Two forces of 5N and 7N act on a aprticle with an angle of `60^(@)` between them. Find the resultant force.

Answer

To find the resultant force of two forces acting at an angle, we can use the formula for the magnitude of the resultant vector. Here’s a step-by-step solution:

### Step 1: Identify the forces and the angle
We have two forces:
- \( F_1 = 5 \, \text{N} \)
- \( F_2 = 7 \, \text{N} \)
The angle between them is \( \theta = 60^\circ \).

### Step 2: Write down the formula for the resultant force
The magnitude of the resultant force \( R \) can be calculated using the formula:
\[ R = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos \theta} \]

### Step 3: Substitute the values into the formula
Now, substitute the values of \( F_1 \), \( F_2 \), and \( \theta \) into the formula:
\[ R = \sqrt{(5)^2 + (7)^2 + 2 \cdot 5 \cdot 7 \cdot \cos(60^\circ)} \]

### Step 4: Calculate the squares and cosine
Calculate \( F_1^2 \) and \( F_2^2 \):
\[ 5^2 = 25 \]
\[ 7^2 = 49 \]
Now calculate \( \cos(60^\circ) \):
\[ \cos(60^\circ) = \frac{1}{2} \]
Now substitute these values back into the equation:
\[ R = \sqrt{25 + 49 + 2 \cdot 5 \cdot 7 \cdot \frac{1}{2}} \]

### Step 5: Simplify the expression
Calculate \( 2 \cdot 5 \cdot 7 \cdot \frac{1}{2} \):
\[ 2 \cdot 5 \cdot 7 \cdot \frac{1}{2} = 5 \cdot 7 = 35 \]
Now substitute this back into the equation:
\[ R = \sqrt{25 + 49 + 35} \]

### Step 6: Add the values inside the square root
Now add the values:
\[ R = \sqrt{109} \]

### Step 7: Calculate the square root
Now calculate \( \sqrt{109} \):
\[ R \approx 10.44 \, \text{N} \]

### Step 8: Find the angle of the resultant force
To find the angle \( \alpha \) of the resultant force with respect to \( F_1 \), we can use the cosine formula:
\[ \cos \alpha = \frac{F_1}{R} \]
Substituting the values:
\[ \cos \alpha = \frac{5}{10.44} \]
Now calculate \( \alpha \):
\[ \alpha = \cos^{-1}\left(\frac{5}{10.44}\right) \approx 35^\circ \]

### Final Result
The magnitude of the resultant force is approximately \( 10.44 \, \text{N} \) and the angle with respect to the 5 N force is approximately \( 35^\circ \).

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