resultant gravitational force

[chocolatelover]

Hi everyone,

Could someone please help me with this problem?
The problem statement, all variables and given/known data
Three uniform spheres of mass m1=2.00kg, 4.00kg, and m2=5.50 kg are placed at the corners of a right triangle. Calculate the resultant gravitational force on the 4.00 kg object, assuming the spheres are isolated from the rest of the Universe.

2. Relevant equations
Fg=m1m2/r^2


3. The attempt at a solution

F32=-6.67X10^-11Nm^2/kg(5.5kg)(4kg)/4^2= -9.2 X 10^-11i

F13= -6.67X10^-11Nm^2/kg^2(4)(2)/3^2= 5.9X10^-11j

Thank you very much

[Doc Al]

Please describe--even better, provide a diagram of--the triangle and its dimensions and the relative locations of the masses.

Recall that forces are vectors. You find the resultant by adding up the individual force vectors (add them like vectors, not plain numbers).

[chocolatelover]

Thank you

The triangle is placed in the second quadrent, where m3 is at (0,0) m2 is at (-4,0)m and m1 is at (0,3)

Does this look correct?
F32=-6.67X10^-11Nm^2/kg(5.5kg)(4kg)/4^2= -9.2 X 10^-11i

F13= -6.67X10^-11Nm^2/kg^2(4)(2)/3^2= 5.9X10^-11j

Thank you

[Doc Al]

I didn't check your arithmetic, but your setup looks good. Now find the magnitude of the resultant force.

[chocolatelover]

Thank you

I just need it in terms of i and j. So, wouldn't that be correct the way it is? F32 is negative and F13 is positive, right?

Thank you

[Doc Al]

The signs of your final answers are correct. But I don't know what you mean by F32 (versus F23). Is that the force on 3 from 2? Realize that you want the force on mass 3, and that F32 = -F23. (Just be consistent.)

[chocolatelover]

Thank you very much

Regards