Superposition Gravataional Forces

[koab1mjr]

Homework Statement

There are three point particles in fixed positions in an xy plane. Particle A lies at the origin and particle B lies 150 degrees from positive A (reference is positive x axis). The distance between P_a and P_b is .5m. The mass of P_a is 6.00g and the mass of P_b is 12.0g. The position of the third particle C is not known but the mass is 8.00g. The net forces acting on particle A (only gravataional) magnittude is 2.77*10^-14 at an angle of -163.8 degrees. Find the x and y coordiantes of particle C

Homework Equations

Need the gravatational force equation F_g = Gm₁m₂/r²

The Attempt at a Solution

This problem is straight forward. Knowing that the force from particle B on particle A plus the coresponding force from particle C must equal the resulting force. So to solve I will break it up into x and y componenet and find the quadanents. So i am working along with the following

F_{net on particle A} cos(-163.8) = F_{gravity AB} cos(150) + F_{gravity AC}COS(theta)

So I subtract the first term on the right from the one on the left. And here is where things go to hell. I have some x component of a force. on the left. I have on the right two unknowns I do not know the distance between A and C, and I do not know x position of particle see which i want to know. I am stuck on how to proceed. I am sure this becomes a pythagorean thing but not sure. Any help much appreciated.

 

[rl.bhat]

If you draw the forces in a co-ordinate system, you can see that the particle c lies in the fourth quadrant. Now take the X and Y components of F(AB), F(AC) and F(net)
X-component of F(net) = X-component of F(AB) +X-component of F(AC). Similarly for Y-components. From these you can find the angle made by F(AC) with X -axis, and the rest of the things.

 

[thomate1]

Hello, it seems like you are trying to solve for the x and y coordinates of particle C using the forces acting on particle A. While this approach may work, it might be easier to use the superposition principle in this case.

The superposition principle states that the net force on a particle is equal to the vector sum of all the individual forces acting on it. In this case, we have three forces acting on particle A: the force from particle B, the force from particle C, and the resultant force. By using the superposition principle, we can set up the following equations:

Fnet = Fgravity AB + Fgravity AC

Fnet = 2.77*10^-14 at an angle of -163.8 degrees

Fgravity AB = Gm1m2/r^2

Fgravity AC = Gm1m3/r^2

Since we know the masses and the distance between particles A and B, we can solve for the magnitude of Fgravity AB. However, we don't know the distance between particles A and C or the angle at which particle C is located. To solve for these unknowns, we can use trigonometry and set up a system of equations.

Let's start by finding the x and y components of Fnet and Fgravity AB:

Fnetx = 2.77*10^-14 cos(-163.8) = -2.77*10^-14

Fnety = 2.77*10^-14 sin(-163.8) = -1.8*10^-14

Fgravity ABx = Fgravity AB cos(150) = 2.4*10^-14

Fgravity ABy = Fgravity AB sin(150) = 1.2*10^-14

Now, we can set up the following equations:

Fnetx = Fgravity ABx + Fgravity ACx

Fnety = Fgravity ABy + Fgravity ACy

We know the values for Fnetx and Fnety, and we can solve for Fgravity ACx and Fgravity ACy using the equations above. Once we have those values, we can use trigonometry to find the distance between particles A and C and the angle at which particle C is located. This will give us the x and y coordinates for particle C.

I hope this helps! Remember to always use the superposition principle when dealing with multiple forces acting on a