# Law of gravitation and net gravitational force

[SOLVED] Law of gravitation

In Figure 13-34, a square of edge length 15.0 cm is formed by four spheres of masses m1 = 5.00 g, m2 = 4.00 g, m3 = 1.50 g, and m4 = 5.00 g. In unit-vector notation, what is the net gravitational force from them on a central sphere with mass m5 = 2.10 g?

I know that since the mass of 1 and 4 are equal they cancel out and i know that the radius is .075*Square root(2). which gives me .106m. I use the equation Gm2m5/r25^2+Gm3m5/r35^2 and i get the answer 6.85e-14 which is not the right answer. I know that the masses are .004, .0015, and .0021 since they are in grams. Can some one help me

Doc Al
Mentor
(1) You didn't provide the figure, but I assume that m1 and m4 are at adjacent corners not opposite corners. So they don't cancel out.
(2) The forces are vectors and must be added as such. Direction counts!

sorry Doc Al M1 and M\$ are on opposite corners here is the picture given

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Doc Al
Mentor
OK. Realize that the forces due to m2 and m3 point in opposite directions.

that should mean i should subtract their forces but i tried that and that did not work i got a value of 3.11e-14 which is also incorrect

Doc Al
Mentor
I get the same answer. But that's the magnitude of the force. They asked for the force in unit vector notation, which means they want x and y components.

But how would i do that I know how to go from unit vector notation but i do not understand how to do the opposite

Doc Al
Mentor
Just find the x and y components of the net force. What direction is that force? What angle does it make with the x-axis?

It is moving towards mass 2 and the angle should be 45

Doc Al
Mentor
It is moving towards mass 2 and the angle should be 45
Right. So what are the x and y components of a vector at an angle of 45 degrees to the horizontal?

so i set up the sin 45=o/h and solve so i get hsin 45=0 and i get 2.64e-14 but that is not the correct answer. i do the same thing with cos and get 1.63e-14. where should i go from here

Doc Al
Mentor
If a force F makes an angle $\theta$ with the x-axis, its components are:

$$F_x = F \cos \theta$$

$$F_y = F \sin \theta$$

Isn't that what i did, i just wrote it in different terms

Doc Al
Mentor
Isn't that what i did, i just wrote it in different terms
How did you get different answers? ($\sin 45 = \cos 45$.)

ooo i am in radian mode

Thank you Doc Al for your help