https://file.aiquickdraw.com/mathbot/file/1730070399903n5q5su6x.jpg

jelaslan secara rinci dan ringkas dan lakukan perhitungan

This problem involves a uniform ladder \( AB \) with mass \( M = 30 \, \text{kg} \) and length \( L = 4 \, \text{m} \), leaning against a frictionless wall at an angle \( \alpha \) with the floor, which has friction. We need to derive the formula for the minimum coefficient of static friction \( \mu_s \) required between the ladder and the floor to keep the ladder in equilibrium.

### Solution Outline and Calculation

1. **Free-Body Diagram and Forces:**
   - **Weight of the ladder** (\( W = Mg = 30 \cdot 9.8 = 294 \, \text{N} \)), acting downward at the ladder's center of gravity, located at \( L/2 \) from point \( A \).
   - **Normal force from the wall** (\( F_w \)) acting horizontally at point \( A \), since the wall is frictionless.
   - **Normal force from the floor** (\( N \)) acting vertically upward at point \( B \).
   - **Frictional force at the floor** (\( f = \mu_s N \)) acting horizontally at point \( B \).

2. **Conditions for Equilibrium:**
   - **Horizontal equilibrium**: The horizontal components must balance.
     \[
     F_w = f = \mu_s N
     \]
   - **Vertical equilibrium**: The vertical components must balance.
     \[
     N = Mg
     \]

3. **Torque Equilibrium:**
   Taking torques around point \( B \) (where the ladder touches the floor) eliminates \( f \) and \( N \), simplifying calculations:
   
   - **Torque due to \( Mg \)**:
     \[
     \tau_{Mg} = Mg \cdot \frac{L}{2} \cos \alpha
     \]
   - **Torque due to \( F_w \)**:
     \[
     \tau_{F_w} = F_w \cdot L \sin \alpha
     \]
   
   Setting the torques equal for rotational equilibrium:
   \[
   Mg \cdot \frac{L}{2} \cos \alpha = F_w \cdot L \sin \alpha
   \]
   Simplifying, we get:
   \[
   F_w = \frac{Mg}{2} \cot \alpha
   \]

4. **Substitute \( F_w = \mu_s N \):**
   Since \( F_w = \mu_s N \) and \( N = Mg \), substitute these values:
   \[
   \mu_s Mg = \frac{Mg}{2} \cot \alpha
   \]
   Dividing both sides by \( Mg \), we get:
   \[
   \mu_s = \frac{1}{2} \cot \alpha
   \]

### Final Answer:
The minimum coefficient of static friction \( \mu_s \) required is:
\[
\mu_s = \frac{1}{2} \cot \alpha
\]

This formula gives the minimum friction needed between the ladder and the floor as a function of the angle \( \alpha \) to prevent slipping.

Would you like more details on each step or have any questions?

---

### Additional Questions:
1. How would the minimum coefficient of friction change if the ladder's length or mass changed?
2. What happens to \( \mu_s \) if the angle \( \alpha \) becomes very small or very large?
3. How would the solution change if the wall also had friction?
4. Can you calculate \( \mu_s \) for specific values of \( \alpha \), for example, \( \alpha = 45^\circ \)?
5. What are the implications of a frictionless wall on the equilibrium conditions?

### Tip:
In problems involving static equilibrium, taking torque around a point where multiple unknown forces act can simplify calculations by eliminating those forces from the torque equation.

Sebuah tangga AB serba sama bermassa M = 30 kg dan panjangnya L = 4 m bersandar pada sebuah dinding yang licin dengan sudut α terhadap tangga, sedangkan lantai tidak licin. Agar tangga dapat bersandar pada dinding dalam keadaan setimbang mantap, nyatakan rumusan untuk menyatakan nilai minimum koefisien gesek statik antara tangga dan lantai yaitu μs sebagai fungsi sudut α.

Learn how to calculate the minimum coefficient of static friction needed for a 4-meter ladder with a mass of 30 kg to stay in equilibrium against a frictionless wall. This problem explores static equilibrium, force analysis, and torque, suitable for Grade 12 physics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation