Jumping off a chair (average force)

Problem

Estimate the average force acting on your feet when you jump down from a chair of height $50\;\text{cm}$ and land, cushioning the fall in $\sim 0{,}2\;\text{s}$. Compare it with your weight. What happens if you land stiffly, in $\sim 0{,}02\;\text{s}$?

Solution

This is a Fermi-style estimate: we assume a body mass $m \approx 70\;\text{kg}$ and $g \approx 9{,}81\;\text{m/s}^2$.

Step 1 — Speed at impact. Free fall from $h = 0{,}50\;\text{m}$ gives, by energy conservation (or $v^2 = 2gh$),

$v = \sqrt{2 g h} = \sqrt{2 \cdot 9{,}81 \cdot 0{,}50} \approx \sqrt{9{,}81} \approx 3{,}1\;\text{m/s}$

Step 2 — Impulse needed to stop. On landing, the momentum goes from $p = m v$ to zero. The impulse (change in momentum) that must be supplied by the legs to stop the body has magnitude

$\abs{\Delta p} = m v = 70 \cdot 3{,}1 \approx 2{,}2 \cdot 10^2\;\text{N}\cdot\text{s} \sim 200\;\text{N}\cdot\text{s}$

Step 3 — Average force (soft landing, $\Delta t \approx 0{,}2\;\text{s}$). The average force decelerating the body is obtained by dividing the impulse by the stopping time:

$\bar F \approx \frac{\abs{\Delta p}}{\Delta t} = \frac{2{,}2 \cdot 10^2}{0{,}2} \approx 1{,}1 \cdot 10^3\;\text{N} \sim 1000\;\text{N}$

$\ev{\abs{\Delta p} \sim 200\;\text{N}\cdot\text{s} \qquad \bar F \sim 1000\;\text{N}}$

Comparison with weight. The weight is $P = mg = 70 \cdot 9{,}81 \approx 690\;\text{N}$. The force on the feet is thus about $1{,}5$–$2$ times the weight: bending the legs spreads the deceleration over a relatively long time and the effort stays tolerable. (The total normal force is $\bar F + P$, but the order-of-magnitude estimate remains $\sim 10^3\;\text{N}$.)

Stiff landing ($\Delta t \approx 0{,}02\;\text{s}$). The same impulse concentrated into a time ten times shorter produces a force ten times greater:

$\bar F \approx \frac{2{,}2 \cdot 10^2}{0{,}02} \approx 1 \cdot 10^4\;\text{N} \sim 15\;\text{times the weight}$

That is why one always lands by bending the knees: lengthening $\Delta t$ is the only way to reduce the force for a given impulse to be absorbed. Landing with straight legs can cause fractures precisely because of this force peak.

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Topics: Dynamics Concepts: Newton’s second law · Impulse Skills: Fermi estimates