Friction

Ah, friction. Friction is the universal scapegoat, the culprit any good physicist blames for all experimental error. Friction is the reason the world has engineers rather than just physicists. Friction can be a drag. And now that I’ve exhausted my repertoire of sardonic comments on the irksome menace that is friction, we should probably learn a thing or two about it.

Friction is the force that opposes relative motion by converting kinetic energy into heat and sometimes sound and deformation. It is traditionally described by what is known as the Coulomb approximation, which states that F_f=mu F_n, where F_f is the frictional force, mu is the (dimensionless) coefficient of friction (most commonly mu_s, the coefficient of static friction, or mu_k, the coefficient of kinetic friction), and F_n is the normal or support force provided by the surface an object is sliding on. The coefficient of static or kinetic friction is dependent on many factors including the two materials in question, the environmental temperature, relative speed, atmospheric conditions, and other things. Therefore it must be determined experimentally; when we work with it in labs we will either be determining it or looking it up in a reference source.

Often we are concerned with the work done by friction. Since W=Int_c F dot ds and F_f=mu F_n, W_f=mu Int_c F_n dot ds along the path C traveled by the object in question. Since the work done by friction is path-dependent, we say that friction is a nonconservative force. If a force F is identified as conservative, that means any of three equivalent things:

  • oInt_c F dot ds=0 along any closed path C
  • F is irrotational; that is, curl F=0
  • The force may be written as F=grad Phi for some potential function Phi

Conservative forces, then, are those that conserve mechanical energy, such as gravity, the electromagnetic force, and Hookian spring forces.

As an example, take a mass m=5.00 kg sliding with initial velocity m=10.0 m/s on a surface with mu_k=0.250. We consider the following questions:

  1. How much energy is dissipated (i.e., how much work is done) by friction as the object comes to a stop?
  2. How far will the object slide before it reaches a stop?
  3. How long will the object take to come to a stop?

To answer the first question, we note that W=delta E, so the work done by friction is the energy dissipated. We consider that the mass has initial kinetic energy K_i=(1/2)mv_i^2=(1/2)(5.00 kg)(10.0 m/s)^2=250. J and must end with kinetic energy K_f=0. J, so W_f=delta K=-250. J.

To answer the second question, we notate as delta x the distance the object slides as it comes to a stop. We first note that F_n is constant and so we can pull it out of the integral W_f=mu Int_c F_n dot ds leaving W_f=mu F_n dot Int_c ds or W_f=mu F_n dot delta s. Then we calculate F_n=m a_g = (5.00 kg)(-9.80 m/s^2)=-49.0 N. We plug in the values to get -250. J=0.250(-49.0 N)delta x. Solving for delta x we have delta x = 20.4 m.

To answer the final question, we consider that F net=ma. We know that F net=F_f=mu F_n, so we plug in F_f=0.250(-49.0 N) to get F_f=-12.3 N. Plugging in for the mass, we have -12.3 N=(5.00 kg)a for a=-2.45 m/s^2. Now a=dv/dt and we know that delta v=-10.0 m/s, which gives -2.45 m/s^2=(-10.0 m/s)/delta t. Then delta t=4.08 s.

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