Net gravitational force on an object

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The discussion focuses on calculating the net gravitational force on an object by analyzing the x and y components of forces exerted by two other masses. The gravitational force from object 1 on object 0 is determined using the formula \(\frac{Gm^2}{d^2}\) with a direction vector of <-1, 0>. For object 2, the distance is calculated as \(r=\sqrt{d^2+ l^2}\), leading to a gravitational force of \(\frac{2Gm^2}{d^2+ l^2}\) with a direction vector of . The resultant force combines these components, correcting an earlier expression to include the proper normalization factor. This detailed analysis illustrates the method for determining the net gravitational force acting on an object in a multi-body system.
Jaccobtw
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Homework Statement
A test object of mass m is placed at the origin of a two dimensional coordinate system. An object 1, of the same mass, is at (d, 0), and an object 2, of mass 2m, is at (-d, l). What is the magnitude of the vector sum of the gravitational forces exerted on the test object by the other two objects?
Relevant Equations
F(gravity) = G * (M * m)/ (r^2)
G = 6.6738 * 10^-11 N * m^2
IMG_0083.jpg


I suppose we can just find the net x components and y components and then go from there.
Σ Fx = F(mass 1) - Fx(mass 2)
G* (m^2)./d^2) - something

I'm not sure how to express the component forces of the 2nd mass
 
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Object 1 is at (d, 0). The distance from object 0 to object 1 is r=d. The direction vector from object 1 to object 0 is <-d, 0> the unit vector in that direction is <-1, 0>. The magnitude of the gravitational force object 1 exert on object 0 is \frac{Gm^2}{d^2} and the vector force is \frac{Gm^2}{d^2}\left&lt;-1, 0\right&gt;. Since you are asking about object 2 I presume you got that.

Object 2 is at (-d, l). The distance from object 0 to object 2 is r=\sqrt{d^2+ l^2}. The direction vector from object 2 to object 0 is <d, -l>. The magnitude of the gravitational force object 1 exerts on object 0 is \frac{2Gm^2}{d^2+ l^2} and the vector force is \frac{2Gm^2}{d^2+ l^2}\left&lt;d, -l\right&gt;.

The resultant is the sum of those, \left&lt;\frac{-Gm^2}{d^2}, 0\right&gt;+ \left&lt;\frac{2dGm^2}{d^2+ l^2}, \frac{-2lGm^2}{d^2+ l^2}\right&gt;.
 
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HallsofIvy said:
The resultant is the sum of those, \left&lt;\frac{-Gm^2}{d^2}, 0\right&gt;+ \left&lt;\frac{2dGm^2}{d^2+ l^2}, \frac{-2lGm^2}{d^2+ l^2}\right&gt;.
Of course you meant to write
"The resultant is the sum of those, \left&lt;\frac{-Gm^2}{d^2}, 0\right&gt;+ \left&lt;\frac{2dGm^2}{(d^2+ l^2)^{3/2}}, \frac{-2lGm^2}{(d^2+ l^2)^{3/2}}\right&gt;."
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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