To find the magnitude of the normal force acting on the construction worker, we need to break down the forces involved, particularly the worker's weight.

The weight of the worker acts vertically downward, and the normal force is perpendicular to the roof surface. The component of the worker’s weight perpendicular to the roof will be opposed by the normal force.

Here’s how to calculate the normal force step by step:

1. Weight of the worker (W): $W = 900 \, \text{N}$ (vertically downward).
2. Slope of the roof (θ): The roof is sloped at an angle of $25^\circ$.

The component of the worker’s weight perpendicular to the slope is what contributes to the normal force. This component is given by:

$W_{\perp} = W \cdot \cos(\theta)$

Where:

• $W = 900 \, \text{N}$ (weight of the worker),
• $\theta = 25^\circ$ (slope angle),
• $W_{\perp}$ is the perpendicular component of the weight (equal to the normal force).

Calculation:

$W_{\perp} = 900 \cdot \cos(25^\circ)$

Using the cosine of $25^\circ$, which is approximately $0.906$:

$W_{\perp} = 900 \cdot 0.906 \approx 815.4 \, \text{N}$

Thus, the magnitude of the normal force is approximately 815.4 N.

Would you like more details or have any questions about this process?

---

Here are 5 related questions you might want to explore:

1. What is the frictional force acting on the worker if the coefficient of friction between the worker and the roof is known?
2. How would the normal force change if the slope of the roof increased?
3. How would we calculate the parallel component of the worker's weight along the roof?
4. What would be the impact of adding additional weight (e.g., a tool) to the worker's normal force?
5. How would the worker's weight change if we were to measure it on a different planet with a different gravitational acceleration?

Tip: When working with inclined planes, always remember to resolve forces into components perpendicular and parallel to the surface to analyze the situation accurately.