Gravitational force and constants

[juggalomike]

Homework Statement

Three identical masses of 570 kg each are placed on the x axis. One mass is at x_1 = -10.0 cm, one is at the origin, and one is at x_2 = 43.0 cm.

What is the magnitude of the net gravitational force F_grav on the mass at the origin due to the other two masses?

Take the gravitational constant to be G = 6.67×10−11 N * m^2/kg^2.

Homework Equations

F=G(m1m2/R^2)

F=F1-F2

The Attempt at a Solution

-(6.67*10^-11)*(570^2)/(-.1)^2 - (6.67*10^-11)*(570^2)/(.43)^2 =-0023 N

What am i doing wrong?

 

[thebigstar25]

maybe you want to reconsider the direction of each force .. (you should do vector subtraction) ..

 

[juggalomike]

maybe you want to reconsider the direction of each force .. (you should do vector subtraction) ..

tried it, still telling me the answer is not 0...

 

[thebigstar25]

why should the answer be zero?

 

[juggalomike]

why should the answer be zero?

The answer is not zero, however because my number is so small it is considering it to be 0

 

[Dick]

Homework Statement

Three identical masses of 570 kg each are placed on the x axis. One mass is at x_1 = -10.0 cm, one is at the origin, and one is at x_2 = 43.0 cm.

What is the magnitude of the net gravitational force F_grav on the mass at the origin due to the other two masses?

Take the gravitational constant to be G = 6.67×10−11 N * m^2/kg^2.

Homework Equations

F=G(m1m2/R^2)

F=F1-F2

The Attempt at a Solution

-(6.67*10^-11)*(570^2)/(-.1)^2 - (6.67*10^-11)*(570^2)/(.43)^2 =-0023 N

What am i doing wrong?

The answer is small. But in the expression you gave you are adding both forces in the negative direction. The sign should be different between them. One should be pulling in the positive direction and the other in the negative. I think this was thebigstar25's point.

 

[thebigstar25]

thanks Dick for explaning my point in a better way .. I think now juggalomike can solve the problem ..