To calculate the magnitude of the normal force acting on the screwdriver, we need to consider the forces involved when an object is on an inclined plane.

Key Concepts:

1. The normal force acts perpendicular to the surface of the ramp.
2. The weight of the screwdriver is $mg$, where $m$ is the mass (0.171 kg) and $g$ is the acceleration due to gravity (approximately 9.8 m/s²).
3. The weight can be decomposed into two components:
   • One parallel to the ramp: $mg \sin(\theta)$
   • One perpendicular to the ramp: $mg \cos(\theta)$

Since the normal force balances the perpendicular component of the weight, we have: $\text{Normal force} = mg \cos(\theta)$

Given:

• Mass of the screwdriver $m = 0.171 \, \text{kg}$
• Angle of incline $\theta = 57.5^\circ$
• Acceleration due to gravity $g = 9.8 \, \text{m/s}^2$

Calculations:

$\text{Normal force} = 0.171 \times 9.8 \times \cos(57.5^\circ)$

First, calculate the cosine of 57.5°: $\cos(57.5^\circ) \approx 0.544$

Now, multiply the values: $\text{Normal force} = 0.171 \times 9.8 \times 0.544 \approx 0.912 \, \text{N}$

Final Answer:

The magnitude of the normal force acting on the screwdriver is approximately 0.912 N.

---

Would you like further details or have any other questions?

Here are some related questions you might find useful:

1. How does the angle of the incline affect the normal force?
2. How can you calculate the frictional force on the screwdriver?
3. What is the parallel component of the weight on the ramp?
4. How would the normal force change if the mass of the screwdriver were increased?
5. What role does the coefficient of friction play in the motion of the screwdriver?

Tip: When calculating forces on an incline, always decompose the weight into components parallel and perpendicular to the surface to find the normal and frictional forces effectively.