What Is the Net Gravitational Force on One Mass in a Square Configuration?

  • Thread starter Thread starter godtripp
  • Start date Start date
  • Tags Tags
    Gravitation Net
AI Thread Summary
The discussion centers on calculating the net gravitational force on one mass in a square configuration of four identical 800 kg masses placed at the corners of a 10 cm square. The initial calculations involve determining the gravitational forces between the masses, with the contributor using the formula F = (GM^2)/(d^2) and considering the distances involved. However, it is pointed out that the net force must account for vector directions, leading to cancellations between forces F1 and F2. The correct net gravitational force is ultimately identified as 8.2 x 10^-3 N, highlighting the importance of superposition in vector calculations. The contributor acknowledges the oversight in their initial approach.
godtripp
Messages
53
Reaction score
0
Here's the question.

"Four identical masses of 800 kg each are placed at the corners of a square whose side length is 10.0 cm. What is the net gravitational force on one of the masses due to the other three?"

for convenience, I replaced mass with "M" and the distance with "d"

So, I figure that Net Force is equal to the sum of the gravitational forces between the three masses.

So if we call the force between the vertical mass and horizontal mass F1 and F2 respectively then F1=F2

Calculating out F1 gives me F= (GM^2)/(d^2)

And by Pythagorean the distance to the horizontal mass is d\sqrt{2}

so F3= (GM^2)/(2d^2)

So net force FN = 2F1+F3

or (5GM^2)/(2d^2).

However this is not the proper answer...


Proper answer is 8.2 x 10^-3 N

Can someone help me with this please? Thank you!
 
Physics news on Phys.org
Hi godtripp,

godtripp said:
Here's the question.

"Four identical masses of 800 kg each are placed at the corners of a square whose side length is 10.0 cm. What is the net gravitational force on one of the masses due to the other three?"

for convenience, I replaced mass with "M" and the distance with "d"

So, I figure that Net Force is equal to the sum of the gravitational forces between the three masses.

Remember that here the net force is a vector sum, so it will be the vector sum of the three individual forces.

So if we call the force between the vertical mass and horizontal mass F1 and F2 respectively then F1=F2

Their magnitudes are equal, but their directions are different.

Calculating out F1 gives me F= (GM^2)/(d^2)

And by Pythagorean the distance to the horizontal mass is d\sqrt{2}

so F3= (GM^2)/(2d^2)

So net force FN = 2F1+F3

When you perform the vector sum, there will be some cancellation occurring between F1 and F2 (so you cannot simply add the magnitudes together). Do you see what to do?

or (5GM^2)/(2d^2).

However this is not the proper answer...


Proper answer is 8.2 x 10^-3 N

Can someone help me with this please? Thank you!
 
Thank you so much for your reply. I totally neglected that the net force would have superposition.

Thanks again
 
godtripp said:
Thank you so much for your reply. I totally neglected that the net force would have superposition.

Thanks again

Sure, glad to help!
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top