66 Falling Injuries

We started out this chapter by claiming that we would eventually be able to analyze the forces on the body during an impact with a hard surface after a fall. We have reached that point. Let’s do it.

Everyday Examples: Forces during a Fall

An 80kg person falls 0.80 m from a hospital bed onto a concrete floor. First-off, how much time do they have to reach out and grab something?

In the previous chapter we found an equation for calculating the fall time when starting from rest:

(1)   \begin{equation*} t =  \sqrt{\frac{-2\bold{x_i}}{\bold{a}}} \end{equation*}

Entering our values:

(2)   \begin{equation*} t =  \sqrt{\frac{-2(0.80\,\bold{m})}{-9.8\,\bold{m/s/s}}}=0.404\,\bold{s} \end{equation*}

If you do the lab at the end of this Unit you will find that 0.4 s is near the limit of human reaction time. If reaction time was impaired for any reason, which is common in hospital patients, it’s likely that the person would hit the ground without grabbing something to slow down.

How fast will the person be moving when they hit the floor (what is the impact speed)? Using another kinematic equation:

(3)   \begin{equation*} \bold{v_f} = \bold{v_i} + \bold{a}\Delta t \end{equation*}

And entering our values:

(4)   \begin{equation*} \bold{v_f} = 0 + (-9.8\,\bold{m/s/s})(0.404\,\bold{s}) = -3.96 \,\bold{m/s} \end{equation*}

The comes out negative as expected because they are moving downward. The hard floor will bring them to a stop in just a fraction of a second, even if they are able to get their arms out to hit first. A reasonable estimate would be about 0.2 s (more on this in the next unit). What is the person’s average during impact?

Using the same equation as before:

(5)   \begin{equation*} \bold{v_f} = \bold{v_i} + \bold{a}\Delta t \end{equation*}

But now solving for acceleration:

(6)   \begin{equation*} \bold{a} = \frac{\bold{v_f} - \bold{v_i}}{\Delta t} \end{equation*}

And entering our values:

(7)   \begin{equation*} \bold{a} = \frac{0 - (-3.96\bold{m/s})}{0.2\,\bold{s}} = 19.8\,\bold{m/s/s} \end{equation*}

Now we are ready to calculate the average on the person. We’ll start from :

(8)   \begin{equation*} \bold{F_{net}} = m\bold{a} \end{equation*}

Entering our values:

(9)   \begin{equation*} \bold{F_{net}} = (80\,\bold{kg})(19.8\,\bold{m/s/s}) =1584\,\bold{N} \end{equation*}

Finally, what force does the floor apply (as a ) to the person’s back to achieve that net force, despite their weight?

We recognize that the the is the result of the upward normal force plus the downward .

(10)   \begin{equation*} \bold{F_{net}} = \bold{F_{N}} + \bold{F_g} \end{equation*}

We solve for the normal force:

(11)   \begin{equation*} \bold{F_{N}} = \bold{F_{net}} - \bold{F_g} \end{equation*}

Now we need to calculate the weight, keeping in mind that is negative because it is downward:

(12)   \begin{equation*} \bold{F_g} = -mg = -(80\,\bold{kg})(9.8\,\bold{m/s/s}) = -784\,\bold{N} \end{equation*}

Finally entering values for net force and weight to get the normal force:

(13)   \begin{equation*} \bold{F_{N}} = 1584\,\bold{N} - (-784\,\bold{N}) = 2368\,\bold{N} \end{equation*}

That is more than three times the body weight. We will see in the next chapter that the peak force is actually much greater than the average force during impacts like this, so in fact this situation is actually worse than our calculations indicate. Now we see why patient falls must be avoided.

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