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Chapter 11: Problem 13

Nail tips exert tremendous pressures when they are hit by hammers because they exert a large force over a small area. What force must be exerted on a nail with a circular tip of 1.00 mm diameter to create a pressure of \(3.00 \times 10^{9} \mathrm{N} / \mathrm{m}^{2}\) ? (This high pressure is possible because the hammer striking the nail is brought to rest in such a short distance.)

Short Answer

Expert verified
The force exerted on the nail is \( 3.00 \times 10^{9} \) \( \mathrm{N/m}^{2} \) \( \times \) \( \pi \times (0.0005)^{2} \) \( \mathrm{m}^{2} \) which equals \( 2,356.19 \) Newtons.

Step by step solution

01

Calculate the area of the nail tip

First, calculate the area of the nail tip, which is a circle, using the formula \( A = \pi r^2 \) where \( A \) is the area, \( \pi \) is a constant (approximately 3.14159), and \( r \) is the radius of the circle. The diameter is given as 1.00 mm, so the radius is half of that, which is 0.50 mm or \( 0.50 \times 10^{-3} \) meters.
02

Convert radius to meters

Convert the radius from millimeters to meters to match the units of pressure. \( 0.50 \times 10^{-3} \) meters is \( 0.0005 \) meters.
03

Plug values into the area formula

Now plug in the value of \( r \) into the area formula: \( A = \pi \times (0.0005)^{2} \) and calculate the area in square meters.
04

Calculate the force needed

Use the pressure formula \( P = \frac{F}{A} \) where \( P \) is the pressure, \( F \) is the force, and \( A \) is the area. Rearrange the formula to solve for \( F \) as \( F = P \times A \) and plug in the values for \( P \) and \( A \) to calculate the force in Newtons.

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Key Concepts

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

Pressure and Force Relationship
Understanding the pressure and force relationship is crucial in physics, especially when analyzing situations like a nail being hit by a hammer. Pressure is defined as the amount of force applied per unit area and is expressed in Pascals (Pa), which are Newtons per square meter (\text{N/m}^2).

The formula to describe this relationship is: \[ P = \frac{F}{A} \] where \( P \) represents pressure, \( F \) is the force applied, and \( A \) is the area over which the force is distributed. When a large force is exerted on a small area, such as the tip of a nail, the resulting pressure is high. This can be visualized by thinking about lying on a bed of nails. The force of your body weight is distributed over a large number of nail tips (large area), so the pressure on any one nail is relatively low, and you don't get hurt. However, if you were to step onto a single nail (small area), the pressure would be extremely high and could potentially pierce your foot.
Area Calculation for Circles
The area of a circle is a foundational concept in geometry that is essential for solving a variety of physics problems, such as calculating pressures. To calculate the area of a circle, the following formula is used: \[ A = \pi r^2 \] where \( A \) stands for the area, \( r \) for the radius, and \( \pi \) is the mathematical constant approximately equal to 3.14159. The radius of a circle is the distance from the center of the circle to any point on its edge. It's important to note that the diameter of a circle is twice the radius, so if you know the diameter, you can find the radius by simply dividing it by two.

For example, if the diameter of the nail tip is 1.00 mm, the radius would be 0.50 mm. In most physics problems, however, it's necessary to convert this measurement into meters to stay consistent with the SI unit system. Hence, after converting millimeters to meters, the formula returns the area in square meters, which can then be used in further calculations.
Unit Conversion
Unit conversion is a vital skill in physics to ensure that all variables are in compatible units before performing calculations. It's commonly needed because measurements can be taken in various units that are not conducive to the formulas used in physics.

The basic principle of unit conversion is to use conversion factors that describe the relationship between different units. For example, there are 1000 millimeters in a meter. So, to convert the radius from millimeters to meters, you multiply by the conversion factor of \(10^{-3}\), which effectively moves the decimal point three places to the left. Keeping units consistent is essential to obtain correct results. An incorrect unit conversion can have a dramatic effect on the answer, potentially leading to errors in understanding or application of the physics concept at hand. This is particularly important in pressure calculations, where the area must be in square meters (\text{m}^2) to match the units for pressure (\text{N/m}^2).

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Most popular questions from this chapter

A proton moves at \(7.50 \times 10^{7} \mathrm{~m} / \mathrm{s}\) perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius \(0.800 \mathrm{~m}\). What is the field strength?

A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it in a graduated cylinder. What is the density of a 240-g rock that displaces \(89.0 \mathrm{cm}^{3}\) of water? (Note that the accuracy and practical applications of this technique are more limited than a variety of others that are based on Archimedes' principle.)

Explain why the magnetic field would not be unique (that is, not have a single value) at a point in space where magnetic field lines might cross. (Consider the direction of the field at such a point.)

Heroes in movies hide beneath water and breathe through a hollow reed (villains never catch on to this trick). In practice, you cannot inhale in this manner if your lungs are more than \(60.0 \mathrm{cm}\) below the surface. What is the maximum negative gauge pressure you can create in your lungs on dry land, assuming you can achieve \(-3.00 \mathrm{cm}\) water pressure with your lungs \(60.0 \mathrm{cm}\) below the surface?

(a) A 75.0-kg man floats in freshwater with 3.00\% of his volume above water when his lungs are empty, and \(5.00 \%\) of his volume above water when his lungs are full. Calculate the volume of air he inhales- -called his lung capacity-in liters. (b) Does this lung volume seem reasonable?
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