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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

Four acrobats of mass \(75.0 \mathrm{~kg}, 68.0 \mathrm{~kg}, 62.0 \mathrm{~kg}\), and \(55.0 \mathrm{~kg}\) form a human tower, with each acrobat standing on the shoulders of another acrobat. The \(75.0-\mathrm{kg}\) acrobat is at the bottom of the tower, (a) What is the normal force acting on the \(75-\mathrm{kg}\) acrobat? (b) If the area of each of the \(75.0-\mathrm{kg}\) acrobat's shoes is \(425 \mathrm{~cm}^{2}\), what average pressure (not including atmospheric pressure) does the column of acrobats exert on the floor? (c) Will the pressure be the same if a different acrobat is on the bottom?

A rubber ball filled with air has a diameter of \(25.0 \mathrm{~cm}\) and a mass of \(0.540 \mathrm{~kg}\). What force is required to hold the ball in equilibrium immediately below the surface of water in a swimming pool?

Be To lift a wire ring of radius \(1.75 \mathrm{~cm}\) from the surface of a container of blood plasma, a vertical force of \(1.61 \times\) \(10^{-2} \mathrm{~N}\) greater than the weight of the ring is required. Calculate the surface tension of blood plasma from this information.

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.

The four tires of an automobile are inflated to a gauge pressure of \(2.0 \times 10^{5} \mathrm{~Pa}\). Each tire has an area of \(0.024 \mathrm{~m}^{2}\) in contact with the ground. Determine the weight of the automobile.
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