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Force required to change the direction

  1. Nov 11, 2013 #1

    adjacent

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    1. The problem statement, all variables and given/known data
    I haven't studied forces acting on angles yet.
    An object of mass 1kg is moving horizontally at a constant speed 10m/s.What is the the force required vertically to turn it 40 degrees to vertical?

    Not a home work question


    2. Relevant equations
    :confused:


    3. The attempt at a solution
    :confused:
    As I said,I haven't studied this yet.This is a homework like question. :smile:
     
  2. jcsd
  3. Nov 11, 2013 #2

    HallsofIvy

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    Your question is not complete. Any amount of force, no matter how small will eventually change the direction by any desired angle. You need to specify a time interval.

    (The force acting vertically is not enough. In order that there be no change in speed, the force must act perpendicular to the velocity vector so the force must change direction as the object does.)

    (Surely you are not thinking that the direction will change instantaneously[/b] by 40 degrees? The path of the path of the object will curve to that direction.)
     
  4. Nov 11, 2013 #3

    adjacent

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    I see.Change in speed is not a matter.What's the equation relating force,time and angle?
     
  5. Nov 11, 2013 #4
    I am not sure what you mean by the above (change in speed is not a matter).
    If you assume constant speed, you could calculate the change in momentum between the two states: moving horizontally and moving at an angle of 40 degrees.
    This change in momentum will give you the impulse of the force:
    Δmv=F*Δt
    where v and F are vectors and F is the average force.
     
  6. Nov 11, 2013 #5
    Its better to think in terms of momentum.
    F=dp/dt
    F=dmv/dt (well, classically speaking...)
    F=ma
    Now to find a relation between F, t, θ and p.
    F-force t-time θ-angle p-momentum
    Take the initial direction of p (ie. p(0)) along x axis ([itex]\hat{i}[/itex]))
    Let force act in the x-y plane and [itex]\hat{i}[/itex] be along x axis.
    F= a[itex]\hat{i}[/itex]+b[itex]\hat{j}[/itex]
    Now at any time t-
    p(t)=at+p[itex]\hat{i}[/itex]+bt[itex]\hat{j}[/itex]
    ∴tan(θ)=bt/(at+p)
    θ=tan-1(bt/(at+p))
    But equations like these are never used or necessary and usually derived to suit the needs.
     
  7. Nov 11, 2013 #6

    adjacent

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    What is a and b?
     
  8. Nov 11, 2013 #7
  9. Nov 11, 2013 #8

    adjacent

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    There is no force on the x axis.It's moving with a constant velocity.A force is applied on the y axis
     
  10. Nov 11, 2013 #9
    Then a=0.
    tan(θ)=bt/(p)
    At θ=90° tanθ does not exist, therefore it can never turn to 90°
    This happens because as there is no force along x axis the momentum along it does not change. Then it follows that there is always some velocity acting along x axis.
     
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