A square of edge length 19.0 cm is formed by four spheres of masses m1 = 6.00 g, m2 = 3.50 g, m3 = 1.00 g, and m4 = 6.00 g. In unit-vector notation, what is the net gravitational force from them on a central sphere with mass m5 = 2.20 g?
_N i + _N j
Physics - bobpursley, Wednesday, February 14, 2007 at 7:13am
You have to put the gravitational equation in vector form and add.
Ok, so I took the formula: F=G*((m1m2)/r^2)
F1=(6.67e-11)*((.006)(.0022))/(.19^2)
F1=2.44e-14 N
F2=(6.67e-11)*((.0035)(.0022))/(.19^2)
F2=1.42e-14 N
F3=(6.67e-11)*((.001)(.0022))/(.19^2)
F3=4.06e-15 N
F4=same as F1
So I addes all of these up and got:
Fnet=6.71e-14 N but this is the wrong answer and I am not sure how to break this one up into vector components. The example from my book never broke it up into i and j, it only gave 1 answer.
Please help!!
The distance between the central sphere and the corner spheres is not 0.19 m. It is 0.19 sqrt 2.
A vector addition is needed to do this problem. Each component force is directed at 45 degrees to the x and y axes defined by the square. The problem asks for unit vector notation. This implies something like "i and j" components, with i and j often written in boldface type after each component. This would look like one answer.
The F1 and F3 forces act in opposite directions, and so do the F2 and F4 forces. That means you have to do a subtraction of those pairs of forces.
Ok, thanks a lot!!! I got it :)