Question

The total cross-sectional area of the load-bearing calcified portion of the two forearm bones (radius and ulna) is approximately 2.3 cm2. During a car crash, the forearm is slammed against the dashboard. The arm comes to rest from an initial speed of 80 km/h in 5.8 ms. If the arm has an effective mass of 3.0 kg, what is the compressional stress that the arm withstands during the crash?

Answer 1

To solve this problem, it is necessary to use the concepts related to the Force given in Newton's second law as well as the use of the kinematic equations of movement description. For this case I specifically use the acceleration as a function of speed and time.

Finally, we will describe the calculation of stress, as the Force produced on unit area.

By definition we know that the Force can be expressed as

F= ma

Where,

m= mass

a = Acceleration

The acceleration described as a function of speed is given by

$a = \frac{\Delta v}{\Delta t}$

Where,

$\Delta v =$ Change in velocity

$\Delta t =$Change in time

The expression to find the stress can be defined as

$\sigma=\frac{F}{A}$

Where,

F = Force

A = Cross-sectional Area

Our values are given as

$v= 80km/h\\t=5.8*10^3s\\m = 3kg \\A = 2.3*10^{-4}m^2$

Replacing at the values we have that the acceleration is

$a = \frac{\Delta v}{\Delta t}$

$a = \frac{80km/h(\frac{1h}{3600s})(\frac{1000m}{1km})}{5.8*10^3}$

$a = 3831.41m/s^2$

Therefore the force expected is

$F = ma\\F = 3*3831.41m/s^2 \\F = 11494.25N$

Finally the stress would be

$\sigma = \frac{F}{A}$

$\sigma = \frac{11494.25N}{2.3*10^{-4}}$

$\sigma = 49.97*10^6 Pa = 49.97Mpa$

Therefore the compressional stress that the arm withstands during the crash is 49.97Mpa