3 A small steel ball B is attracted by two bar magnets. One magnet exerts a force of $$ 0,9 \mathrm{~N} $$ to the west and the other magnet exerts a force of $$ 1,2 \mathrm{~N} $$ to the south. The steel ball does not move. 3.1 Draw a free body diagram of the forces acting at $$ B $$. 3.2 Make a rough sketch of the vector diagram and use this to calculate the resultant force on the steel ball. 3.3 What is the magnitude and direction of the force which keeps the steel ball from moviry? $$ \qquad $$ 14

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Let's analyze the problem step by step.

### **3.1 Free Body Diagram of the Steel Ball $$ B $$**

A free body diagram (FBD) visually represents all the forces acting on an object. For the steel ball $$ B $$, there are three primary forces to consider:

1. **Force from the West Magnet ($$ F_{\text{West}} $$)**:  
   - **Magnitude**: $$ 0.9 \, \text{N} $$
   - **Direction**: Towards the **West**

2. **Force from the South Magnet ($$ F_{\text{South}} $$)**:  
   - **Magnitude**: $$ 1.2 \, \text{N} $$
   - **Direction**: Towards the **South**

3. **Equilibrium Force ($$ F_{\text{Eq}} $$)**:  
   - **Magnitude**: To be determined
   - **Direction**: Opposite to the resultant of $$ F_{\text{West}} $$ and $$ F_{\text{South}} $$, which will be towards the **Northeast**

**Diagram Description:**

- Draw a dot to represent the steel ball $$ B $$.
- From $$ B $$:
  - Draw an arrow pointing West labeled $$ 0.9 \, \text{N} $$.
  - Draw an arrow pointing South labeled $$ 1.2 \, \text{N} $$.
  - Draw an arrow pointing Northeast labeled $$ F_{\text{Eq}} $$ (the equilibrium force).

*Since a visual diagram can't be provided here, ensure that in your sketch, the two given forces are perpendicular to each other (West and South), and the equilibrium force balances them diagonally.*

### **3.2 Vector Diagram and Resultant Force Calculation**

To find the resultant of the two forces acting on the steel ball, we can use the **Pythagorean theorem** since the forces are perpendicular.

1. **Given Forces:**
   - $$ F_{\text{West}} = 0.9 \, \text{N} $$ (West)
   - $$ F_{\text{South}} = 1.2 \, \text{N} $$ (South)

2. **Resultant Force ($$ R $$):**
   $$
   R = \sqrt{F_{\text{West}}^2 + F_{\text{South}}^2} = \sqrt{(0.9)^2 + (1.2)^2} = \sqrt{0.81 + 1.44} = \sqrt{2.25} = 1.5 \, \text{N}
   $$
   
3. **Direction of the Resultant ($$ \theta $$):**
   $$
   \theta = \tan^{-1}\left(\frac{F_{\text{South}}}{F_{\text{West}}}\right) = \tan^{-1}\left(\frac{1.2}{0.9}\right) \approx 53.13^\circ \, \text{South of West}
   $$
   
   *(Note: Since both forces are towards the West and South, the resultant points towards the Southwest. However, for equilibrium, the balancing force will be Northeast.)*

### **3.3 Magnitude and Direction of the Equilibrium Force**

Since the steel ball $$ B $$ is in equilibrium (it does not move), the **sum of all forces acting on it must be zero**. This means the equilibrium force $$ F_{\text{Eq}} $$ must be equal in magnitude and opposite in direction to the resultant of the two given forces.

1. **Magnitude of $$ F_{\text{Eq}} $$:**
   $$
   F_{\text{Eq}} = R = 1.5 \, \text{N}
   $$
   
2. **Direction of $$ F_{\text{Eq}} $$:**
   - Opposite to the resultant direction, which is Northeast.
   - The angle $$ \theta $$ relative to the East axis is:
     $$
     \theta = 53.13^\circ \, \text{North of East}
     $$
     
   *(This is because the equilibrium force must counteract the Southwest direction of the resultant, pointing Northeast.)*

**Final Answer:**

3.3 **The force that keeps the steel ball from moving has a magnitude of 1.5 N and acts in a direction 53° north of east.**
The steel ball is kept from moving by a force of 1.5 N acting 53° north of east.

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