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The problem is asking for the force **FR** and its direction **αR** that acts on the axis of the fixed pulley due to a hanging mass **m = 18 kg**. The force is applied through a rope which passes over the pulley at an angle of **60°** relative to the horizontal.

### Step-by-step solution:

1. **Force due to the hanging mass**:
   The weight of the hanging mass \( m = 18 \, \text{kg} \) creates a downward force:
   \[
   F_g = m \cdot g
   \]
   where \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).

   \[
   F_g = 18 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 176.58 \, \text{N}
   \]
   This is the tension force in the vertical part of the rope.

2. **Tension in the rope**:
   The tension in the rope is the same throughout (neglecting friction), so the tension is **176.58 N** in both the vertical and the angled section.

3. **Resolve forces at the pulley**:
   The pulley experiences two forces: one vertical and one angled at **60°**. These forces can be represented as vectors, and the resultant force will act on the axis of the pulley.

   The force components are:
   - Vertical component \( F_{\text{vertical}} = 176.58 \, \text{N} \)
   - Angled component \( F_{\text{angled}} = 176.58 \, \text{N} \) at **60°** to the horizontal.

4. **Resolve the angled force into horizontal and vertical components**:
   The angled force can be split into horizontal and vertical components:
   \[
   F_{\text{horizontal}} = F_{\text{angled}} \cdot \cos(60°) = 176.58 \cdot \cos(60°) = 88.29 \, \text{N}
   \]
   \[
   F_{\text{vertical,angled}} = F_{\text{angled}} \cdot \sin(60°) = 176.58 \cdot \sin(60°) = 152.89 \, \text{N}
   \]

5. **Total force components on the pulley**:
   Now sum the vertical components:
   \[
   F_{\text{total,vertical}} = F_{\text{vertical}} + F_{\text{vertical,angled}} = 176.58 + 152.89 = 329.47 \, \text{N}
   \]
   The horizontal component is just:
   \[
   F_{\text{total,horizontal}} = F_{\text{horizontal}} = 88.29 \, \text{N}
   \]

6. **Resultant force on the pulley**:
   The resultant force **FR** is found using the Pythagorean theorem:
   \[
   F_R = \sqrt{F_{\text{total,vertical}}^2 + F_{\text{total,horizontal}}^2}
   \]
   Substituting the values:
   \[
   F_R = \sqrt{329.47^2 + 88.29^2} = \sqrt{108554.14 + 7795.06} = \sqrt{116349.2} = 341.04 \, \text{N}
   \]

7. **Direction of the resultant force**:
   The direction **αR** is given by:
   \[
   \alpha_R = \tan^{-1}\left(\frac{F_{\text{total,vertical}}}{F_{\text{total,horizontal}}}\right)
   \]
   Substituting the values:
   \[
   \alpha_R = \tan^{-1}\left(\frac{329.47}{88.29}\right) = \tan^{-1}(3.73) = 74.92^\circ
   \]
   So the direction is approximately **74.92°** from the horizontal.

### Final Answer:
- **Resultant force**: \( F_R \approx 341.04 \, \text{N} \)
- **Direction**: \( \alpha_R \approx 74.92^\circ \)

Would you like further details or have any questions?

### Related Questions:
1. How do you calculate the force exerted by a pulley with friction?
2. What changes if the angle of the inclined rope is different?
3. How is the tension in a rope affected if the pulley has mass?
4. What happens if the rope is not massless in such a system?
5. How do you calculate the mechanical advantage of a pulley system?

### Tip:
Always resolve forces into their horizontal and vertical components when dealing with inclined planes or pulleys to simplify the problem.

Mit welcher Kraft FR und in welcher Richtung αR wird die Achse der festen Seilrolle belastet?

Calculate the resultant force and its direction acting on the axis of a fixed pulley due to a hanging mass of 18 kg. Learn step-by-step how to resolve forces, apply trigonometric functions, and use the Pythagorean theorem to determine the force magnitude and direction.

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