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 :)
To find the net gravitational force on the central sphere in unit-vector notation, you first need to calculate the magnitudes of the individual gravitational forces exerted by each of the four surrounding spheres.
The gravitational force equation is: F = G * ((m1 * m2) / r^2), where G is the gravitational constant (6.67e-11 N * m^2 / kg^2), m1 and m2 are the masses of the two objects, and r is the distance between their centers.
Let's calculate the magnitudes of the forces:
F1 = (6.67e-11) * ((0.006) * (0.0022)) / (0.19 * sqrt(2))^2
F1 = 2.443e-14 N
F2 = (6.67e-11) * ((0.0035) * (0.0022)) / (0.19 * sqrt(2))^2
F2 = 1.425e-14 N
F3 = (6.67e-11) * ((0.001) * (0.0022)) / (0.19 * sqrt(2))^2
F3 = 4.061e-15 N
F4 = F1 (assuming it's the same as F1, since it's using mass m1)
Now, to determine the net force, we need to account for the vector components.
The forces F1 and F3 act along the negative i direction, whereas forces F2 and F4 act along the positive i direction.
So, we subtract the magnitudes of forces F1 and F3 from the magnitudes of forces F2 and F4 to determine the net force components in the i direction:
Net Force in i-direction = (- F1 - F3) + (F2 + F4)
Net Force in i-direction = - (2.443e-14 N) - (4.061e-15 N) + (1.425e-14 N) + (2.443e-14 N)
Net Force in i-direction = 0
The net force in the i-direction is zero, meaning that the forces in the positive and negative i-directions cancel each other out.
For the j-direction, the forces F1 and F2 act along the negative j direction, whereas forces F3 and F4 act along the positive j direction.
So, we subtract the magnitudes of forces F1 and F2 from the magnitudes of forces F3 and F4 to determine the net force components in the j direction:
Net Force in j-direction = (-F1 - F2) + (F3 + F4)
Net Force in j-direction = - (2.443e-14 N) - (1.425e-14 N) + (4.061e-15 N) + (2.443e-14 N)
Net Force in j-direction = 3.636e-15 N
Therefore, the net gravitational force acting on the central sphere in unit-vector notation is:
Net Force = 0 i - (3.636e-15 N) j
So the net gravitational force in unit-vector notation is (0 N)i + (-3.636e-15 N)j.