Three uniform spheres are located at the corners of an equilateral triangle. Each side of the triangle has a length of 1.41 m. Two of the spheres have a mass of 2.66 kg each. The third sphere (mass unknown) is released from rest. Considering only the gravitational forces that the spheres exert on each other, what is the magnitude of the initial acceleration of the third sphere?
I'd like to add my attempted solution to this but unfortunately I don't even know where to begin. The question doesn't say they're attached or if some are higher than the other or sdjgdsdsfsdkfg. Stress.
:(
I'd like to add my attempted solution to this but unfortunately I don't even know where to begin. The question doesn't say they're attached or if some are higher than the other or sdjgdsdsfsdkfg. Stress.
:(
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assume the spheres are on a level plane and not attached; that makes the most sense for this question
the idea is to find the net force acting on the unknown mass sphere
so consider this configuration..they are all equal but this will help visualize it
M
2.66 2.66
and let's make believe that's really an equilateral triangle.
now, each 2.66kg mass exerts a force on M; each force has both an x and y component (x is horizontal, y up and down)
the x components of the force cancel, leaving only y components pulling M down the middle line of the triangle
now, you know that the total grav force between two masses is
F = Gm1/m2/r^2
now, we know that r is 1.41 m since that is the length of each side
but we also need to find the y component of this force, in other words we want
y component of F = F sin(theta)
we know the hypotenuse is 1.41 m, we need to find the length of the central line (from the midpoint of the base to M)
we know half the base is 1.41/2 = 0.705m, and we use the pythagorean theorem to find y, the height of the line:
1.41^2 = y^2 + 0.705^2
y = 1.22
from which we learn that sin(theta) = 1.22/1.41 = 0.866
getting back to forces...the y component of each force on M is then
F sin (theta) = G (2.66kg x M) *0.866/(1.41m^2)
there are two such forces, so the total force on M is:
2G(2.66M)(0.866)/(1.41^2)
now, to find the acceleration of M, we know this total force = Ma, so that
a = 2G(2.66kg)(0.866)/(1.41m)^2
G=6.67x10^-11
grind and find
hope this helps
the idea is to find the net force acting on the unknown mass sphere
so consider this configuration..they are all equal but this will help visualize it
M
2.66 2.66
and let's make believe that's really an equilateral triangle.
now, each 2.66kg mass exerts a force on M; each force has both an x and y component (x is horizontal, y up and down)
the x components of the force cancel, leaving only y components pulling M down the middle line of the triangle
now, you know that the total grav force between two masses is
F = Gm1/m2/r^2
now, we know that r is 1.41 m since that is the length of each side
but we also need to find the y component of this force, in other words we want
y component of F = F sin(theta)
we know the hypotenuse is 1.41 m, we need to find the length of the central line (from the midpoint of the base to M)
we know half the base is 1.41/2 = 0.705m, and we use the pythagorean theorem to find y, the height of the line:
1.41^2 = y^2 + 0.705^2
y = 1.22
from which we learn that sin(theta) = 1.22/1.41 = 0.866
getting back to forces...the y component of each force on M is then
F sin (theta) = G (2.66kg x M) *0.866/(1.41m^2)
there are two such forces, so the total force on M is:
2G(2.66M)(0.866)/(1.41^2)
now, to find the acceleration of M, we know this total force = Ma, so that
a = 2G(2.66kg)(0.866)/(1.41m)^2
G=6.67x10^-11
grind and find
hope this helps