Kvant Physics Problem 63

Verified: yes
Verdicts: PASS + PASS
Solve time: 8m57s
Source on kvant.digital

Problem

F63. In order not to slip on an icy hill, a person runs down it. Why is this advisable?

Setup and Assumptions

The physical system consists of a person moving on an icy slope inclined at an angle $\alpha$ to the horizontal. The coefficient of static friction between the soles of the shoes and the ice is $\mu$.

The question asks why running down the slope is safer than attempting to stand or walk slowly on it.

The motion is analyzed during a single supporting phase when one foot is in contact with the ground. Air resistance is neglected. The coefficient of friction is assumed to be small, corresponding to icy conditions. During the contact interval, the acceleration of the center of mass is assumed to be directed parallel to the slope. Between supporting phases, the runner is airborne and no contact force acts.

Physical Principles

The condition for the absence of slipping is that the tangential force required at the contact point must not exceed the maximum static friction force,

$|F_t|\le F_{\max},$

where

$F_{\max}=\mu N,$

and $N$ is the normal reaction of the surface.

Newton's second law applies both parallel and perpendicular to the slope. During a supporting interval, if the center of mass has instantaneous acceleration $a$ down the slope, then

$mg\sin\alpha-F_t=ma,$

where $m$ is the person's mass and $g$ is the acceleration due to gravity.

Perpendicular to the slope,

$N=mg\cos\alpha,$

because the acceleration is taken parallel to the slope.

These equations describe only the intervals when the foot is in contact with the ice. During the airborne part of the stride there is no contact, so there is neither friction nor any possibility of slipping.

Derivation

Suppose first that a person is standing on the slope without accelerating. In that case $a=0$, and Newton's second law along the slope gives

$F_t=mg\sin\alpha.$

The condition for equilibrium is

$mg\sin\alpha\le\mu mg\cos\alpha,$

$\tan\alpha\le\mu.$

If the slope is steeper than permitted by this inequality, the person cannot remain at rest and begins to slide.

Now consider a person running downward. During a supporting phase, the body is allowed to have a nonzero instantaneous acceleration $a$ down the slope. Newton's second law gives

=m(g\sin\alpha-a).$$The no-slip condition becomes$$m(g\sin\alpha-a)\le\mu mg\cos\alpha.$$After dividing by $m$,$$g\sin\alpha-a\le\mu g\cos\alpha.$$Solving for $a$ gives$$a\ge g(\sin\alpha-\mu\cos\alpha).$$This relation refers to a particular contact interval and not to the average motion over many steps. In steady running the average acceleration over a complete stride is zero, but the acceleration during each support phase varies with time. Whenever the center of mass is accelerating downward along the slope, the tangential force required from the supporting foot is smaller than the force required for static equilibrium. The limiting value$$a=g\sin\alpha$$gives$$F_t=0.$$This corresponds to the situation in which gravity alone produces the instantaneous acceleration along the slope. Such a value is not maintained continuously while the foot remains on the ground, but it shows that larger downward acceleration during contact requires less friction. Running differs from standing because contact with the ice is intermittent. During the airborne parts of the stride the body moves under gravity with no contact and hence no danger of slipping. During the brief supporting intervals the foot can supply only a relatively small tangential force, since the body is already moving downward and need not be held in equilibrium. The average acceleration over a complete stride may vanish, but the friction requirement during each short contact interval is substantially smaller than the friction needed to remain stationary. ## Result The friction force required during a supporting interval with instantaneous downhill acceleration $a$ is$$F_t=m(g\sin\alpha-a).$$For a stationary person,$$F_t=mg\sin\alpha.$$Whenever the supporting phase satisfies $a>0$,$$F_t<mg\sin\alpha.$$The condition for secure footing during that contact interval becomes$$a\ge g(\sin\alpha-\mu\cos\alpha).$$No numerical data are given in the problem, so no numerical value can be calculated. The physical answer is$$\boxed{\text{Running downhill is advisable because contact with the ice is only intermittent, and during the short supporting intervals the body is allowed to accelerate downhill, so the friction required from the foot is smaller than that needed to remain at rest.}}$$## Sanity Checks The expression$$F_t=m(g\sin\alpha-a)$$has units of force, because both $g$ and $a$ have units of $\mathrm{m/s^2}$ and multiplication by $m$ gives newtons. If $a=0$, the formula reduces to$$F_t=mg\sin\alpha,$$which is the result for a person standing on the slope. If$$a=g\sin\alpha,$$then$$F_t=m(g\sin\alpha-g\sin\alpha)=0.$$This limiting case represents motion whose instantaneous acceleration along the slope is produced entirely by gravity. The expression$$F_t=mg\sin\alpha+ma$$would imply that increasing the downhill acceleration increases the required tangential force. Substituting this expression into Newton's second law,$$mg\sin\alpha-F_t =mg\sin\alpha-(mg\sin\alpha+ma) =-ma,$$shows that it would correspond to an acceleration directed up the slope, contrary to the assumed motion. Hence the correct sign is the minus sign in$$F_t=mg\sin\alpha-ma.$$