2 }$

$\sf F_{R} = \sqrt{2} \ F$

$\sf F_{net} = F_R + F_{AC}$

$\sf Distance\ between\ A\ and\ C = \sqrt{a^2 + a^2}$

$\sf Distance\ between\ A\ and\ C = \sqrt{2a^2}$

$\sf Distance\ between\ A\ and\ C = \sqrt{2} \ a$

$\sf F_{AC} = \dfrac{Gm^2}{(\sqrt{2}a)^2}$

$\sf F_{AC} = \dfrac{Gm^2}{2a^2}$

$\sf F_{R} = \dfrac{\sqrt{2} \ Gm^2}{a^2}$

$\sf F_{net} = \dfrac{\sqrt{2} \ Gm^2}{a^2} + \dfrac{Gm^2}{2a^2}$

$\sf Take \ \dfrac{\ Gm^2}{a^2}\ as\ common.$

$\sf F_{net} = \dfrac{Gm^2}{a^2} \left({ \sqrt{2} }+ \dfrac{1}{2} \right)$

FORCE ACTING ON ANY ONE OF THE BODY WILL EQUAL TO THE NET FORCE.

$\sf Net \ force = \dfrac{Gm^2}{a^2} \left({ \sqrt{2} }+ \dfrac{1}{2} \right)$

Answer 2

QUESTION:

If four identical particles of mass m are kept at the four vertices of a sqaure of side length a the gravitational force of attraction on anyone of the particle is.

ANSWER:

\sf F_{net} = \dfrac{Gm^2}{a^2} \left({ \sqrt{2} }+ \dfrac{1}{2} \right)Fnet​=a2Gm2​(2​+21​)

GIVEN:

Four identical particles of mass m are kept at the four vertices of a sqaure.

Side length of the square is a.

TO FIND:

The gravitational force of attraction on anyone of the particle.

EXPLANATION:

DIAGRAM:


Let we calculate the force on the particle C.
\boxed{ \bold{ \large{ \gray{F =\dfrac{Gmm}{r^2}}}}}F=r2Gmm​​
\sf F_{CD} = F_{CB} = \dfrac{Gm^2}{a^2} = FFCD​=FCB​=a2Gm2​=F
\sf We \ want \ to \ \ find \ the \ resultant \ of \ F_{CD} \ and \ F_{CB}We want to  find the resultant of FCD​ and FCB​
\sf F_{R} = \sqrt{(F_{CD})^2 + (F_{CB})^2 + 2F_{CD}F_{CB}Cos\ \theta}FR​=(FCD​)2+(FCB​)2+2FCD​FCB​Cos θ​
\sf F_{R} = \sqrt{F^2 + F^2 + 2F(F)Cos\ 90^{ \circ} }FR​=F2+F2+2F(F)Cos 90∘​
As the each angles of the square is 90°.
\sf F_{R} = \sqrt{2F^2 }FR​=2F2​
\sf F_{R} = \sqrt{2} \ FFR​=2​ F
\sf F_{net} = F_R + F_{AC}Fnet​=FR​+FAC​
\sf Distance\ between\ A\ and\ C = \sqrt{a^2 + a^2}Distance between A and C=a2+a2​
\sf Distance\ between\ A\ and\ C = \sqrt{2a^2}Distance between A and C=2a2​
\sf Distance\ between\ A\ and\ C = \sqrt{2} \ aDistance between A and C=2​ a
\sf F_{AC} = \dfrac{Gm^2}{(\sqrt{2}a)^2}FAC​=(2​a)2Gm2​
\sf F_{AC} = \dfrac{Gm^2}{2a^2}FAC​=2a2Gm2​
\sf F_{R} = \dfrac{\sqrt{2} \ Gm^2}{a^2}FR​=a22​