Chapter 13, Problem 010 is partially correct. Try again. Two dimensions. In the

Question

Chapter 13, Problem 010 is partially correct. Try again. 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- 16 cm) 5d (a) N (b) N

Explanation / Answer

coordinates of the points are:

A:(0,0) m

B:(0,0.16) m

C: (-0.24,0) m

let D be placed at (x,y)

force on A due to B:

vector along the force=(0,0.16)-(0,0)=(0,0.16)

distance=0.16 m

unit vector=(0,0.16)/0.16=(0,1)

force magnitude=G*mA*mB/d^2

=G*mA*2*mA/d^2

=2*G*mA^2/d^2

force in vector notation=F1=(2*G*mA^2/d^2)*(0,1)

force on A due to C:

vector along the force=(-0.24,0)-(0,0)=(-0.24,0)

distance=0.24 m

unit vector=(-0.24,0.)/0.24=(-1,0)

force magnitude=G*mA*mC/(1.5*d)^2

=G*mA*3*mA/(1.5*d)^2

=1.33*G*mA^2/d^2

force in vector notation=F2=(1.33*G*mA^2/d^2)*(-1,0)

total force on A due to B and C=F1+F2

=(G*mA^2/d^2)*(-1.33,2)

force on A due to D should be opposite to this so that total force is zero.

hence force on A due to D=(G*mA^2/d^2)*(1.33,-2)

magnitude of force=(G*mA^2/d^2)*sqrt((1.33^2)+2^2)

=2.4019*G*mA^2/d^2

let distance of D from A be d1.

then force magnitude=G*mA*mD/dq^2

==>2.4019*G*mA^2/d^2=G*mA*4*mA/d1^2

==>d1=1.2905*d

direction of force=(1.33,-2)/sqrt(1.33^2+2^2)

=(0.55374,-0.83269)

hence x and y coordinates are:

1.2905*d*(0.55374,-0.83269)

=(11.434,-17.193) cm

so x coordinate of D=11.434 cm

y coordinate of D=-17.193 cm

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