91Ó°ÊÓ

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.

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 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.

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.