Ativity 2 Torces Two forces act on an object. The first is a horizontal 50 N force. second is a 40 N force acting at an angle to the hosizontal. Use a graphical method drawn to a scale of $$ 1 \mathrm{~mm}: 1 \mathrm{~N} $$ to find resultant if the 40 N force is inclined at: $$ 30^{\circ} $$ to the horizontal $$ 60^{\circ} $$ to the horizontal.

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To determine the resultant force when two forces act on an object using a graphical method, we'll break down the process step-by-step. Although we can't draw the vectors here, we'll use numerical calculations to simulate the graphical approach based on the given scale of $$1 \, \text{mm} : 1 \, \text{N}$$.

### Given:
1. **Force $$ \mathbf{F}_1 $$:** 50 N (horizontal)
2. **Force $$ \mathbf{F}_2 $$:** 40 N at an angle to the horizontal
   - **Case 1:** $$30^\circ$$ to the horizontal
   - **Case 2:** $$60^\circ$$ to the horizontal

### Objective:
Find the magnitude of the **resultant force $$ \mathbf{R} $$** for both cases.

### Method:
We'll use the **Law of Cosines** to calculate the magnitude of the resultant force when two vectors are at a certain angle to each other.

$$
R = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos(\theta)}
$$

where:
- $$ R $$ = magnitude of the resultant force
- $$ F_1 $$ and $$ F_2 $$ = magnitudes of the two forces
- $$ \theta $$ = angle between the two forces

### Calculations:

#### **Case 1: $$40 \, \text{N}$$ at $$30^\circ$$ to the horizontal**

1. **Identify the Angle Between Forces:**
   - $$ \theta = 30^\circ $$

2. **Apply the Law of Cosines:**
   $$
   R = \sqrt{50^2 + 40^2 + 2 \times 50 \times 40 \times \cos(30^\circ)}
   $$
   $$
   \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660
   $$
   $$
   R = \sqrt{2500 + 1600 + 4000 \times 0.8660}
   $$
   $$
   R = \sqrt{2500 + 1600 + 3464}
   $$
   $$
   R = \sqrt{7564} \approx 87 \, \text{N}
   $$

#### **Case 2: $$40 \, \text{N}$$ at $$60^\circ$$ to the horizontal**

1. **Identify the Angle Between Forces:**
   - $$ \theta = 60^\circ $$

2. **Apply the Law of Cosines:**
   $$
   R = \sqrt{50^2 + 40^2 + 2 \times 50 \times 40 \times \cos(60^\circ)}
   $$
   $$
   \cos(60^\circ) = 0.5
   $$
   $$
   R = \sqrt{2500 + 1600 + 4000 \times 0.5}
   $$
   $$
   R = \sqrt{2500 + 1600 + 2000}
   $$
   $$
   R = \sqrt{6100} \approx 78.1 \, \text{N}
   $$

### Summary of Results:
- **When the 40 N force is inclined at $$30^\circ$$:**
  - **Resultant Force $$ R \approx 87 \, \text{N} $$**

- **When the 40 N force is inclined at $$60^\circ$$:**
  - **Resultant Force $$ R \approx 78.1 \, \text{N} $$**

### Graphical Interpretation:
If you were to draw this graphically:
1. **Scale:** $$1 \, \text{mm} = 1 \, \text{N}$$
2. **Draw $$ \mathbf{F}_1 $$:** Draw a horizontal line 50 mm long.
3. **Draw $$ \mathbf{F}_2 $$:** From the end of $$ \mathbf{F}_1 $$, draw a line at the specified angle (30° or 60°) 40 mm long.
4. **Resultant $$ \mathbf{R} $$:** Draw a line from the origin (start of $$ \mathbf{F}_1 $$) to the end of $$ \mathbf{F}_2 $$. The length of this line represents the magnitude of the resultant force.

Using this method with accurate measurements will give you the resultant forces calculated above.
When a 50 N horizontal force and a 40 N force at 30° or 60° to the horizontal act on an object, the resultant forces are approximately 87 N and 78.1 N, respectively.

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