Calculate Speed from Magnetic Pull Force

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Discussion Overview

The discussion revolves around calculating the speed of an object based on the magnetic pull force applied to it, particularly in the context of building a gauss gun. Participants explore the relationship between force, mass, and acceleration, while also addressing the complexities introduced by magnetic forces and coil designs.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant seeks an equation to determine the speed of an object based on the force applied, indicating a lack of clarity in their understanding.
  • Several participants reference Newton's Second Law (F=ma) to explain the relationship between force, mass, and acceleration.
  • There is a discussion about the practical application of these principles in the context of building a gauss gun, with one participant sharing their experience with various coil designs.
  • One participant attempts to calculate speed using a force of 200 pounds applied to a 5-gram object, converting units but receiving corrections regarding the mass unit conversion.
  • Another participant notes the complexity of calculations when magnetic forces are involved and suggests researching solenoids for a better understanding.
  • Equations for magnetic force and railgun operation are presented, but some participants express confusion over the variables involved.
  • One participant shares their struggles with energy loss and heat in their designs, indicating a learning curve in understanding the electrical aspects of their project.

Areas of Agreement / Disagreement

Participants generally agree on the application of Newton's Second Law but express differing levels of understanding regarding the calculations and complexities of magnetic forces. The discussion remains unresolved as participants navigate through various equations and concepts without reaching a consensus.

Contextual Notes

There are limitations in the discussion regarding the clarity of variables in the equations presented, as well as the assumptions made about the systems being analyzed. The participants also acknowledge their varying levels of expertise in physics and electrical engineering.

Who May Find This Useful

This discussion may be useful for individuals interested in physics, particularly those exploring concepts related to force, acceleration, and magnetic systems in experimental projects like gauss guns.

ironman13405
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Okay i am new to this site so i figured this would be the best place to post this. I need an equation for my project i am working on.

I need an equation to find the speed of a object based off how much force is applied to it.

If this isn't clear enough i am welcome to give more information. Let me know
 
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F=ma
 
dacruick said:
F=ma

Could you explain? Yes i am very new to this..
 
The force(F) times the time for which it acts(t) is equal to the mass of the object(M) times the change of velocity of the object....Ft=M times velocity change.This is Newtons second law(as is F=Ma where a is acceleration)
 
Okay, sorry for the abrupt answer. This is Newton's Second Law. It states that mass of the object, multiplied by the acceleration of the object, is the Force applied.

But in a more practical sense, in the context that you want to use it this is essence.

If you apply a certain force to an object of mass 'm', the acceleration of that object will be the Force divided by mass. Your acceleration is measure in metres per second squared. So say your acceleration is 10 m/s². That means that your speed increases by 10 metres / second, per second. Or, every second that passes you have an increase of speed of 10 metres per second.

What is it that you are trying to do exactly?
 
dacruick said:
What is it that you are trying to do exactly?

I trying to show my friend how unlikely the idea he has is unlikely to work. Cause we both are going against each other (In order to get things done faster cause we want to beat each other) to make a gauss gun. I've made many coil designs some of which do amazing. But i have wasted a lot of money due to the lack of physics understandning. I am starting to get more into, and its helping greatly
 
coil designs as in springs? because if you have a spring Newton's second law has a change
 
No coils as in eletric.

So tell me if i got this right let's say i where to apply 200 pounds of force to a 5 gram object.

there are .2248 pounds in one Newton so that would be 889.67972 N then you would divide by 5 grams to make 177m/s and there are .44m/s in one MPH.. so thus it would be 395 MPH.. right?
 
mass is measured in SI units of kilograms. So youre going to have to divide that 5 by 1000. which will give your answer a couple more 0's
 
  • #10
but i forgot that your force is magnetic. you have current running through a coil and you want to figure out the force? you would have to do research on something called a solenoid then. which is going to make your calculations more complicated.
 
  • #11
I found the equation for for magnetic force

F=A*N*I (((mu0*N*)/(*(C)^2)) +(1/CL)

or you can do its for railgun

step one H = (N*I)/M

Step two B= (H*mu0)/A

Step three F= B*L*I

Step four speed = ((F*T)/W) *3.33
 
  • #12
Im not sure what all of your variables are, but the equations look good. give it a shot and see if your answer is plausible
 
  • #13
It gets more and more confusing the more i travel down the rabit hole. I am collecting data mostly from

http://www.intalek.com/AV/Railguns-Tim-Ventura-1994-2.pdf

My design is very apart from his but they follow the same concept. I am getting a major headache haha. a lot of my issues are the same repeating ones over and over again. Energy lost due to heat, the melting of wire due to heat, and etc. Let alone i don't know too much about eletrics... i am slowly finding out more and more.. i want to get it done haha. I have unlimited resources too, cause we are working through a school program, so it helps
 
  • #14
yeah electrics arent my forte either. I don't think i can help you any further. but good luck definitely
 

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