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Vector form of position, velocity, acceleration, and force

  1. Aug 16, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]x=\left(\begin{array}{cc}\frac{3}{5}t^2+2t\\10t^2+1\end{array}\right) , 0\leq t \leq 5 \; \text {and} \; F=\left(\begin{array}{cc}3\\1\end{array}\right)[/tex]

    a. find v in vector form

    b. find mass

    c. when t = 5, there is addition of [tex]F=\left(\begin{array}{cc}t\\0\end{array}\right)[/tex]. Find the acceleration when t = 6

    d. find v when t = 6

    2. Relevant equations
    F = ma


    3. The attempt at a solution
    a. [tex]v=\left(\begin{array}{cc}\frac{6}{5}t+2\\20t\end{array}\right)[/tex]


    b. [tex]|F|=\sqrt{3^2+1^2}=\sqrt{10}[/tex]

    [tex]a=\left(\begin{array}{cc}\ 6/5 \\20\end{array}\right)[/tex]

    [tex]|a|=\sqrt{\left(\frac{6}{5}\right)^2+20^2}\approx 20.04[/tex]

    [tex]m=\frac{|F|}{|a|}=\frac{\sqrt{10}}{20.04}\approx 0.158\; kg[/tex]


    c. [tex]F \; \text{total}=\left(\begin{array}{cc}3\\1\end{array}\right)+\left(\begin{array}{cc}t\\0\end{array}\right) = \left(\begin{array}{cc}3\\1\end{array}\right)+\left(\begin{array}{cc}6\\0\end{array}\right)=\left(\begin{array}{cc}9\\1\end{array}\right)[/tex]

    [tex]|F \; \text{total}|=\sqrt{9^2+1^2}=\sqrt{82}[/tex]

    [tex]|a|=\frac{|F|}{m}\approx 57.31 \;ms^{-2}[/tex]


    d. [tex]v=\left(\begin{array}{cc}46/5\\120\end{array}\right)[/tex]

    [tex]|v|=\sqrt{\left(\frac{46}{5}\right)^2+120^2}\approx 120.35 \;ms^{-1}[/tex]


    Do I get it right ?

    Thx
     
  2. jcsd
  3. Aug 16, 2009 #2

    rock.freak667

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    yep looks correct to me.
     
  4. Aug 16, 2009 #3

    kuruman

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    This problem troubles me. Newton's Second Law is a vector equation

    F = ma and implies that

    Fx=max and Fy=may

    from which we get that

    m = Fx/ax = Fy/ay

    I agree that a = (6/5, 20) so if F = (3, 1), this simply doesn't work.
     
  5. Aug 16, 2009 #4
    Thx rock.freak667

    Another question : Is it possible if

    [tex]x=\left(\begin{array}{cc}\frac{3}{5}t\\10t^2+ 1\end{array}\right)[/tex]

    So, the velocity :

    [tex]v=\left(\begin{array}{cc}\frac{3}{5}\\20t\end{array}\right)[/tex]

    i.e. the velocity is constant in x-direction and depends on t in y-direction?

    I think it's possible such in projectile motion, where the velocity in x-direction is constant and changing in y-direction. But I'm not sure....

    Thx :)

    EDIT :
    Sorry, I just read kuruman's post. Yes that makes sense and now I'm confused....or maybe F = ma doesn't imply that Fx=max and Fy=may ? (just guessing)
     
  6. Aug 16, 2009 #5

    kuruman

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    You are correct. It is entirely possible to have one component depend on time and not the other. As you say, this is the case with projectile motion

    [tex]
    v = \left(\begin{array}{cc}v_{0x}\\v_{0y}-gt\end{array}\right).
    [/tex]

    F = ma is a vector equation. It says that "the vector on the left is the same as the vector on the right". When are two vectors the same? When their x components are the same and their y components are the same. Whoever authored this problem over-constrained it so that the bottom line is inconsistent with Newton's Second Law.
     
  7. Aug 16, 2009 #6

    kuruman

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    Addendum to my previous post

    This is a bad problem.
     
  8. Aug 16, 2009 #7
    Hi kuruman

    So it should be :

    [tex]m=\frac{F_x}{a_x}=\frac{F_y}{a_y}=\frac{|F|}{|a|}\; ????[/tex]

    Another question :
    Maybe it is also possible to have acceleration that is constant in x-direction and changing in y-direction? If so, the force will also constant in x-direction and changing in y-direction?

    Thx
     
  9. Aug 16, 2009 #8

    kuruman

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    You are correct on both accounts. But the ratio

    Fx/ax should be equal to to Fy/ay and equal to |F|/|a| at all times, no matter what the time dependence of the individual components is.

    Stated differently:

    The angle between the acceleration vector and the x-axis is
    arctan(20/(6/5))=33.7o

    The angle between the force vector and the x-axis is at all times
    arctan(1/3) = 18.4o

    Conclusion: The acceleration vector does not point in the same direction as the force. What does one make of this?

    If the given force is the only one acting on the mass, then it is a bad problem because F is the net force and must be in the same direction as a. If F is not the net force and there is another force that tips the acceleration vector relative to the given force, the problem mentions no such force or asks you to find it. Still bad problem.
     
    Last edited: Aug 16, 2009
  10. Aug 16, 2009 #9
    Hi kuruman

    After reading your explanation, I agree that this is bad problem

    Thx a lot for your help ^^
     
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