There are many forces at work during running.   There are probably too many to name.  But here are some.  There is air drag, which is very important at the highly competitive levels, especially for long distance running.  There is the constant pounding of the feet and shock on the muscles, connective tissue, and skeleton of the runner.  And there are the forces that come into play from the change in the course like turns, banked turns, rises, and falls in elevation.  I defer the forces involved in the physiology of locomotion because those sorts of things are covered in biophysics, biomechanics, and some medical physics texts.  I will concentrate on the forces involved between the runner and the track.

 

There are a couple of loosely related questions/responses on centripetal and centrifugal forces, curved roads, turning, and leaning on a motorcycle. [note added by MadSci Admin: These previous answers can be found, and many many others, too, by using our search engine]

 

Re: gravity and centripital force of a rotating planet

http://www.madsci.org/posts/archives/may98/893685609.Ph.r.html

 

Re: Why are curves on roads banked?

http: //www.madsci.org/posts/archives/may99/927343264.Eg.r.html

 

Re: What devices help race cars take turns at faster speeds? And how?

htt p://www.madsci.org/posts/archives/oct2000/971203102.Ph.r.html

 

Re: What forces enable motorcycle to go through turns at the highest speeds?

http://www.madsci.org/posts/archives/apr2000/955980971.Ph.r.html

 

 

 

In general, a turn can be modeled as a part of a circle, or many smaller circles, during which there must be a centripetal force to stay on the course

 

Force_centripetal= - mass * speed²/radius_turn (radial)

 

where speed is the tangential speed of the runner and radius_turn is the radius_turn of the current circular turn, mass is the mass of the runner, and the negative sign indicates that the force (vector) is directed toward the center of the circle.

 

The equation basically means that a tighter turn requires more force or lower speed.  The force to make the turn comes from the runners legs pushing on the track.  This force is limited by the amount of friction between the shoe and the surface.   This is true for skaters, motorcycles, and cars as well, except that the contact is between skates and surface (blade and ice for ice skating), or the tires and the road.  

 

The force of friction is

 

Force_friction = u_friction * Normal force

 

where u_friction is the coefficient of friction when the runner or vehicle is in moving contact with the road.  It is the contact force between the shoe or tire and the road.  The normal force is the force applied perpendicularly to the surface.  If the surface is flat then this is the weight of the runner or vehicle.

 

If the friction is low then the racer could lose grip with the surface and fly off the curve.  This effect is exaggerated in short or long track ice speed skating.  If the skater loses an edge then the friction drops tremendously, and the poor skater cannot turn and usually wipes out into the wall or off the track.  In other words, there is an equipment limit to how hard a racer can push.  Usually this limit is above the athlete’s strength limit but if the track is slick then it could be possible to push hard enough to just slip.

 

 

 

Banking the turns lowers the friction force required to maintain the turn because the mass of the moving object contributes to the centripetal acceleration.  This is similar to an incline plane problem or going up or down a hill.  A previous response I submitted has a drawing of the situation for a bicycle going up hill.

 

 

 

Reference Frame Axes

Re: effects of wind(air resistance), gravity, surfaces on speed

htt p://www.madsci.org/posts/archives/apr2000/955809806.Ph.r.html

 

If you think of the runner on the banked curve as a block on an incline plane the component of the weight that is trying to move down a bank of angle Theta is given by

  

Weight_{down bank} = mass * gravity * sin (theta)

 

On a banked curve, the normal force is the weight down into the bank:

 

Normal force = mass * gravity * cos (theta)

 

where theta is the angle the bank makes with the horizontal.  You can see this in a diagram of an incline plane.

 

If the friction is enough to keep the runner (or motorcycle, or car) from sliding, then the Weight _{down bank} will allow for faster motion around the banked turn because the lean reduces the force required to make the turn.

 

For the following derivation, let’s assume for the moment that the runner doesn’t lean, but that the angle of the bank creates a contribution that comes from the weight of the runner.

 

 

We need to break this down into components and the easiest, or at least most instructive frame of reference is the track. 

Note, this is not a statics problem.  I We need the sum of forces to equal to the mass times the acceleration desired to keep running the circle  (mass * speed²/radius_turn).  In the case of a static problem, the force we want is zero.

 

Axis Normal to Track(away from track is positive):

Sum F_N = + (mass * speed²/radius_turn) * sin(theta)

 

Axis Parallel to Track toward center of turn:

Sum F_{down bank} = + (mass * speed²/radius_turn) * cos(theta)

 

 

Adding up the force components:

 

Sum F_N = + (mass * speed²/radius_turn) * sin(theta) = - mass * gravity * cos(theta) + Push from Track

 

 

Sum F_{down bank} = + (mass * speed²/radius_turn) * cos(theta) = + mass * gravity * sin(theta) + Friction from shoes

These two equation need to be solved to maintain lane and keep running the banked curve.

 

 

The Push from the track is adjustable.  In otherwords, the track will push back as much is needed:

 

Push From Track = - (mass * speed²/radius_turn) * sin(theta)  + mass * gravity * cos (theta),

 

where positive is upward.   Remember, the push from the track cannot be negative!  It gives a sort of critical speed based on the push from the track being negative:

 

speed_{max from normal force} = SQRT [(gravity/radius) * cotangent(theta)].

 

Note that if the track is flat, then there is no limitation on speed based on the normal forces from this equation.  So it really doesn’t blow up exactly.

 

Substituting the normal force into the frictional force equation we have

 

Friction from shoes = u_friction* mass*gravity * cos(theta).

 

And when this is substituted into the equation for force along the direction of the track toward the center of the curve,

 

Sum F_{down bank} = + (mass * speed²/radius_turn) * cos(theta) = + mass * gravity * sin(theta) + u_friction* mass*gravity * cos(theta).

 

This equation indicates that as friction goes down, the racer needs to increase the lean, perhaps beyond the angle of the track.

 

 

Okay, now if we let the racer lean into the turn then we can change the equation to:

 

 

Sum F_{down bank} = + (mass * speed²/radius_turn) * cos(theta) = + mass * gravity * sin(phi) + u_friction* mass*gravity * cos(theta),

 

where phi is the total angle of leaning by the racer.  I am assuming the frictional component is still related to the angle theta because the weight of the racer is still down into the track and the racer would have to adjust his/her balance.

 

Similar derivation could be used for cross country or long distance road racing with changes in course pitch.

 

In the above discussion, I ignored thinking about the torque involving the center of mass of the runner and leaning into a turn.   Provided the runner keeps his/her center of mass within the frame of his/her support.  A book that does a pretty good job of explaining torque in human motion is Physics, Dance, and the Pas de Deux by Kenneth Laws and Cynthia Harvey.  I recommend it highly for anyone interested in the physics of human athletics.

 

 

I hope this helps,

 

Sincerely,

 

Tom “Lean, Mean, or Somewhere In Between” Cull