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A window washer pushes his scrub brush up a vertical window at constant speed by applying a force F⃗ as shown in (Figure 1). The brush weighs 10.5 N and the coefficient of kinetic friction is μk = 0.120. Calculate the magnitude of the force F⃗ . The angle is 53.1

To solve the problem, we need to analyze the forces acting on the scrub brush. The forces include:

1. The force \( \vec{F} \) applied at an angle of \( 53.1^\circ \).
2. The weight of the brush \( W = 10.5 \, \text{N} \) acting vertically downward.
3. The normal force \( N \) exerted by the window.
4. The kinetic friction force \( f_k = \mu_k N \).

### Steps to Solve:

1. **Resolve the force \( \vec{F} \) into components:**
   - The horizontal component is \( F_x = F \cos(53.1^\circ) \).
   - The vertical component is \( F_y = F \sin(53.1^\circ) \).

2. **Set up the equilibrium conditions:**
   - Since the brush is moving at constant speed, the sum of the forces in both the x-direction and y-direction must be zero.

   **In the x-direction:**
   \[
   N = F \cos(53.1^\circ)
   \]

   **In the y-direction:**
   \[
   F \sin(53.1^\circ) = W + f_k
   \]

3. **Substitute the kinetic friction force:**
   - The friction force is \( f_k = \mu_k N = 0.120 \times N \).
   - Using the expression for \( N \), we get:
     \[
     f_k = 0.120 \times F \cos(53.1^\circ)
     \]

4. **Substitute \( f_k \) into the y-direction equation:**
   \[
   F \sin(53.1^\circ) = 10.5 + 0.120 \times F \cos(53.1^\circ)
   \]

5. **Solve for \( F \):**

Let's calculate this step-by-step.The magnitude of the force \( \vec{F} \) that the window washer needs to apply is approximately \( 14.43 \, \text{N} \).

### Would you like further details on this calculation or have any other questions?

### Related Questions:
1. How would the required force change if the coefficient of kinetic friction were increased?
2. What would happen if the angle were different, say 45° instead of 53.1°?
3. How would the force change if the weight of the brush were increased to 15 N?
4. What if the force \( \vec{F} \) were applied horizontally instead of at an angle?
5. Can we calculate the normal force exerted by the window?

### Tip:
When dealing with forces, always break down vectors into their components and apply equilibrium conditions for both horizontal and vertical directions to simplify the calculations.

A window washer pushes his scrub brush up a vertical window at constant speed by applying a force F⃗ as shown. The brush weighs 10.5 N and the coefficient of kinetic friction is μk = 0.120. Calculate the magnitude of the force F⃗. The angle is 53.1°.

Learn how to calculate the force required to push a window washer's scrub brush up a vertical window at constant speed. The problem involves a 53.1° angle, a brush weight of 10.5 N, and a coefficient of kinetic friction μk = 0.120. Detailed step-by-step trigonometric and kinetic analysis included. Suitable for students in Grades 10-12.

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