Problem 60 - Physics

Question

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?

Step-by-Step Solution

Answer

The force required is 65,940 N.

Step 1: Identify Given Values

Extract the values given in the problem: radius of the hole \( r = 1.00 \times 10^{-2} \text{ m} \), thickness of the sheet \( t = 3.0 \times 10^{-3} \text{ m} \), and the shearing stress \( \tau = 3.5 \times 10^{8} \text{ Pa} \).

Step 2: Understand the Shearing Stress Formula

Shearing stress (\(\tau\)) is given by the formula:\[ \tau = \frac{F}{A} \]where \( F \) is the force applied, and \( A \) is the area over which the force is applied.

Step 3: Calculate the Area of the Circular Section

The area \(A\) of the circular section is the lateral area of the cylindrical hole punched out. It can be determined using:\[ A = 2 \pi r t \]Substitute the values:\[ A = 2 \pi (1.00 \times 10^{-2} \text{ m}) (3.0 \times 10^{-3} \text{ m}) \]

Step 4: Compute the Area Value

Calculate the area using the formula from the previous step:\[ A = 2 \pi \times 1.00 \times 10^{-2} \times 3.0 \times 10^{-3} = 1.884 \times 10^{-4} \text{ m}^2 \]

Step 5: Solve for the Force

Rearrange the stress formula to solve for force \( F \):\[ F = \tau \times A \]Substitute \( \tau = 3.5 \times 10^{8} \text{ Pa} \) and \( A = 1.884 \times 10^{-4} \text{ m}^2 \):\[ F = 3.5 \times 10^{8} \times 1.884 \times 10^{-4} = 65,940 \text{ N} \]

Step 6: Present the Force Required

The force required \( \overrightarrow{\mathbf{F}} \) to punch a hole through the metal sheet is \( 65,940 \text{ N} \).

Key Concepts

Force Calculation

Calculating the force required to punch a hole in a metal sheet involves understanding how different variables in a problem interplay. The force (\( F \)) necessary to perform this action is governed by the relationship between shearing stress and the area over which this stress acts. This makes the force not just a simple straight calculation.To determine the correct force:

• Identify the shearing stress (\( \tau \)) required for the material. Think of this as the force per unit area needed to break the intermolecular bonds in the material.
• Calculate the area (\( A \)) of the surface where the stress acts. This area is typically the surface area of the lateral section of the hole punched.

Combine these with the stress formula to find the applied force, ensuring your calculations maintain consistent units throughout. Understanding these basics helps define the force needed for any engineering task involving shearing stress.

Circular Section Area

When assessing the force required to punch through a metal sheet, calculate the area of the circular section involved. This is pivotal as the area directly influences the amount of force needed. In this exercise, the circular section refers to the lateral surface area of the cylindrical hole resulting from the punch.To find this area:

• Use the formula \( A = 2 \pi r t \), where \( r \) is the radius of the hole and \( t \) is the thickness of the metal sheet. This gives the lateral surface, not just the cross-sectional area of the circle.
• Application: For a radius of \( 1.00 \times 10^{-2} \text{ m} \) and a thickness of \( 3.0 \times 10^{-3} \text{ m} \), substitute into the formula and solve.

This concept highlights how geometry directly impacts engineering calculations and ensures the precision of the applied method.

Shearing Stress Formula

The shearing stress formula is crucial in determining the force needed to punch through a metal sheet. Shearing stress (\( \tau \)) signifies the internal forces exploit the material's sliding or cutting along a plane.As per the formula:\[ \tau = \frac{F}{A} \]

• \( F \) is the applied force that causes the punching action.
• \( A \) is the precise area over which the force is distributed, important in ensuring uniform stress application.

To apply this formula:

• Rearrange it to solve for force, resulting in \( F = \tau \cdot A \)
• Substitute the known values of stress and area to get accurate results.

This understanding is essential in materials engineering and applications, including anything from cutting metal sheets to crafting precise components.

Metal Sheet Thickness

The thickness of a metal sheet is an essential factor in calculating the force needed for punching. Think of it as the measure of resistance offered by the material to shearing.When factoring thickness:

• It defines part of the area calculation, as the thickness contributes to the lateral area of the circular hole to be punched.
• The greater the thickness, the more material needs to be sheared, increasing the stress and hence the force required.

In this case:

• Given a thickness of \( 3.0 \times 10^{-3} \text{ m} \), it influences both the biological and mathematical parameters of the project.
• Ensure the thickness is consistently used with other dimensions for precise results.

Understanding thickness helps in accurately predicting the force required and designing the tools and processes necessary for shearing operations.