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Chapter 9: Problem 12

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?

Short Answer

Expert verified
The answer will depend on the comparison made in Step 5. If the calculated stress is less than the maximum compressional stress, then the arm is likely to withstand the crash, otherwise, it will not.

Step by step solution

01

Convert Initial Speed to m/s

The initial speed is given in km/hr. We need to convert this into m/s because the mass of the arm is given in Kg and time in seconds. This is necessary for further calculations. To convert km/hr into m/s, divide the speed by 3.6. So, \(v = 80 \mathrm{~km/hr} = \frac{80}{3.6} \mathrm{~m/s}\).
02

Calculate Acceleration

Acceleration is defined as change in velocity per unit time. During the crash, the arm comes to rest so its final velocity is 0. Therefore, acceleration (\(a\)) can be calculated using the formula \(a = \frac{(Final Velocity - Initial Velocity)}{Time} = \frac{(0 - v)}{t}\). Where \(v\) is initial speed and \(t\) is time taken to come to rest.
03

Calculate Force

Once we have the acceleration, we can calculate the force using Newton's Second Law, where force equals mass times acceleration (\(F=m \times a\)). Here, \(m\) is the effective mass of the arm and \(a\) is the acceleration calculated in the previous step.
04

Calculate Stress

Stress can be calculated using the equation \(Stress = \frac{Force}{Area}\), where area is the total cross sectional area of the two forearm bones. Substitute the force calculated in the previous step into this equation to get the stress.
05

Compare the Calculated Stress with the Maximum Compressional Stress

Compare the stress calculated in the previous step with the given maximum compressional stress. If the calculated stress is less than the maximum compressional stress then the arm is likely to withstand the crash, otherwise, it will not.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acceleration
When an object changes its speed, it experiences acceleration. In the context of a car crash, when the arm comes to a stop, it undergoes negative acceleration, also known as deceleration. To calculate this, we first need the initial speed in meters per second (m/s). This conversion is important because the given speed is in kilometers per hour (km/h). Divide the speed by 3.6 to get it in m/s. For example:
  • Initial speed, \(v = 80 \text{ km/h} = \frac{80}{3.6} \text{ m/s}\).
To find acceleration, use the formula:
  • \(a = \frac{\text{Final Velocity} - \text{Initial Velocity}}{\text{Time}}\)
  • Given that the arm comes to rest, final velocity = 0.
  • Time \(t = 5.0 \text{ ms} = 0.005 \text{ seconds}\).
  • So, \(a = \frac{0 - v}{t}\).
The negative sign indicates deceleration, highlighting how the arm quickly slows to a stop.
Force Calculation
Understanding force requires using Newton's Second Law, which states that force equals mass times acceleration. The effective mass of the forearm in our problem is given as 3.0 kg.
To calculate force, first find the acceleration from the previous step. Then, apply it to Newton's formula:
  • \(F = m \times a\)
  • Where \(m = 3.0 \text{ kg}\)
  • \(a\) is the acceleration value calculated earlier.
This force is the reason our arm comes to a stop rapidly during the collision, and it's crucial for understanding the impact on the forearm bones. By calculating this force, we determine how much pressure will be applied to the bones during the crash.
Cross-Sectional Area
Cross-sectional area helps understand how force is distributed across an object, specifically essential when evaluating stress on materials. In our exercise, the bone's cross-sectional area is given as 2.4 cm虏. To find stress:
  • Use the equation \(\text{Stress} = \frac{\text{Force}}{\text{Area}}\).
  • Area here is crucial as it influences stress concentration. A smaller area results in higher stress.
The maximum compressional stress that the bone can handle is \(16 \times 10^{7} \text{ Pa}\). By calculating the stress from the force obtained previously, we can compare it to this maximum value to examine if the arm is likely to withstand the crash. If the calculated stress is lower than this value, the bones are less likely to break.

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Most popular questions from this chapter

What diameter needle should be used to inject a volume of \(500 \mathrm{~cm}^{3}\) of a solution into a patient in \(30 \mathrm{~min}\) ? Assume the length of the needie is \(2.5 \mathrm{~cm}\) and the solution is elevated \(1.0 \mathrm{~m}\) above the point of injection. Further, assume the viscosity and density of the solution are those of pure water, and that the pressure inside the vein is atmospheric.

The distortion of Earth's crustal plates is an example of shear on a large scale. A particular crustal rock has a shear modulus of \(1.5 \times 10^{10} \mathrm{~Pa}\). What shear stress is involved when a \(10-\mathrm{km}\) layer of this rock is sheared through a distance of \(5.0 \mathrm{~m}\) ?

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The average human has a density of \(945 \mathrm{~kg} / \mathrm{m}^{3}\) after inhaling and \(1020 \mathrm{~kg} / \mathrm{m}^{3}\) after exhaling. (a) Without making any swimming movements, what percentage of the human body would be above the surface in the Dead Sea (a lake with a water density of about \(1230 \mathrm{~kg} / \mathrm{m}^{3}\) ) in each of these cases? (b) Given that bone and muscle are denser than fat, what physical characteristics differentiate "sinkers" (those who tend to sink in water) from "floaters" (those who readily float)?

A hypodermic syringe contains a medicine with the density of water (Fig, \(\mathrm{P} 9.47)\). The barrel of the syringe has a cross-sectional area of \(2.50 \times 10^{-5} \mathrm{~m}^{2}\). In the absence of a force on the plunger, the pressure everywhere is \(1.00 \mathrm{~atm} . \mathrm{A}\) force \(\mathbf{F}\) of magnitude \(2.00 \mathrm{~N}\) is exerted on the plunger, making medicine squirt from the needle. Determine the medicine's flow speed through the necdle. Assume the pressure in the needle remains equal to \(1.00\) atm and that the syringe is horizontal.
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