50 Locomotion Forces

Walking with Friction

In the previous chapter we used the Impulse-Momentum Theorem to analyze how force, time, mass, and change in velocity are related during falling and landing. We can use the same strategies to analyze forces associated with everyday locomotion.

Everyday Example: Responding to a Code Blue

Jolene is walking with a speed of a 2.5 mph down the hospital corridor when a code blue is called over the intercom. She stops then turns around and starts walking the other direction, toward the room where the code was called, at a speed of 4.0 mph.  If Jolene’s mass is 61 kg and she tried to make that change in motion very quickly, for example in just 0.75 s,  how large of a force must she receive from the ground?

We will start with the impulse momentum theorem for an object with constant mass:

(1)   \begin{equation*} \Delta t \bold{F_{ave}} = m \bold{\Delta v} \end{equation*}

We want to solve for the force applied to Jolene so we divide both sides by the time interval:

(2)   \begin{equation*} \bold{F_{ave}} = m \frac{\bold{\Delta v}}{\Delta t} \end{equation*}

We have only been given speeds in the problem, but in order to analyze velocity we need to define direction of motion. Let’s assign Jolene’s initial direction as the negative direction so her initial velocity was -2.5 mph. In that case her final velocity was +4.2 mph. We can calculate her change in velocity as:

(3)   \begin{equation*} \bold{\Delta v} = \bold{v_f}-\bold{v_i} = 4.0\,\bold{mph}  - (- 2.5\,\bold{mph}) = 6.5 \,\bold{mph} \end{equation*}

If we convert our answer to units of m/s we get: 6.5 mph = 2.9 m/s. (You can check this yourself using unit analysis or an online unit converter). Now we can enter the mass, change in velocity, and  time interval into the impulse momentum equation:

(4)   \begin{equation*} \bold{F_{ave}} = m \frac{\bold{\Delta v}}{\Delta t} = (61 \,\bold{kg})\frac{(2.9 \,\bold{m/s})}{0.75 \,\bold{s}} = 240\,\bold{N} \end{equation*}

To make the desired change in motion Jolene must experience a 240 horizontal frictional force from the floor.

Newton’s Third Law of Motion

Looking back at the example above, we could analyze the larger system that includes both Jolene as one internal component and the building + Earth as a second internal component. In that case the frictional forces between Jolene and the floor are internal to the system and we can say the system is isolated. Therefore the system momentum can not change so the net impulse on the system must be zero and we know the Building + Earth must have experienced the same impulse as Jolene, but in the opposite direction. In other words, during the time that Jolene received a force from the floor, she must have also exerted an equal force back onto the floor in the opposite direction. This result is summarized by Newton's Third Law of Motion, which states that when system A applies a force to system B, then System B must also apply an equal and opposite force back on System A. 

The equal and opposite forces referenced in Newton’s Third Law are known as third law pair forces (or third law pairs).

Other Third Law pair forces include:

  • The Earth pulls down on you due to gravity and you pull back up on the Earth due to gravity.
  • A falling body pushing air out of its way and air resistance pushing back on the body.
  • You pull on a rope and the rope pulls back against your hand via tension.
  • You push on the wall, and the wall pushes back with a normal force.
  • A rocket engine pushes hot gasses out the back, and the gasses push back on the rocket in the forward direction.
  • You push your hand along the wall surface, and the wall pushes back on your hand due to kinetic friction.
  • You push your foot against the ground as you walk, and the floor pushes back against your food due to friction (static if your foot doesn’t slip, kinetic if it does).

You may have noticed that in each of the cases above there were two objects listed. This is because Newton’s Third Law pairs must act on different objects.  Therefore, Third Law pair forces cannot be drawn on the same free body diagram and can never cancel each other out.  (Imagine if they did act on the same object, then they would always balance each other out and no object could ever have a net force, so no object could ever accelerate!)

Reinforcement Exercises

Everyday Examples

At some point in your life you may have tried to change direction too quickly on a slick floor and slipped. Let’s check to make sure that wouldn’t happen to Jolene in the previous example. We know that the friction force between Jolene and the floor must be 240 N, so we will check to see if the maximum static frictional force can be that large. If not, then Jolene would slip. In that case kinetic friction would eventually bring her to a stop, but not as quickly and possibly not before she lost her balance.

First we start with the equation for max static friction force:

    \begin{equation*} F_{f,s} = \mu_s F_N \end{equation*}

Looking up friction coefficients we find that the rubber-concrete friction coefficient is typically 0.6 or greater. Assuming the floor is level and only Jolene’s horizontal motion is changing, then the normal force must be balancing Jolene’s weight (if these were not balanced then she would have a vertical change in velocity as well). In that case we substitute her weight for the normal force:

    \begin{equation*} F_{f,s} = \mu_s mg \end{equation*}

Finally we enter our values:

    \begin{equation*} F_{f,s} = 0.6(61\,\bold{kg})(9.8\,\bold{m/s/s}) = 358\bold{N} \end{equation*}

The max static frictional force is 358 N and the quick turn described in the previous example only requires 240 N of frictional force so Jolene could easily make the turn around without slipping.



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