Two dimensions. In the figure, three point particles are fixed in place in an xy

Question

Two dimensions. In the figure, three point particles are fixed in place in an xy plane. Particle A has mass mA = 3 g, particle B has mass 2.00mA, and particle C has mass 3.00mA. A fourth particle D, with mass 4.00mA, is to be placed near the other three particles. What (a) x coordinate and (b) y coordinate should particle D be placed so that the net gravitational force on particle A from particles B, C, and D is zero (d = 25 cm)?

Two dimensions. In the figure, three point particles are fixed in place in an xy plane. Particle A has mass mA = 3 g, particle B has mass 2.00mA, and particle C has mass 3.00mA. A fourth particle D, with mass 4.00mA, is to be placed near the other three particles. What (a) x coordinate and (b) y coordinate should particle D be placed so that the net gravitational force on particle A from particles B, C, and D is zero (d = 25 cm)?

Explanation / Answer

Let the mass of A be m,so masses of B,C and D will be 2m,3m,4m repectively.and let mass D is placed at (x,y) so that net gravitational force acting on it is zero.as the gravitational force is attractive in nature it will be towards A due to A and so on.

Fnet = FD--A + F D--B + FD--C =0

as gravitational force is given by =GM1M2 / R2

SO above equation can be written as

Fnet = GMD( MA / RA2 + MB / RB2 + MC / RC2 )

( MA / RA2 + MB / RB2 + MC / RC2 ) = 0

m ( 1 / RA2 + 2 / RB2 + 3 / RC2 ) =0

here RA2 =(X-0)2 + (Y-0)2

RA2 =(X-0)2 + (Y-0)2

RB2 =(X-0)2 + (Y-0.25)2

RC2 = (X-0.375)2 + (Y-0)2

so equation will come

1 / + 2 / RB2 + 3 / RC2 =0

1/X2+Y2 + 2/X2 + (Y-0.25)2 + 3/ (X-0.375)2 + Y2 =0

X=0.125 Y=0.0833

which are the centroid of right angle triangle at which net gravitational force will be equal to zero.

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