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Minimal Tension angle

  1. Jan 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Body of mass m is being pulled by a string so that it moves horizontally at constant speed.
    Friction coefficient between surface and body is u.Find an angle (alpha) between a string and horizontal surface for which tension in string is minimal and find that tension.

    2. Relevant equations
    weight w=mg ; n=normal force
    friction force=f=un
    acceleration due x-axis=0
    tension T

    3. The attempt at a solution
    [tex]
    \sum F_y=Tsin(\alpha)+n+(-w)=0; \\
    T=\frac{(mg-n)}{sin(\alpha)} \\
    \sum F_x=Tcos(\alpha)+(-f)=0 \\
    \sum F_x=Tcos(\alpha)+(-un)=0; \: Tcos\alpha=un ; \\
    T=\frac{umg}{cos\alpha} \\
    \frac{(mg-n)}{sin(\alpha)}=\frac{umg}{cos\alpha} \\
    tan\alpha=\frac{mg-n}{umg}
    [/tex]

    Now I'm not sure what to do next.I need to find minimal angle so derivative of the last expression should be zero_Or should I differentiate arctan?I'm confused.
     
  2. jcsd
  3. Jan 12, 2015 #2

    rude man

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    Homework Helper
    Gold Member

    I can't follow what you wrote. What are f and n?

    You might consider: express Fx as a function of Fy and constants where T2 = Fx2 + Fy2.

    Then, express Fx and Fy as functions of constants and θ.
    Then find minimum of T in the usual way.
    Warning: the math is a bit messy but the answer is beautifully simple.
     
  4. Jan 12, 2015 #3

    mfb

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    2016 Award

    Staff: Mentor

    Don't cancel the tension if you are interested in it. You are not interested in the normal force, so you can get rid of this, that gives you an expression with tension and angle in it.
     
  5. Jan 13, 2015 #4
    @rude man , Note what i wrote under 2. : f - friction force, n - normal force
    @mfb I'm not sure how to connect these

    T=umg/cosα
    T=(mg-n)/sinα

    If i just differentiate T=umgcosx, I get nonsensical angle
    [tex]
    T'=-\frac{umg}{cos^2(\alpha)}
    [/tex]

    So this to be minimum(zero) , angle should be 90 degrees which makes no sense
     
  6. Jan 13, 2015 #5

    mfb

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    Staff: Mentor

    The normal force is not mg, you have to consider this for the horizontal part (you did that correctly for the vertical part).

    You have two equations with three unknown parameters - tension, angle, and normal force. This does not have a unique solution. To find the angle of minimal tension, it would be nice to have an equation that depends on tension and angle only (there you can follow the standard approach with the derivative), but not on the unknown normal force any more. You can solve one equation for the normal force and plug it into the other to get that.
     
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