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Magnet's attraction/rejection Force

  1. Aug 23, 2005 #1
    I dont need an exact formola, but somthing for me to know how hard it is to make a really big rejection force.

    What are the parameters for magnet's force?
    How do I find out their value?
    How can I put them together in an equation, to know what the force is?

    thnks, grasia, spassiva and so on..
     
  2. jcsd
  3. Aug 23, 2005 #2
    related force question

    I am interested in this as well as a general formula for determing force. I am wanting to use a electro magnet to push something.

    Thanks for any info.
     
  4. Aug 23, 2005 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The general law is similar to "gravity": the magnetic force between two magnets is given by [tex]F= \frac{M_1M_2}{r^2}[/tex]
    where M1 and M2 are the "strengths" of the two magnets and r is the distance beween them. If M1 and M2 have the same sign, this is positive and is a repulsion. If they have different signs, this is negative and is an attraction.
     
  5. Aug 23, 2005 #4
    Thanks i have another question

    Just want to let you know I am VERY inexperenced with physics so some of my questions might seem dumb. If I want to move something with a mass of one kilogram one mile how do I figure out how much force I need? Also do you know any good web sites where I could find out more?
     
    Last edited: Aug 23, 2005
  6. Aug 23, 2005 #5

    jtbell

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    Staff: Mentor

    You need to know how much friction or other force is opposing the motion. If there's no friction at all and no other opposing forces, you need only a small brief force to start the object moving, then a lot of patience (but no force) while the object "coasts" one mile under Newton's First Law of Motion!
     
  7. Aug 23, 2005 #6
    What if there is just a little bit of friction say.... if it was going through the air? Is there a formula for figureing it out? :confused:
     
    Last edited: Aug 23, 2005
  8. Aug 23, 2005 #7
    You draw your free body diagram, you will see that the force you need to move the object will need to overcome the friction force. So the same approach is taken as in the first case, the force you need to move the 1kg object the distance is a little bit more than your friction force, as in the last post, this assumes you will have patience, since the weight will eventually reach the destination due to Newtons First law of Motion.

    Regards,

    Nenad
     
  9. Aug 23, 2005 #8
    Free Body Diagram?? As I said I am VERY inexpirenced! Let me give a bigger picture of what I am trying to figure out.

    I am wanting to use an electomagnet to repell an object of about 8.42 GRAMs for a distance of approximately 100 yards. It will be traveling roughly horizontal to the ground, starting at approximately 1.3 meters above the ground, so:

    I am trying to figure out how much force I need to create to propell the object (lets assume a sphere for the moment to keep air resistance simple) 100 yard before it drops more than 10cm.

    This is the formula I am trying to figure out.
     
  10. Aug 23, 2005 #9
    HallsofIuy,
    I already new that formola, but I also asking for some explanation about the value of the "strengths"..
     
  11. Aug 23, 2005 #10

    jtbell

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    Staff: Mentor

    OK, now we have a specific scenario to talk about. So this is sort of an electromagnetic gun, right? I assume the "barrel" of the gun is relatively short, so that for most of its flight, the object is an ordinary projectile, subject only to gravity and air resistance.

    In that case, you've got basically a two-stage problem: (1) find out how fast the object needs to be moving when it leaves the gun, so that it will drop 10cm after travelling 100m horizontally; (2) find out how much force the gun needs to exert on the object while it's inside the gun, given information like the length of the gun's "barrel" and how much friction acts on the object while it's traveling through the gun.

    If you didn't have to worry about air resistance, step (1) would be a fairly simple exercise in two-dimensional projectile motion. But for an 8.42-gram object travelling 100m, air resistance is going to be very significant. The force of air resistance depends on the object's speed, which is going to decrease rapidly because of the air resistance. So the force of air resistance won't be constant. There's no simple formula for this situation. If I had to work this out myself, I'd write a computer program to simulate the motion numerically in small steps, and then run it several times with different initial velocities, until I hit on the right velocity to produce a 10-cm drop.

    Hmmm.... 8.42 grams sounds like the weight of a bullet of some standard size. Maybe you can find tables somewhere for rifle-shooting, that give the required muzzle velocity in order to produce a specified drop at a specified distance. There might even be an approximate formula for this, probably based on actual measurements and some graph-fitting.

    And we haven't even gotten to step (2) yet!
     
  12. Aug 23, 2005 #11
    :eek: Thanks, um this sounds complicated probobly I should try somthing els :grumpy:
     
    Last edited: Aug 23, 2005
  13. Aug 23, 2005 #12
    Or...

    I could pick a point to start to experiment and see what issues I run into, get some numbers to put into SOME type of formula along the way.

    Thanks for all your replies...I know I don't know hardly anything about physics, you have been a big help thinking through it. :smile:
     
  14. Nov 12, 2010 #13
    hi to all ,, i need some information
    i want to lifting up 5 kg with magnet from 50 cm distance
    i know its hard but can i do that at all?
    i can put a second magnet above may object and lifting that with first magnet .
    my object is 5kg
    can i do that with permanent magnet or electromagnet?? how??
    please help me ..
    i really need this.
    tanx alot
    im waiting.
     
  15. Nov 18, 2011 #14
    Sorry but the strenth of the magnet needed is too strong, more information is needed to work out how strong but if you are needing to lift it by 50cm the magnet would not be consumer grade and not a permanent magnet!
     
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