Four point mass bodies of masses as shown in figure are placed at the verticles of a square $A B C D$ where $A B = B C = C D = D A = 1.0 m$ .find the net gravitational force on the body at $A$ . Given $G = 6.6 \times 10^{- 11} N m^{2} K g^{- 2}$

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Solution

$F_{A B} = \frac{G \times 4 \times 2}{(1)^{2}}$
$= 8 G$
$FAC=dfracG×2×(√2)2$
$= \frac{4 G}{2} = 2 G$
$F_{A D} = \frac{G \times 4 \times 2}{(1)^{2}}$
$= 8 G$
Resultant of $F_{A R}$ and $F_{A D}$
$R_{i} = \sqrt{F_{A B}^{2} + F_{A D}^{2}} = \sqrt{(8 G) + (8 G)^{2}}$
$\sqrt{128 G^{2}} = 8 \sqrt{2} G$
Net force= $R_{1} + F_{n c}$
$= 8 \sqrt{2 G} + 2 G$
$(8 \sqrt{2} + 2) G$
$(8 \times 1.414 + 2) \times 6.60 \times 10^{- 11}$
$(11.312 + 2) \times 6.6 \times 10^{- 11}$

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Four point mass bodies of masses as shown in figure are placed at the verticles of a square ABCD where AB=BC=CD=DA=1.0m.find the net gravitational force on the body at A. Given G=6.6×10−11Nm2Kg−2

FAB=G×4×2(1)2=8GFAC=dfracG×2×(√2)2=4G2=2GFAD=G×4×2(1)2=8GResultant of FAR and FADRi=√F2AB+F2AD=√(8G)+(8G)2√128G2=8√2GNet force=R1+Fnc=8√2G+2G(8√2+2)G(8×1.414+2)×6.60×10−11(11.312+2)×6.6×10−11

Four point mass bodies of masses as shown in figure are placed at the verticles of a square $A B C D$ where $A B = B C = C D = D A = 1.0 m$ .find the net gravitational force on the body at $A$ . Given $G = 6.6 \times 10^{- 11} N m^{2} K g^{- 2}$