Finding the net gravitational force with Vector Notation

AI Thread Summary
The discussion focuses on calculating the net gravitational force on a central sphere (m5) due to four surrounding spheres with given masses. The attempted solution uses the gravitational force formula but raises questions about the values and calculations used, particularly the distance (r) and mass values. Participants highlight the need for clarity on how the distance was determined and suggest including all relevant forces from each sphere. Additionally, there are concerns about the calculations being incomplete and the importance of a diagram for better understanding. The conversation emphasizes the need for accurate application of gravitational force principles in vector notation.
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Information Given: In the figure, a square of edge length 17.0 cm is formed by four spheres of masses m1 = 4.70 g, m2 = 2.90 g, m3 = 0.800 g, and m4 = 4.70 g.

Question: In unit-vector notation, what is the net gravitational force from them on a central sphere with mass m5 = 2.90 g? Attempted Solution:F=k*m1*m2/r^2
r= (sqrt (0.1m)^2 + (0.1m)^2)=0.1414m
F25=(6.67*10^-11)(0.0290)/0.1414m
F25=1.36796322*10^-11

F35=(1.36796322*10^-11)/4=3.14990806*10...

Resultant: F25-F35=(1.36796322*10^-11)-(3.141990806...

Horizontal:( 2.77917256*10^-13) cos(45)= 1.4596046*10^-13 N

Vertical: (2.77917256*10^-13) sin (45)= 1.2422855*10^-13 N
Answer(Incorrect):1.46e-13 i + 1.24e-13 j N
I don't understand I checked this several times yet it's incorrect!
Help is greatly appreciated
Thank You!
 
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Ella1777 said:
Information Given: In the figure, a square of edge length 17.0 cm is formed by four spheres of masses m1 = 4.70 g, m2 = 2.90 g, m3 = 0.800 g, and m4 = 4.70 g.

Question: In unit-vector notation, what is the net gravitational force from them on a central sphere with mass m5 = 2.90 g?Attempted Solution:F=k*m1*m2/r^2
r= (sqrt (0.1m)^2 + (0.1m)^2)=0.1414m
Where did the 0.1m values come from? Did you draw a sketch of the setup?
F25=(6.67*10^-11)(0.0290)/0.1414m
F25=1.36796322*10^-11
Where did the 0.0290 value come from?
F35=(1.36796322*10^-11)/4=3.14990806*10...

Resultant: F25-F35=(1.36796322*10^-11)-(3.141990806...

Horizontal:( 2.77917256*10^-13) cos(45)= 1.4596046*10^-13 N

Vertical: (2.77917256*10^-13) sin (45)= 1.2422855*10^-13 N
Assuming that "F25" means the force of m2 on m5, then what happened to the calculations for the other masses? Shouldn't there be F15, F25, F35, and F45?
 
A diagram would reduce the confusion.
1. In calculating F25, why did you use r in the denominator? Should it not be r2?
2. In the numerator, you seem to use only one of the masses, instead of the product of the two masses.
 

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