# Gravitational net force of zero

• UrbanXrisis
In summary, to balance a 200kg mass and a 500kg mass, the 50kg mass must be placed between the 200kg and 500kg masses so that the 50kg mass experiences a net force of zero.
UrbanXrisis
A 200 kg mass and a 500kg mass are separated by 0.4m. At what position can a 50kg mass be placed so as to experience a net force of zero (other than infinity)?

here's what I did:

I need both forces to equal each other.

$$F_{200}=G\frac{200m}{x^2}=G\frac{500m}{y^2}=F_{500}$$
$$\frac{200}{x^2}=\frac{500}{y^2}$$
$$200y^2=500x^2$$
$$x+y=0.4$$
$$x=0.4-y$$
$$500(0.4-y)^2=200y^2$$
$$500(0.16-0.8y+y^2)=200y^2$$
$$200y^2=80-400y+500y^2$$
$$1=\frac{0.4}{y^2}-\frac{2}{y}+2.5$$
$$y=0.245m,x=0.155m$$

did I do this correctly? was there a faster way?

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Your question isn't clear,are you trying to balance the entire system? How many dimensions here, 1 or 2? Do you want just the 50kg to experience no force? It looks like your talking about gravitational forces among the 3, right?

well, I copied the book. What I got was that I had to place the 50kg mass in between the 200kg and 500kg masses so that the 50kg mass experiences a net force of zero.

then you want the pull from left to equal the pull from the right. Though I didnt check your math, the method is right and your answer seems logical.

Are the CORES of the masses separated by 0.4m or just surface to surface? Anyway if it is the cores, don't bother with two variables. Say one will be x, the other 0.4-x.

Look:

GM1/(0.4-x)^2=GM2/x^2

So

M1/(0.4-x)^2=M2/x^2

So

M1x^2=M2(0.4-x)^2

SO

(M1^0.5)x=M2^0.5(0.4-x)

-((M1^0.5)x/M2^0.5)+0.4=x
-((M1^0.5)/M2^0.5)+0.4/x=1
-((M1^0.5)/M2^0.5)-1=-0.4/x
x=0.4/((M1^0.5)/M2^0.5)-1

It would have been faster if you knew this formula!

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Since the radius of each mass isn't provided, you are forced to assume that it is core to core.

In this case my guess was good :)

(M1^0.5)x=M2^0.5(0.4-x)

-((M1^0.5)x/M2^0.5)+0.4=x
-((M1^0.5)/M2^0.5)+0.4/x=1
-((M1^0.5)/M2^0.5)-1=-0.4/x
x=0.4/((M1^0.5)/M2^0.5)-1

It would have been faster if you knew this formula!

Theres a mistake in the algebra somewhere there. The answer should have + 1, not -1.

-((M1^0.5)/M2^0.5)+0.4/x=1

so

0.4/x =1 + ((M1^0.5)/M2^0.5)
x = 0.4 / ( 1+ ((M1^0.5)/M2^0.5))

UrbanXrisis said:
$$200y^2=80-400y+500y^2$$
So far, so good. (Assuming these masses can be treated as particles.) Simplifying and rewriting this in standard form for a quadratic:
$$15y^2 -20y + 4 = 0$$
Which yields two solutions, only one of which is relevant to this problem.
$$1=\frac{0.4}{y^2}-\frac{2}{y}+2.5$$
I don't know why you wrote it this way.
$$y=0.245m,x=0.155m$$
did I do this correctly? was there a faster way?
Perfectly correct.

It would have been slightly faster to immediately take the (positive) square root (as Werg22 showed), which eliminates the need to solve a quadratic:
$$\frac{200}{x^2}=\frac{500}{y^2} \; \Longrightarrow \; \frac{2}{x^2}=\frac{5}{y^2}$$
becomes:
$$\frac{\sqrt{2}}{x}=\frac{\sqrt{5}}{y}$$

Sorry guys! It is actually because I forgot the 1 was an individual term... I tought it was part of the parenthesis when it is not.

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## 1. What is the concept of "gravitational net force of zero"?

The gravitational net force of zero, also known as mechanical equilibrium, refers to a state in which the total force acting on an object is balanced and the object remains at rest or in a constant state of motion. This means that the forces pulling on the object in different directions cancel each other out, resulting in a net force of zero.

## 2. How does the gravitational net force of zero affect objects?

Objects in a state of gravitational net force of zero will remain in their current state of motion, whether it is at rest or moving at a constant velocity. This is because there is no unbalanced force acting on the object to change its motion.

## 3. What are some real-life examples of objects experiencing gravitational net force of zero?

Some examples include a book resting on a table, a person standing on the ground, and a car traveling at a constant speed on a level road. In all of these situations, the forces acting on the object are balanced, resulting in a net force of zero.

## 4. How is the gravitational net force of zero related to Newton's First Law of Motion?

The gravitational net force of zero is a manifestation of Newton's First Law, which states that an object at rest will remain at rest and an object in motion will continue in motion with a constant velocity, unless acted upon by an unbalanced force. In the case of gravitational net force of zero, there is no unbalanced force acting on the object to change its state of motion.

## 5. Can the gravitational net force of zero ever be broken?

In a closed system, the gravitational net force of zero cannot be broken. However, in an open system where external forces can act on the object, the equilibrium can be disrupted and the object's state of motion can change. For example, a book resting on a table will experience a gravitational net force of zero, but if someone were to push the book off the table, the equilibrium would be broken and the book would start to fall due to the force of gravity.

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