Thank you for providing the angle. Let's go through the calculation with the given angle of $58^\circ$ to determine the normal force acting on the block.

Step-by-Step Solution:

1. Given:

   • Angle of the incline: $\theta = 58^\circ$
   • Mass of the block: $m$ (since it's not provided, we'll leave the mass as $m$)
   • Gravitational acceleration: $g = 9.8 \, \text{m/s}^2$

2. Gravitational Force ($mg$):

   • The weight of the block is given by $F_g = mg$.

3. Components of the Gravitational Force:

   • Perpendicular to the incline (Normal force direction): $F_{\perp} = mg \cos(\theta)$
   • Parallel to the incline: $F_{\parallel} = mg \sin(\theta)$

4. Normal Force $F_N$:

   • The normal force $F_N$ is equal to the perpendicular component of the gravitational force. $F_N = mg \cos(58^\circ)$

5. Calculate the Cosine:

   • Find $\cos(58^\circ)$: $\cos(58^\circ) \approx 0.5299$
   • Therefore, the normal force is: $F_N \approx mg \times 0.5299$
   • This simplifies to: $F_N \approx 0.5299 \, mg$

Conclusion:

The normal force acting on the block is approximately $0.5299 \times mg$. If you know the mass of the block, you can multiply it by $0.5299 \times 9.8 \, \text{m/s}^2$ to find the numerical value of the normal force.

Do you need any further details, or have any questions?

Related Questions:

1. How does the normal force change if the angle of the incline increases to $70^\circ$?
2. What would be the effect on the normal force if the mass of the block doubles?
3. How would you find the frictional force if a coefficient of friction is given?
4. What is the net force acting on the block along the incline if friction is negligible?
5. How would the normal force change if the block were moving up the incline with a constant velocity?

Tip: Always remember that the normal force on an inclined plane is dependent on the cosine of the angle of the incline. As the angle increases, the normal force decreases.