Two forces A and B are acting at a point . The least and greatest resultants of these forces are 5 N and 10 N respectively . If each of these forces is decreased by 2 N , then what will be the resultant of these new forces acting at an angle of `30^(@)` to each other ?

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the original forces A and B
Given that the least resultant \( R_{min} = 5 \, \text{N} \) and the greatest resultant \( R_{max} = 10 \, \text{N} \), we can use the following relationships:

1. \( R_{max} = A + B \)
2. \( R_{min} = |A - B| \)

From these equations, we can express the two forces A and B.

### Step 2: Set up the equations
From the maximum resultant:
\[ A + B = 10 \, \text{N} \quad (1) \]

From the minimum resultant:
\[ |A - B| = 5 \, \text{N} \quad (2) \]

This gives us two cases to consider:
- Case 1: \( A - B = 5 \, \text{N} \)
- Case 2: \( B - A = 5 \, \text{N} \)

### Step 3: Solve for A and B
**Case 1:**
From equation (2):
\[ A - B = 5 \]
Adding equations (1) and (2):
\[ (A + B) + (A - B) = 10 + 5 \]
\[ 2A = 15 \]
\[ A = 7.5 \, \text{N} \]

Substituting \( A \) back into equation (1):
\[ 7.5 + B = 10 \]
\[ B = 2.5 \, \text{N} \]

**Case 2:**
From equation (2):
\[ B - A = 5 \]
Adding equations (1) and (2):
\[ (A + B) + (B - A) = 10 + 5 \]
\[ 2B = 15 \]
\[ B = 7.5 \, \text{N} \]

Substituting \( B \) back into equation (1):
\[ A + 7.5 = 10 \]
\[ A = 2.5 \, \text{N} \]

Thus, the forces are:
- \( A = 7.5 \, \text{N} \) and \( B = 2.5 \, \text{N} \) or vice versa.

### Step 4: Decrease the forces by 2 N
Now, we decrease both forces by 2 N:
- New \( A = 7.5 - 2 = 5.5 \, \text{N} \)
- New \( B = 2.5 - 2 = 0.5 \, \text{N} \)

### Step 5: Calculate the resultant of the new forces
The resultant \( R \) of two forces \( A \) and \( B \) acting at an angle \( \theta \) can be calculated using the formula:
\[ R = \sqrt{A^2 + B^2 + 2AB \cos(\theta)} \]

Substituting \( A = 5.5 \, \text{N} \), \( B = 0.5 \, \text{N} \), and \( \theta = 30^\circ \):
- \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)

Thus,
\[ R = \sqrt{(5.5)^2 + (0.5)^2 + 2 \cdot 5.5 \cdot 0.5 \cdot \cos(30^\circ)} \]
\[ R = \sqrt{30.25 + 0.25 + 2 \cdot 5.5 \cdot 0.5 \cdot \frac{\sqrt{3}}{2}} \]
\[ R = \sqrt{30.5 + 5.5 \cdot 0.5 \cdot \sqrt{3}} \]
\[ R = \sqrt{30.5 + 2.75\sqrt{3}} \]

Calculating \( R \):
Using \( \sqrt{3} \approx 1.732 \):
\[ R \approx \sqrt{30.5 + 2.75 \cdot 1.732} \]
\[ R \approx \sqrt{30.5 + 4.75} \]
\[ R \approx \sqrt{35.25} \]
\[ R \approx 5.93 \, \text{N} \]

### Final Answer
The resultant of the new forces acting at an angle of \( 30^\circ \) to each other is approximately \( 5.93 \, \text{N} \).

---