Problem 18
Question
The total cross-sectional area of the load-bearing calcified portion of the two forearm bones (radius and ulna) is approximately \(2.4 \mathrm{~cm}^{2}\). During a car crash, the forearm is slammed against the dashboard. The arm comes to rest from an initial speed of \(80 \mathrm{~km} / \mathrm{h}\) in \(5.0 \mathrm{~ms}\). If the arm has an effective mass of \(3.0 \mathrm{~kg}\) and bone material can withstand a maximum compressional stress of \(16 \times 10^{7} \mathrm{~Pa}\), is the arm likely to withstand the crash?
Step-by-Step Solution
Verified Answer
No, the bone is not likely to withstand the stress of the crash. The calculated compressional stress (-5.56 x 10^8 Pa) exceeds the maximum stress that the bone material can withstand (1.6 x 10^8 Pa).
1Step 1: Convertion of initial speed to m/s
First, the initial speed of the arm is given in km/h. To work in consistent units, this speed needs to be converted to meters per second. This can be done by using the conversion factor: \(1 \mathrm{\ km / h} = 0.27778 \mathrm{\ m / s}\). Thus, the initial speed in m/s is: \(80 \mathrm{\ km / h} \times 0.27778 \mathrm{\ m / s / km / h} = 22.22 \mathrm{\ m / s}\) .
2Step 2: Calculation of Deceleration
Next, calculate the deceleration the arm experiences when it comes to rest during the collision. Since the arm is brought to rest, the final speed is 0. Deceleration (or acceleration, in this context) can be calculated using the formula \(a = (v_{f} - v_{i}) / t\), where \(v_{f}\) is the final speed, \(v_{i}\) is the initial speed, and t is the time. Thus, \(a = (0 \mathrm{\ m/s} - 22.22 \mathrm{\ m/s}) / 5.0 \times 10^{-3} \mathrm{s} = -4444.4 \mathrm{m/s^2}\), where the negative sign indicates deceleration.
3Step 3: Calculation of Force
Now that we have deceleration, we can calculate the force the deceleration exerts on the arm using Newton's second law: \(F = m \times a\), where m is the mass of the arm. Thus, the force is: \(F = 3.0 \mathrm{\ kg} \times -4444.4 \mathrm{\ m / s^2} = -13333.2 \mathrm{\ N}\). The negative sign merely indicates that the force is exerted in the opposite direction of the initial motion.
4Step 4: Calculation of Stress
Next, the stress on the bones of the arm can be calculated using the formula \(S = F / A\), where S is the stress, F is the force, and A is the cross-sectional area of the bones. So, the stress here is \(S = -13333.2 \mathrm{\ N} / 2.4 \times 10^{-4} \mathrm{\ m^2} = -5.56 \times 10^{8} \mathrm{\ Pa}\).
5Step 5: Comparison of calculated stress and maximum allowable stress
Finally, compare the calculated stress with the maximum stress the bone can withstand, which is given as \(1.6 \times 10^{8} \mathrm{\ Pa}\). Since the calculated stress exceeds the maximum allowable stress, the bone would not withstand the impact.
Other exercises in this chapter
Problem 15
Bone has a Young's modulus of \(18 \times 10^{9} \mathrm{~Pa}\). Under compression, it can withstand a stress of about \(160 \times 10^{6} \mathrm{~Pa}\) before
View solution Problem 16
A high-speed lifting mechanism supports an \(800-\mathrm{kg}\) object with a steel cable that is \(25.0 \mathrm{~m}\) long and \(4.00 \mathrm{~cm}^{2}\) in cros
View solution Problem 21
(a) Calculate the absolute pressure at the bottom of a fresh-water lake at a depth of \(27.5 \mathrm{~m}\). Assume the density of the water is \(1.00 \times 10^
View solution Problem 23
A collapsible plastic bag (Fig. P9.23) contains a glucose solution. If the average gauge pressure in the vein is \(1.33 \times 10^{3} \mathrm{~Pa}\), what must
View solution