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calculating required coefficient of friction

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  • calculating required coefficient of friction

    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.
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