What’s The Maximum Traction A Runner, Who Can Exert A Force Of Twice His Body Weight, Can Generate On A Rough

What’s the maximum traction a runner, who can exert a force of twice his body weight, can generate on a rough horizontal surface? Leave your answer in terms of the runners mass, the coefficient of friction. What’s the corresponding angle at which he must push against the ground?

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3 Responses to “What’s The Maximum Traction A Runner, Who Can Exert A Force Of Twice His Body Weight, Can Generate On A Rough”

  1. Dr D on January 16th, 2010 at 7:39 pm

    Hillarious.
    He pushes down with a force 2W at an angle ? with the vertical.
    Total downward force = vertical component of push + weight
    = 2W*cos? + W = W*(2cos? + 1)
    That’s the normal reaction, N.
    Horizontal component of push, T = 2W*sin?
    The question is whether this horizontal component can be supported by the friction.
    Max frictional force = ?*N
    = ?W(2cos? + 1)
    So essentially
    T = 2W*sin? <= ?W(2cos? + 1)
    And we wish to maximize T which involves maximizing sin?
    This is accomplished when
    2*sin? = ?(2cos? + 1)
    sin? – ?cos? = ?/2
    You could write the LHS as
    R*sin(? – ?)
    where R = sqrt(1 + ?^2)
    ? = atan(?)
    sqrt(1 + ?^2) * sin[? - atan(?)] = ?/2
    ? = asin[?/{2*sqrt(1 + ?^2)}] + atan(?)
    Tmax = 2*W*sin?
    If ? = 1, ? = 65.7 degrees,
    Tmax = 1.823*W

  2. (?) on January 17th, 2010 at 2:30 am

    the best angle so the traction is the highest is an angle at which u don’t feel the carried object’s weight and the friction coefficient is the lowest…..i would say like 45 degrees from the ground so that the mass of the pulled object is half the real weight of the object but yet again it depends on the friction if u want us to do ur homework at least give us some data to work with i would be happy to do it.

  3. linglong on January 17th, 2010 at 4:33 am

    Traction, friction force, is T = kN; where N = W cos(theta) + F and k is the friction coefficient, N is the normal force of the runner, W is the runner’s static weight, F is the impact force as the runner’s foot hits the track, and theta is the angle of incline (hopefully 0 degrees on a track).
    The traction is distributed over two feet (no not distance), but we presume they alternate on the track as the person runs. So all the runner’s weight, plus impact force, is on just one foot at a time.
    I am presuming “a force of twice his body weight” means N = 2W = W cos(theta) + F. In which case T = k(2W) = 2kmg; where m is the runner’s mass and g = 9.81 m/sec^2 or 32.2 ft/sec^2 depending on the units used for mass.

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