The figure shows three point particles fixed in place in an xy plane. Particle A

Question

The figure shows three point particles fixed in place in an xy plane. Particle A has mass mA. particle B has mass 2mA. and particle C has mass 3mA. A fourth particle D, with mass 4mA, is to be placed near the other three particles In terms of die distance d, what are the x and y coordinates should particle D be placed so that the net gravitational force on particle A from particles B. C. and D is zero? In terms of d and mA, what is the gravitational potential energy of the system before particle D has been put in place?

Explanation / Answer

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gravitational force on A = force due to B+ force dueto C +force due to D

                        0          = FAB+ FAC+ FAD

Let D is placed at (x,y) coordinate

FAB= G(mA)(2mA)/d2 j (j means along y axis)

FAC= G(mA)(3mA)/(1.5d)2i = 4GmA2/3d2 (-i) (- i means along negative x axis)

FAD= G(mA)(4mA)/x2 i + G(mA)(4mA)/y2 j

Now sum of all forces in x direction as well as in y directions are 0.

G(mA)(4mA)/x2i + 4GmA2/3d2 i = 0

x = 3d i

Also,

G(mA)(4mA)/y2j + G(mA)(2mA)/d2 j = 0

y = d/2 (-j)

so D is placed at (3 d, - d/2) position.

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