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The 35kg crate illustrated in Figure 3 is initially at rest on a horizontal surface which the coefficient of kinetic friction is µ = 0.25. The crate is subjected to an external force of 250N. Determine the vertical and horizontal forces acting on the crate.


To solve the problem of determining the vertical and horizontal forces acting on the 35 kg crate, we can break down the forces into their components based on the angles given.

### Given Data:
- Mass of the crate, \( m = 35 \, \text{kg} \)
- External force applied, \( F = 250 \, \text{N} \)
- Angle of force with the horizontal, \( \theta = 25^\circ \)
- Coefficient of kinetic friction, \( \mu = 0.25 \)
- Gravitational acceleration, \( g = 9.81 \, \text{m/s}^2 \)

### Steps to Solve:

1. **Calculate the weight of the crate:**
   \[
   W = mg = 35 \times 9.81 \, \text{N} = 343.35 \, \text{N}
   \]
   The weight acts vertically downward.

2. **Resolve the applied force into horizontal and vertical components:**
   - Horizontal component of the force:
     \[
     F_x = F \cos(25^\circ) = 250 \cos(25^\circ)
     \]
   - Vertical component of the force:
     \[
     F_y = F \sin(25^\circ) = 250 \sin(25^\circ)
     \]

3. **Calculate the normal force:**
   The normal force \( N \) is the force exerted by the surface to balance the vertical forces. It is given by:
   \[
   N = W - F_y = 343.35 \, \text{N} - 250 \sin(25^\circ)
   \]

4. **Calculate the frictional force:**
   The frictional force \( f_k \) is given by:
   \[
   f_k = \mu N = 0.25 \times N
   \]

5. **Check if the crate moves:**
   Compare the horizontal component of the applied force \( F_x \) with the frictional force \( f_k \). If \( F_x \) is greater than \( f_k \), the crate will move, and the net horizontal force will be \( F_x - f_k \).

Now, let's perform the calculations:

### Calculations:

1. **Weight of the crate:**
   \[
   W = 343.35 \, \text{N}
   \]

2. **Horizontal and Vertical components of the force:**
   \[
   F_x = 250 \cos(25^\circ) \approx 226.39 \, \text{N}
   \]
   \[
   F_y = 250 \sin(25^\circ) \approx 105.69 \, \text{N}
   \]

3. **Normal force:**
   \[
   N = 343.35 - 105.69 \approx 237.66 \, \text{N}
   \]

4. **Frictional force:**
   \[
   f_k = 0.25 \times 237.66 \approx 59.42 \, \text{N}
   \]

5. **Net horizontal force:**
   Since \( F_x > f_k \), the crate will move.
   \[
   F_{\text{net}} = F_x - f_k \approx 226.39 - 59.42 \approx 166.97 \, \text{N}
   \]

### Summary:
- **Horizontal force acting on the crate:** \( F_{\text{net}} = 166.97 \, \text{N} \)
- **Vertical force acting on the crate:** \( N = 237.66 \, \text{N} \)

These forces determine the motion of the crate on the surface.

Would you like to delve into more details or have any questions? Below are some related questions you might find interesting:

1. How does increasing the angle of the applied force affect the normal force?
2. What would happen if the coefficient of friction was higher?
3. How would the forces change if the crate was on an inclined plane?
4. What is the significance of the vertical component of the applied force?
5. How would the problem change if the crate's mass was increased?
6. What would be the net force if the applied force was less than 250N?
7. How can the frictional force be reduced in this scenario?
8. What is the role of kinetic friction in determining the motion of the crate?

**Tip:** Understanding how forces decompose into components is crucial in solving problems involving inclined planes or surfaces with friction.

Understand how to calculate vertical and horizontal forces acting on a 35kg crate subjected to an external force and friction. Detailed solution with calculations involving gravitational force, components of applied force, normal force, and frictional force.

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