Three identical masses of 390 kg each are placed on the x-axis. One mass is at x_1 = −11.0 cm, one is at the origin, and one is at x_2 = 39.0 cm.
What is the magnitude of the net gravitational force 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.
best answer is the most explanatory one
2006-06-28 13:42:20 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Physics
normally its a simple problem but the 3rd mass in the middle complicates the problem for me.
2006-06-28 14:29:05 · update #1
Use Newton's universal gravitation equation, then sum the forces.
F = G*m1*m2/r^2
F = force (newtons)
G = gravitational constant (6.6742 x 10^-11)
m1 = mass of object 1 (kg)
m2 = mass of object 2 (kg)
r = distance between objects (meters) ==> don't forget to convert the units
In this case, you do it twice:
F1 = G * 390 * 390 / 0.11^2 = 0.84mN to the left
F2 = G * 390 * 390 / 0.39^2 = 0.067mN to the right
Net Force = 0.84 - 0.067 = 0.772mN to the left
2006-06-29 16:19:51 · answer #1 · answered by tom_2727 5 · 0⤊ 0⤋
Let the mass of the center object be m1. The force by another object m2 on m1 is given by F=G*m1*m2/(r^2), by Newton's Universal Law of Gravitation. Just compute the force by each object on m1 and take the difference.
So, net F=G*m1*(m1/0.11^2 - m2/0.39^2)
F=G*390(390/0.0121-390/0.1521)
=7.7(10-4) N to the left
2006-06-29 08:41:16 · answer #2 · answered by gradient 2 · 0⤊ 0⤋
The mass at the origin has no net effect at the origin (distance = 0). Ignore it. Classic exess data question.
2006-06-28 15:02:22 · answer #3 · answered by STEVEN F 7 · 0⤊ 0⤋
Force = (G * m1*m2)/ s^2
That should make it pretty easy.
2006-06-28 14:04:54 · answer #4 · answered by festavan32 2 · 0⤊ 0⤋
Over my head. hope someone else can answer for you.
Hope
2006-06-28 13:46:46 · answer #5 · answered by Anonymous · 0⤊ 0⤋