Calculating Initial Acceleration of Third Sphere in Gravitational Force Triangle

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
ladolce
Messages
56
Reaction score
0
1. Homework Statement

Three uniform spheres are located at the corners of an equilateral triangle. Each side of the triangle has a length of 3.80 m. Two of the spheres have a mass of 3.30 kg each. The third sphere (mass unknown) is released from rest. Considering only the gravitational forces that the spheres exert on each other, what is the magnitude of the initial acceleration of the third sphere?

2. Homework Equations

F=Gm1m2/d^2
F=ma

3. The Attempt at a Solution

I tried doing it by setting the 2 above equations = to each and so I got:

G(distance b/w 2 spheres)/d^2=a

and I don't know if I'm almost there or not...
 
Physics news on Phys.org
Yes, you are eventually going to have to set two force equations equal to each other, but you have to remember there is a force from EACH of the other spheres, meaning two forces in total.

Try starting here:

Remember that force is a vector.
What do you know about the X components of the two forces on the 3rd sphere?
What about the y components?

After you can answer these, you should be able to find the sum of the forces and then like you did above, remember that the sum of the forces is:

[tex]\Sigma \vec F=m\vec a[/tex]

See how far you can get now. Good Luck!
 
Last edited:
G01 said:
Yes, you are eventually going to have to set two force equations equal to each other, but you have to remember there is a force from EACH of the other spheres, meaning two forces in total.

Try starting here:

Remember that force is a vector.
What do you know about the X components of the two forces on the 3rd sphere?
What about the y components?

After you can answer these, you should be able to find the sum of the forces and then like you did above, remember that the sum of the forces is:

[tex]\Sigma \vec F=m\vec a[/tex]

See how far you can get now. Good Luck!

Thanks so much for helping but I got it yesterday =)
 
OK. Good for you!