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Forces Problem

  1. Sep 16, 2007 #1
    I have 2 problems I just can't seem to get so I'll post both here instead of making 2 threads.
    Both of these are from Halliday & Resnick 5th ed chapter 3 Problems.

    Problem #1.
    1. The problem statement, all variables and given/known data

    A light beam from a satellite-carried laser strikes an object ejected from an accidentally launched ballistic missile. The beam exerts a force of [itex]2.0 * 10^{-5}[/itex] N on the target.
    If the "dwell time" of the beam on the target is 2.4s by how much is the object displaced if it is
    a) a 280-kg warhead
    b) a 2.1-kg decoy?

    (These displacements can be measured by observing the reflected beam)

    2. Relevant equations
    [tex]\triangle x=v_0 t +\frac{1}{2}at^2[/tex]


    3. The attempt at a solution
    The missile experiences some uknown horizontal acceleration/velocity so I will just ignore those..
    It also experiences a downward acceleration of g.
    So does the laser beam?

    I need to find the sum of the forces to find the acceleration.
    [tex]\frac{mg+2.7*10^{-5}}{m}[/tex] for a I get about 9.8 m/s^2 .
    Plugging that into the equation I get a displacement of about 56m but the answer should be some micro-meters..

    Problem #9.
    1. The problem statement, all variables and given/known data

    A chain consisting of five links, each with mass 100g, is lifted vertically with a constant a =2.5m/s^2.
    a) the forces acting between adjacent links
    b) the force F exerted on the top link by the agent lifting the chain
    c) the net force on each link

    2. Relevant equations

    3. The attempt at a solution
    2.5m/s^2 up
    -9.8m/s^2 up

    a total of -7.3m/s^2 up

    for a.
    add up the masses of link+links beneath it and then multiply by 7.3m/s^2 to get a force of:
    for the top most link
    .5kg*7.3m/s^2 = 3.65 N being exerted on the top link.

    Which is completely wrong :/.

    for b.
    find the total mass and multiply by 7.3m/s^2
    F=.5kg*7.3m/s^2=3.65 N
    again completely wrong..

    for c.
    Somehow I got this right...
    lowest link force acting upon it is: .1kg*2.5m/s^2 = .25 N
    link above it: .2kg*2.5m/s^2=.50 N
    ...for all others

    then subtracting the link force - link below it to get a net force of .25 N on each link.

    Any help on either problem is greatly appreciated.
  2. jcsd
  3. Sep 16, 2007 #2
    I'm mainly trying to get problem #9.

    I was working backwards from the answers the book gave.
    for a the lowest # and thus the lowest link in the chain I think.

    has a total force of 1.23N which requires a force of 12.3 m/s^2 (g+2.5) on a mass of 100g.
    Isn't this wrong though? since there's a downward force of .1g N but an upward of .1*2.5 N but somehow the in the book they're adding them...
  4. Sep 16, 2007 #3


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    For the first problem, I don't think you should use mg... we don't really know anything about the missile... it may not be accelerating downwards at g (it may have its own thrust or something). we also don't know that the laser is directed straight upwards...
  5. Sep 16, 2007 #4


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    For problem 9, part b)

    Write this equation out:

    [tex]\Sigma\vec{F} = ma[/tex]
  6. Sep 16, 2007 #5


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    I don't think you are corrrectly applying newton 2 in your free body diagrams. Isolating the bottom link, there are 2 forces acting on it. You have correctly identified the downward force. The upward force is unknown...call it T. Now use newton 2nd law....F_net = ma. a is given, don't mess with it.
  7. Sep 16, 2007 #6
    Yea the first problem is wierd especially the "These displacements can be measured by observing the reflected beam" o.o since it's barely chapter 3 and I have no knowledge of optics so far.

    So sticking to #9.
    for part a, there are 2 forces acting on each link, 1 up and 1 down, I add them together right?
    so for the lowest link:
    .1 kg * 2.5 m/s^2=.25 N
    .1 kg * 9.8 m/s^2=.98 N
    total being 1.23 N

    the second link:
    there are 2 links being pulled up so I add the previous link's force + a new link:
    1.23 N + (.1kg * 2.5 m/s^2 + .2kg * 9.8 m/s^2) = 1.23 N + 1.23 N=2.46

    and so forth for the rest correct?

    for part b)
    I would do almost the same as I did for part a correct? part a I stop adding @ the 4th link since it has force exerted by the 5th which is topmost and the 3rd beneath it.
  8. Sep 16, 2007 #7


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    Not sure what happened above... : .1kg * 2.5 m/s^2 + .2kg * 9.8 m/s^2 is not 1.23... but 2.46 is the correct answer though.
  9. Sep 16, 2007 #8
    sorry did it first as 2ma+2mg=2.46 N
    but split it up into (ma+mg)+(ma+mg)=2.46N

    just forgot to change that .2 to a .1.

    But I'm still curious, why is it that when I add ma and mg they are both positive? a is pointing up but g is pointing down so shouldn't they be subtracting?....
  10. Sep 16, 2007 #9


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    If you're lifting 1 kg object upwards at an acceleration 1.0m/s^2 in a gravityless environment... you only need to exert 1N of force.

    If you're lifting 1 kg object upwards at an acceleration 1.0m/s^2 in a 9.8m/s^2 gravity environment... you'll need to exert a greater force... you're exerting a force to compensate for gravity.

    Fnet = ma
    Fupwards - mg = ma
    Fupwards = ma + mg

    The greater the gravity... the greater the force required to compensate for gravity, and move the object at the same acceleration.
  11. Sep 16, 2007 #10
    ok I think I get it now.
    so when I find ma for .1kg m the net force IS .1*2.5=.25N but that is the total of the force I want - mg so that's why I add the mg!

    Thanks alot!!
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