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Tension car cable Problem

  1. Feb 10, 2005 #1
    The Problem:
    Your car breaks down in the middle of nowhere. A tow truck weighing 4000 lbs. comes along and agrees to tow your car, which weighs 2000 lbs., to the nearest town. The driver of the truck attaches his cable to your car at an angle of 20 degrees to horizontal. He tells you that his cable has a strength of 500 lbs. He plans to take 10 secs to tow your car at a constant acceleration from rest in a straight line along a flat road until he reaches the maximum speed of 45 m.p.h. Can the driver carry out the plan? You assume that the rolling friction behaves like kinetic friction, and the coefficient of the rolling friction between your tires and the road is 0.10.

    I'm not sure where to start. I'm pretty sure I need to find the sum of all the forces in the x direction, then use the angle of 20 degrees to see if the the tension on the rope is more that 500 lbs. Mainly, im just not sure how to incorporate the kinetic friction. If anyone could provide a hint as to how incorporate the kinetic friction, or just let me know if I'm even on the right track, I'd appreciate it. Thanks
  2. jcsd
  3. Feb 10, 2005 #2
    The force from kinetic friction is anti parrallel to the direction of the car's motion.

    The equation for the magnitude of this force

    [tex] F_f = \mu_k n [/tex]

    where n is the normal force and [itex] u_k[/itex] is the coefficient of friction.

    your definitly on the right track, consider newton's second equation for horizontal and vertical components of force on the car.
  4. Feb 10, 2005 #3

    Andrew Mason

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    I assume that the tension limit is 500 lb force (ie the weight of 500 lb = 500*32 = 16000 ftlb of force).

    The forces on the tow cable are friction, gravity and ma. (note: the vertical component of the tension reduces the normal force which reduces friction).

    [tex]F_{acceleration} = ma[/tex]
    [tex]F_{friction} = \mu_kN = \mu_k(mg - Tsin(20))[/tex]

    (1)[tex]Tcos(20)=F_{acceleration} + F_{friction} = ma + \mu_k(mg-Tsin(20))[/tex]

    [tex]Tcos(20) = ma + \mu_kmg - \mu_kTsin(20))[/tex]
    [tex]T(cos(20)+sin(20)\mu_k) = ma + \mu_kmg[/tex]

    (2)[tex]T = \frac{ma + \mu_kmg}{(cos(20)+sin(20)\mu_k)}[/tex]

    [tex]ma = m\Delta v / \Delta t = 2000 * 66/10 = 13200[/tex]lbft/sec^2 (45 mph = 66 ft/sec)

    [tex]T = \frac{13200 + .1*2000*32}{(.94 + .34*.1)}[/tex]

    [tex]T = \frac{13200 + 6400}{.974} = 19770 > 16000[/tex]lb ft/sec^2

    Last edited: Feb 10, 2005
  5. Feb 10, 2005 #4
    Thanks a lot for the explanations!

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