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Chapter 10: Problem 63

A die is designed to punch holes with a radius of \(1.00 \times 10^{-2} \mathrm{m}\) in a metal sheet that is \(3.0 \times 10^{-3} \mathrm{m}\) thick, as the drawing illustrates. To punch through the sheet, the die must exert a shearing stress of \(3.5 \times 10^{8} \mathrm{Pa} .\) What force \(\overrightarrow{\mathbf{F}}\) must be applied to the die?

Short Answer

Expert verified
The force required is approximately 65,940 N.

Step by step solution

01

Understanding Shearing Stress

Shearing stress (\(\sigma\)) is defined as the force (\(F\)) applied parallel to the surface area (\(A\)) over which the force is acting. It can be expressed as, \(\sigma = \frac{F}{A}\).
02

Calculate the Area

To find the area (\(A\)) over which the force must act, consider it as the lateral surface area created by the punch. This is a cylindrical shape with height equal to the thickness of the sheet and the radius of the hole. The lateral area of a cylinder is given by \(A = 2\pi rh\), where \(r = 1.00 \times 10^{-2} \mathrm{m}\) and \(h = 3.0 \times 10^{-3} \mathrm{m}\). Therefore, \[A = 2\pi (1.00 \times 10^{-2})(3.0 \times 10^{-3})\].
03

Plug in the Values

Plug the known values into the area formula. \ \[A = 2 \times 3.1416 \times (1.00 \times 10^{-2}) \times (3.0 \times 10^{-3}) = 1.884 \times 10^{-4} \ \mathrm{m^2}\].
04

Rearranging Shearing Stress Formula

Rearrange the shearing stress formula to solve for the force: \[F = \sigma \times A\], where \\(\sigma = 3.5 \times 10^8 \ \mathrm{Pa}\) and \(A = 1.884 \times 10^{-4} \ \mathrm{m^2}\).
05

Calculate the Force Required

Insert the values for \(\sigma\) and \(A\) into the equation and calculate the force:\[F = (3.5 \times 10^8) \times (1.884 \times 10^{-4}) = 65,940 \ \mathrm{N}\].

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Key Concepts

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

Force Calculation
The first step in understanding how to calculate the force required to punch through the metal sheet is to comprehend what force calculation entails in this context. In general, force is the interaction that, when unopposed, changes the motion of an object. It can be calculated using the formula:
  • Force (\( F \) ) = Stress (\( \sigma \) ) * Area (\( A \) )
In this specific problem, force refers to the amount of interaction necessary for the die to effectively punch through the metal sheet. To find the needed force (\( F \) ), one must already understand the values of the shearing stress (\( \sigma \) ) and the area (\( A \) ) it affects. Here, the stress is given, so calculating the area is our next task.
Cylindrical Area
To calculate the force correctly, understanding the concept of the cylindrical area is crucial. A die punching a hole in a sheet results in a cylindrical area being the effective area. The formula for calculating the lateral area of a cylinder is:
  • \( A = 2\pi rh \)
In our exercise, the radius (\( r \) ) is the radius of the hole, \( 1.00 \times 10^{-2} \ \mathrm{m} \) and the height (\( h \) ), which in this case is the thickness of the sheet, is \( 3.0 \times 10^{-3} \ \mathrm{m} \). By plugging these values into the formula, we find the area:
  • \( A = 2\pi (1.00 \times 10^{-2})(3.0 \times 10^{-3}) = 1.884 \times 10^{-4} \, \mathrm{m^2} \)
This area value tells us the extent over which the force is applied during the punching process.
Material Thickness
Material thickness is an essential component in calculating the force required for the die. Thickness can influence how much force is needed to shear through the material.
In this problem, the thickness of the metal sheet forms the height of the cylindrical area over which shearing occurs, which is \( 3.0 \times 10^{-3} \ \mathrm{m} \).Knowing the thickness helps determine how the stress is distributed and calculated over the area. It's crucial for assessing the amount of shearing stress (\( \sigma \) ) the die needs to apply to shear the sheet effectively. The thicker the material, generally, the greater the force required. This factor directly affects the lateral cylindrical area, which is part of the force calculation.
Stress Formula
The stress formula forms the backbone of the solution for this exercise. Shearing stress (\( \sigma \) ) is calculated as the force (\( F \) ) applied per unit area (\( A \) ). The formula is straightforward:
  • \( \sigma = \frac{F}{A} \)
This relationship helps in rearranging the formula to solve for the force if stress and area are known. Here the values are:
  • Shearing stress \( \sigma = 3.5 \times 10^8 \, \mathrm{Pa} \)
  • Area \( A = 1.884 \times 10^{-4} \, \mathrm{m^2} \)
When these values are inserted into the rearranged equation:
  • \( F = \sigma \times A \)
This computation leads to the required force: \( F = 65,940 \, \mathrm{N} \).Understanding this formula is pivotal as it connects stress, force, and area in practical applications like this problem.

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Most popular questions from this chapter

A vertical spring with a spring constant of \(450 \mathrm{N} / \mathrm{m}\) is mounted on the floor. From directly above the spring, which is unstrained, a \(0.30-\mathrm{kg}\) block is dropped from rest. It collides with and sticks to the spring, which is compressed by \(2.5 \mathrm{cm}\) in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in \(\mathrm{cm}\) ) above the compressed spring was the block dropped?

Objects of equal mass are oscillating up and down in simple harmonic motion on two diff erent vertical springs. The spring constant of spring 1 is 174 N/m. The motion of the object on spring 1 has twice the amplitude as the motion of the object on spring 2. The magnitude of the maximum velocity is the same in each case. Find the spring constant of spring 2.

When responding to sound, the human eardrum vibrates about its equilibrium position. Suppose an eardrum is vibrating with an amplitude of \(6.3 \times 10^{-7}\) m and a maximum speed of \(2.9 \times 10^{-3} \mathrm{m} / \mathrm{s}\) (a) What is the frequency (in Hz) of the eardrum鈥檚 vibration? (b) What is the maximum acceleration of the eardrum?

A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block oscillates back and forth with an angular frequency of \(7.0 \mathrm{rad} / \mathrm{s}\). The drawing indicates the position of the block when the spring is unstrained. This position is labeled " \(x=0 \mathrm{m}\)." The drawing also shows a small bottle located \(0.080 \mathrm{m}\) to the right of this position. The block is pulled to the right, stretching the spring by \(0.050 \mathrm{m},\) and is then thrown to the left. In order for the block to knock over the bottle, it must be thrown with a speed exceeding \(v_{0} .\) Ignoring the width of the block, find \(v_{0}\).

In a room that is 2.44 m high, a spring (unstrained length \(=0.30 \mathrm{m}\) ) hangs from the ceiling. A board whose length is \(1.98 \mathrm{m}\) is attached to the free end of the spring. The board hangs straight down, so that its \(1.98-\mathrm{m}\) length is perpendicular to the floor. The weight of the board \((104 \mathrm{N})\) stretches the spring so that the lower end of the board just extends to, but does not touch, the floor. What is the spring constant of the spring?
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