To find the normal stress $\sigma_n$ acting on a plane inclined at $35^\circ$ (clockwise) to the $x$-axis under the given state of plane stress, we can use the following formula from plane stress transformation:

$\sigma_n = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2} \cos(2\theta) + \tau_{xy} \sin(2\theta)$

Given:

• $\sigma_x = 0$ MPa
• $\sigma_y = 0$ MPa
• $\tau_{xy} = -200$ MPa
• $\theta = 35^\circ$

Now, let's plug in the values:

$\sigma_n = \frac{0 + 0}{2} + \frac{0 - 0}{2} \cos(2 \times 35^\circ) + (-200) \sin(2 \times 35^\circ)$

Simplifying:

$\sigma_n = 0 + 0 + (-200) \sin(70^\circ)$

We can now compute the value of $\sin(70^\circ)$:

$\sin(70^\circ) \approx 0.9397$

Thus:

$\sigma_n = -200 \times 0.9397 \approx -187.94 \text{ MPa}$

So, the normal stress $\sigma_n$ on the plane inclined at $35^\circ$ to the $x$-axis is approximately $-187.94$ MPa.

Would you like a more detailed explanation, or do you have any questions about this solution?

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Here are 5 related questions to deepen your understanding:

1. What is the formula for shear stress on an inclined plane under plane stress conditions?
2. How does the angle of inclination affect the magnitude of normal stress?
3. What would be the normal stress if the plane was inclined at $45^\circ$ instead of $35^\circ$?
4. Can you explain the significance of the principal stresses in this context?
5. What role does the sign of $\tau_{xy}$ play in the stress transformation equations?

Tip: When working with trigonometric functions in stress transformation, always ensure your calculator is set to the correct mode (degrees or radians) based on the angle you are using.