The tensile stress ($\sigma$) on a bar subjected to a tensile load is calculated using the formula:

$\sigma = \frac{F}{A}$

Where:

• $F$ is the tensile force applied to the bar.
• $A$ is the cross-sectional area of the bar.

Given:

• Tensile force, $F = 200 \, \text{N}$
• Cross-sectional dimensions of the bar: $20 \, \text{mm} \times 5 \, \text{mm}$

First, calculate the cross-sectional area ($A$):

$A = 20 \, \text{mm} \times 5 \, \text{mm} = 100 \, \text{mm}^2 = 100 \times 10^{-6} \, \text{m}^2 = 0.0001 \, \text{m}^2$

Now, calculate the tensile stress:

$\sigma = \frac{200 \, \text{N}}{0.0001 \, \text{m}^2} = 2,000,000 \, \text{N/m}^2 = 2 \, \text{MPa}$

Tensile stress:

$\sigma = 2 \, \text{MPa}$

Would you like further details or clarifications on this problem?

Here are 5 related questions:

1. What is the unit of tensile stress?
2. How would the stress change if the cross-sectional area doubled?
3. What happens to the tensile stress if the force is reduced by half?
4. How can tensile stress be reduced in a real-world scenario?
5. What are typical applications where tensile stress is crucial to consider?

Tip: Always ensure that force and area are in consistent units when calculating stress!