View Full Version : calculating required coefficient of friction

06-25-2008, 02:12 AM
Friction force equals the coefficient of friction times the normal force.
By calculating the ratio of shear forces to normal forces, one may calculate
the "required coefficient of friction," which is the coefficient of friction
needed to avoid slipping. From what I've read, on a flat surface, shear
force is the vector sum of anteroposterior and mediolateral ground reaction
force components, and normal is the vertical component. If the substrate is
cylindrical (as in the arboreal trackways that I use), then mediolateral and
vertical components each contribute some to shear and normal forces,
depending on where the limb contacts the cylinder. The anteroposterior
force component contributes only to shear. My question is: does pure torque
also contribute to the calculation of shear force? For example, if a person
steps onto the very center of a force plate and twists to the right, is the
shear force the vector sum of anteroposterior force, mediolateral force, and
the torque within the horizontal plane that does not result from
anteroposterior and mediolateral forces applied off-center? If it's an
arboreal substrate (which is my real question), does a torque around the
long axis of the branch trackway contribute to shear force? My guess is
that the "pure" torque (that torque which results from the limb exerting a
twisting moment, and NOT the torque that results from substrate reaction
forces being applied off-center to the cylinder) SHOULD be included in the
vector sum of shear components of vertical and mediolateral forces and the
anteroposterior force. Please e-mail me with your arguments or suggestions!
Thanks - Andrew Lammers, Dept of Health Sciences, Cleveland State
University, a.Lammers13@csuohio.edu.