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Math Help - Really stuck please help

  1. #1
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    Really stuck please help

    Hey people,
    im really stuck on this and wondered if you could help please.

    Q: A heavy crate of mass m is pulled along a rough horizontal surface at a constant speed by a rope. The coefficient of friction between the crate and surface is u (substituted for mew as i dont know how to insert that)









    A) Show that T = umg/(cos(theta)+usin(theta).

    B) By finding dT/d(theta) and letting dT/d(theta) = 0, show that for a minimum value of T , tan(theta) = u. ( you may assume that, for a maximum value of T, (theta) = 0).

    any help would be really greatful guys as im really stuck.
    thanks.
    Last edited by the_sensai; March 8th 2007 at 03:44 PM. Reason: forgot diagram
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  2. #2
    MHF Contributor
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    Quote Originally Posted by the_sensai View Post
    Hey people,
    im really stuck on this and wondered if you could help please.

    Q: A heavy crate of mass m is pulled along a rough horizontal surface at a constant speed by a rope. The coefficient of friction between the crate and surface is u (substituted for mew as i dont know how to insert that)









    A) Show that T = umg/(cos(theta)+usin(theta).

    B) By finding dT/d(theta) and letting dT/d(theta) = 0, show that for a minimum value of T , tan(theta) = u. ( you may assume that, for a maximum value of T, (theta) = 0).

    any help would be really greatful guys as im really stuck.
    thanks.
    So where is the diagram?
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  3. #3
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    Quote Originally Posted by the_sensai View Post
    Hey people,
    im really stuck on this and wondered if you could help please.

    Q: A heavy crate of mass m is pulled along a rough horizontal surface at a constant speed by a rope. The coefficient of friction between the crate and surface is u (substituted for mew as i dont know how to insert that)









    A) Show that T = umg/(cos(theta)+usin(theta).

    B) By finding dT/d(theta) and letting dT/d(theta) = 0, show that for a minimum value of T , tan(theta) = u. ( you may assume that, for a maximum value of T, (theta) = 0).

    any help would be really greatful guys as im really stuck.
    thanks.
    Still no diagram, but I think I know what's going on here.

    First of all, the Greek letter "mew" (shudders) is spelled "mu." It's not a cat!

    My diagram has a crate being dragged to the right by a rope. The rope makes an angle t (in place of theta) with the horizontal.

    My Free-Body Diagram on the crate has a weight (w) acting directly downward, a normal force (N) acting directly upward, a kinetic friction force (f) acting to the left, and a tension (T) acting upward and to the right at an angle t with the horizontal. I have a +x axis to the right and a +y axis directly upward. The coefficient of kinetic friction is u.

    The speed is constant, so the acceleration is 0 m/s^2. Thus Newton's 2nd in the x and y directions say:
    SumFx = -f + Tcos(t) = 0
    SumFy = N - w + Tsin(t) = 0

    Thus
    N = w - Tsin(t) = mg - Tsin(t)

    Thus
    f = uN = umg - uTsin(t)

    So
    -(umg - uTsin(t)) + Tcos(t) = 0

    -umg + uTsin(t) + Tcos(t) = 0

    T(usin(t) + cos(t)) = umg

    T = umg/(usin(t) + cos(t))

    For the second part, find dT/dt:
    dT/dt = -umg/(usin(t) + cos(t))^2 * (ucos(t) - sin(t))

    Setting dT/dt = 0 gives:
    -umg(ucos(t) - sin(t))/(usin(t) + cos(t))^2 = 0

    So the numerator must be 0 (as long as the denominator isn't 0 for this angle). Thus
    ucos(t) - sin(t) = 0

    u = tan(t)
    as advertised.

    I'll leave it to you to prove if this is a max or min value for T.

    -Dan
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