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Gravitational force among 4 spheres

  • Thread starter gillyr2
  • Start date
45
0
1. Homework Statement
Four 7.5kg spheres are located at the corners of a square of side 0.70 m.
Calculate the magnitude of the gravitational force exerted on one sphere which is in left and down corner of the square by the other three.


2. Homework Equations

F_tot = F_2 + sqrt(2) * F1

3. The Attempt at a Solution

came up with that equation, but it says i am wrong.
can anyone help/
 

Answers and Replies

631
0
It'll be better if you explain your equation (or your solution).
 
45
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well since they are in a square and have equal masses F_1 = F_3 in magnitude. and for our triangle we have 1/sqrt2 are the sides and 1 is the hypotenuse. i got that. but i don't think its right
 
45
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am i way offf? can anybody help
 
Doc Al
Mentor
44,827
1,083
2. Homework Equations

F_tot = F_2 + sqrt(2) * F1
If I understand what you're saying here, this is correct. But it's not the final answer.

What are F_2 and F_1?
 
45
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F1 and F3 are the 90 degree forces and F2 is the 45 degree force. what do i still need to do?
 
631
0
Still your solution solution looks bit wrong. Show the substitution process (the values for F1 ..) and and calculation, so that I can help you better.
 
Doc Al
Mentor
44,827
1,083
F1 and F3 are the 90 degree forces and F2 is the 45 degree force. what do i still need to do?
That's just the schematic of the answer. Now you have to figure out F1 & F2 and plug in the numbers. I assume they want an actual answer for the force in Newtons.
 
45
0
F1 = ((6.67 * 10 ^-11)* 7.5 *7.5 )/ .70^2 = 7.66 * 10^-9

length of hypotnuse
sqrt(.70^2 +.70^2); = .99

F2 = ((6.67 * 10 ^-11)* 7.5 *7.5 )/ .99^2 = 3.83*10^-9

F_tot = 3.83*10^-9 + sqrt(2) * 7.66*10^-9 =

1.46 * 10^-8

PROBLEM SOLVED. I MUST HAVE JUST PLUGGED IN SOME WRONG NUMBERS. THANKS ANYWAY
 

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