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How to find force necessary to propel a 51g object 120 m/s

  1. Nov 13, 2011 #1
    I am building a cannon of sorts, from which a Nerf mini-vortex football needs to be fired at 120 (at least) metres/second from compressed air. What force will be necessary to achieve this? preferably in psi.
    Additional Info:
    Mass of Projectile: ~51g

    Thanks in advance!
     
  2. jcsd
  3. Nov 13, 2011 #2
    If you assume the minimum acceleration needed is 120 m/s^2, you have:

    .51g = .00051 kg

    F=MA so F=(.00051 kg)(120 m/s^2) --> F= (.0612 N)

    Now Force and Pressure are two different things, so you won't be able to convert directly between the two, but Pressure=[itex]\frac{Force}{Area}[/itex] so if you know the cannon's specs you can figure out the pressure required.

    Note 1: This is the minimum pressure and it is assuming that the projectile will be accelerated over a period of 1 second, considering it will most likely be far less time than that, you will have to adjust the pressure accordingly.

    Note 2: If I did my math right, the pressure you would get out of these equations will be in Pascals, not PSI, but any computer converter can translate between the two.
     
  4. Nov 14, 2011 #3
    An important question that needs to be answered is; Over what distance will the object be accelerated? That will determine how much force you will need, along with the desired final velocity.

    Using the two equations for position and velocity as well a known parameters
    x = .5a*(t)^2
    v = a*t
    v = 120m/s
    m = .051kg
    x = (distance to be accelerated over)m

    We can determine the acceleration required

    (t = v/a) -> (x = .5a*(t)^2) = (x = .5a*(v/a)^2)

    x = .5*(v^2)/a
    a = .5*(v^2)/x

    Since F = m*a, then a = F/m
    Putting that in we get, F/m = .5*(v^2)/x
    So, F = .5*m*(v^2)/x

    Force is in Newtons so then convert to pounds; After that you'll want to divide that by the area (in, in^2) that the pressure is being applied over to determine the needed PSI.

    That's the best I can do. You'll will need to tweak the final pressure since I'm assuming that this an ideal situation in my head.
     
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