Gravitation in 2D Homework: Find x,y Coordinates of Particle D

[Gold3nlily]

Homework Statement

Two dimensions. In Figure 13-34, three point particles are fixed in place in an xy plane. Particle A has mass mA = 7 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 = 24 cm)?

Figure:
http://edugen.wiley.com/edugen/courses/crs4957/art/qb/qu/c13/qu_1.14.gif

Homework Equations

F = (GMm)/r²
F_ab + F_ac + F_ad = 0
split into vector components?
G = 6.67e-11

The Attempt at a Solution

X component:
(no F_ab b/c that's all in y component)
F_ac + F_ad,x = 0
F_ac = (GM_am_c)/r_ac²
F_ac = [(0.007)(.021)(6.67e-11)]/ (.36)^2 = 7.57e-14
F_ad,x = - 7.57e-14

So use force to solve for distance (x component of F_ad): r_ad
F_ad = (GM_am_d)/(r_ad)²
r_ad² = [(0.007)(.028)(6.67e-11)]/ - 7.57e-14
r_ad = - 0.457m

Before I type up the y component... What should I do differently?

 

[gneill]

Rather than trying to solve for the r_ad components as you go (which I'm not sure is practical anyways), I would first obtain the net force vector due to B and C on A, then use the magnitude of that force vector to find the scalar distance r for mass D using Newton. That then will be the magnitude of the distance vector. For direction, multiply the magnitude by a unit vector obtained from the previous net force vector (be sure to reverse its direction!).

 

[Gold3nlily]

That then will be the magnitude of the distance vector. For direction, multiply the magnitude by a unit vector obtained from the previous net force vector (be sure to reverse its direction!).

F_ac + F_ab = -F_ad
F_ad = -1.307e^-14 Is this looking right so far?

F_ad = (Gm_am_d)/(r²)
(can't change the direction yet or answer isn't real)
+1.307e^-14 = (6.67e^-11*0.007_a4*0.007_d)/(r²)
r = 0.263

It is the next part that confuses me. I didn't find a vector for the previous net force, I just found the force. How do I do this?

Is the vector something like this: (7.57e^-14)i + (1.13e^-13)j?
I don't think this makes sense and i think my net force for B and C on A would be different if this were the vector.

I thank you for the help. If you could please explain a little more I would appreciate it greatly.

 

[gneill]

If you first find the net force due to masses B and C (or alternatively, the net gravitational acceleration at the origin due to them) in component form, you have the net force vector. The values that I see for that vector's components are:

f = (-7.570 x 10^-14, 1.136 x 10^-13)N

The magnitude of that vector is |f| = 1.365 x 10^-13N

So the "new" mass, mass D, should apply an equal and oppositely directed force. The distance r that D must be placed from A can be obtained using |f| in Newton's law of gravitation with masses A and D.

The vector for the position of D will be r multiplied by a unit vector in the opposite direction of the force vector f. Do you know how to obtain that unit vector?

 

[Gold3nlily]

The vector for the position of D will be r multiplied by a unit vector in the opposite direction of the force vector f. Do you know how to obtain that unit vector?

Thank you!

So the net force vector makes sense to me now. It makes sense that the net force is equal to the magnitude of this vector. (I verified the numbers on my own too).

SO now I am finding "The distance r that D must be placed from A can be obtained using |f| in Newton's law of gravitation with masses A and D."
F_ad = (Gm_am_d)/(r²)
1.365e^-13 = (6.67e-11*0.007a4*0.007d)/(r2)
r = 0.310028

For the next part, I tried this:
Fnet,x * r = -7.57e^-14* 0.310028 = -2.35e^-14
then I change the direction by changing the sign:
F_d,x = 2.35e^-14
Fnet,y * r = 1.13ee^-13* 0.310028 = 3.50^-14
then I change the direction by changing the sign:
F_d,y = -3.50e^-14

How am I doing?

 

[gneill]

I'm not sure what you're hoping to accomplish by multiplying the force components by the radial distance -- the units would be energy (work = force x distance).

A unit vector approach would be my own choice. The unit vector for the net force due to B and C is given by u = f/|f|. That is, divide each component of the net force vector by the magnitude of the overall vector. Now, to reverse the direction of this unit vector (so that it points in the direction you want from the origin to where D is to be located), simply negate the signs of the components. Scale this by your radial distance r and you're done.