Finding the net gravitational force with Vector Notation

[Ella1777]

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!

 

[gneill]

Please retain and use the formatting template that is provided in the edit window to prepare your homework help requests.

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?

 

[Chandra Prayaga]

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.