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The problem in the image involves an 8-kg crate being pulled across a horizontal surface with a 50-N applied force, while there is a 10-N friction force opposing the motion. We are asked to find the normal force (\(F_{\text{norm}}\)) acting on the crate.

### Step-by-step Solution:

1. **Given Data:**
   - Mass of the crate (\(m\)) = 8 kg
   - Gravitational force (\(F_{\text{grav}}\)) = 80 N (this is the weight of the crate, calculated using \(F_{\text{grav}} = m \times g\), where \(g = 9.8 \, \text{m/s}^2\), which gives \(8 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 78.4 \, \text{N}\), approximately 80 N).
   - Applied force (\(F_{\text{app}}\)) = 50 N
   - Friction force (\(F_{\text{fric}}\)) = 10 N
   - The motion is horizontal.

2. **Forces in the Vertical Direction (Y-axis):**
   The crate is not accelerating vertically, so the sum of forces in the vertical direction must be zero:
   \[
   F_{\text{norm}} - F_{\text{grav}} = 0
   \]
   Since \(F_{\text{grav}} = 80 \, \text{N}\), this implies:
   \[
   F_{\text{norm}} = F_{\text{grav}} = 80 \, \text{N}
   \]

Thus, the normal force on the crate is **80 N**.

Would you like more details or further explanation?

### Follow-up questions:
1. How would the normal force change if the surface was inclined?
2. What is the net horizontal force acting on the crate?
3. How would you calculate the acceleration of the crate?
4. What would happen to the friction force if the surface was rougher?
5. How would the situation change if the applied force was at an angle?

**Tip:** Always break down forces into their vertical and horizontal components to analyze motion more clearly.

An 8-kg crate is pulled across a horizontal surface with a 50-N applied force against a 10-N friction force. What is the normal force acting on the crate?

Learn how to calculate the normal force acting on an 8-kg crate pulled across a horizontal surface with a 50-N applied force and a 10-N friction force. Using Newton's laws and the gravitational formula, we find that the normal force is equal to 80 N.

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