/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 A die is designed to punch holes... [FREE SOLUTION] | 91Ó°ÊÓ

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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
Force required is approximately \(1.1 \times 10^{5} \mathrm{~N}\).

Step by step solution

01

Understand the Problem

The problem requires us to calculate the force needed to punch holes in a metal sheet. This process involves shearing, which is cutting through the material. We are provided with the shear stress required and need to find the force exerted by the die.
02

Know the Shear Stress Formula

The formula for shearing stress is given by \( \tau = \frac{F}{A} \), where \( \tau \) is the shear stress, \( F \) is the force, and \( A \) is the area. We need to rearrange this formula to solve for the force \( F \).
03

Calculate the Area

The area \( A \) is the area of the circle being sheared, which is calculated using the formula \( A = \pi r^2 \), where \( r \) is the radius of the hole. Here, \( r = 1.00 \times 10^{-2} \mathrm{~m} \). Calculate the area: \[ A = \pi \times (1.00 \times 10^{-2})^2 \approx 3.14 \times 10^{-4} \mathrm{~m^2}. \]
04

Recalculate Force Using Shearing Stress

Rearrange the shear stress formula \( \tau = \frac{F}{A} \) to find the force \( F \) as \( F = \tau \times A \). Substitute the values: \[ F = 3.5 \times 10^{8} \mathrm{~Pa} \times 3.14 \times 10^{-4} \mathrm{~m^2} \approx 1.1 \times 10^{5} \mathrm{~N}. \]
05

Review and Conclude

Check all calculations to make sure each step is followed correctly and the correct formula is used. The applied force needed by the die is approximately \(1.1 \times 10^{5} \mathrm{~N}\), ensuring the die can exert sufficient force to punch through the sheet.

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Key Concepts

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

Shearing Force
When we talk about shearing force, we refer to a force that causes layers or particles of a material to slide past each other in parallel directions. In simpler terms, it's like using a pair of scissors to cut through paper. The blades apply a force that slides one part of the paper against the other until it splits. Shearing force is highly relevant in engineering and manufacturing, particularly when working with materials that need to be cut or shaped, like metal sheets.

In the context of our exercise, the shearing force is represented by the force exerted by the die on the metal sheet. The die cuts through the sheet by sliding its atoms against each other, overcoming the internal resistive forces of the material, until the sheet is punctured. The amount of shearing force required depends on the material’s properties, such as its shear strength, and the dimensions of the cut, particularly the thickness of the sheet being cut.
Shearing Process
The shearing process is crucial in many industrial applications. Shearing involves using a tool to apply a force on a material, usually metal, in such a way that it results in an offset in its structure or complete separation. Think of cutting a piece of fabric with a sharp blade or shearing wool from a sheep.

When you punch a hole in a metal sheet, the material directly under the punch experiences shear stress, which is essentially stress that tends to cause one part of a material to slide past another. This makes it a highly efficient method of cutting materials quickly and cleanly. The thickness of the material and the required precision also play an essential role in determining the necessary force and the kind of equipment needed for the process.
  • Clean cuts and quick processing are major benefits.
  • The process is used in manufacturing key components for various machines and structures.
  • Accuracy depends on the precision of the equipment used and the properties of the material.
Punching Force Calculation
Calculating the punching force needed to puncture a material is fundamental in designing efficient tools and dies. This calculation allows engineers to determine the size and power of machinery required for specific tasks. The calculation hinges on knowing the shear stress the material can withstand, i.e., how much stress it can endure before it is sheared. To calculate this, we first need to determine the area through which the stress exerts force. In this exercise, that’s the area of the circular section being punched out, calculated using the formula for the area of a circle, which is \[ A = \pi r^2 \], where \( r \) is the radius of the hole.

Once the area is known, we can rearrange the shear stress equation, \( \tau = \frac{F}{A} \), to solve for the force \( F \). Thus, we compute the force needed using \[ F = \tau \times A \], where \( \tau \) is the given shear stress. This helps ensure the die can apply sufficient force to cleanly shear the material into the desired shape or hole, optimizing both energy and material efficiency in industrial applications.

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Most popular questions from this chapter

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?

A square plate is \(1.0 \times 10^{-2} \mathrm{~m}\) thick, measures \(3.0 \times 10^{-2} \mathrm{~m}\) on a side, and has a mass of \(7.2 \times 10^{-2} \mathrm{~kg}\). The shear modulus of the material is \(2.0 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}\). One of the square faces rests on a flat horizontal surface, and the coefficient of static friction between the plate and the surface is \(0.90 .\) A force is applied to the top of the plate, as in Figure \(10-32 a\). Determine (a) the maximum possible amount of shear stress, (b) the maximum possible amount of shear strain, and (c) the maximum possible amount of shear deformation \(\Delta X\) (see Figure \(10-32 b\) ) that can be created by the applied force just before the plate begins to move.

Refer to Interactive Solution \(\underline{10.29}\) at for help in solving this problem. A heavy-duty stapling gun uses a \(0.140-\mathrm{kg}\) metal rod that rams against the staple to eject it. The rod is pushed by a stiff spring called a "ram spring" \((k=32000 \mathrm{~N} / \mathrm{m})\). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses the ram spring by \(3.0 \times 10^{-2} \mathrm{~m}\) from its unstrained length and then releases it. Assuming that the ram spring is oriented vertically and is still compressed by \(0.8 \times 10^{-2} \mathrm{~m}\) when the downwardmoving ram hits the staple, find the speed of the ram at the instant of contact.

An \(86.0-\mathrm{kg}\) climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of \(1.20 \times 10^{3} \mathrm{~N} / \mathrm{m} .\) He accidentally slips and falls freely for \(0.750 \mathrm{~m}\) before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

A car is hauling a \(92-\mathrm{kg}\) trailer, to which it is connected by a spring. The spring constant is \(2300 \mathrm{~N} / \mathrm{m}\). The car accelerates with an acceleration of \(0.30 \mathrm{~m} / \mathrm{s}^{2} .\) By how much does the spring stretch?

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