The resultant of two forces , one double the other in magnitude is perpendicular to the smaller of the two forces. The angle between the two forces is ________?

Explanation

To solve this problem, we need to determine the angle between two forces given that one force is double the other and their resultant is perpendicular to the smaller force. Let's denote the smaller force as \(\vec{A}\) and the larger force as \(\vec{B}\). Given that \(\vec{B} = 2\vec{A}\), we need to find the angle \(\theta\) between \(\vec{A}\) and \(\vec{B}\).

### Step-by-Step Solution:

1. **Define the Forces:**
Let the magnitude of \(\vec{A}\) be \(A\) and the magnitude of \(\vec{B}\) be \(B\). Given that \(B = 2A\).

2. **Resultant Force:**
The resultant force \(\vec{R}\) is given by:
\[ \vec{R} = \vec{A} + \vec{B} \]
Since \(\vec{R}\) is perpendicular to \(\vec{A}\), the dot product \(\vec{R} \cdot \vec{A} = 0\).

3. **Dot Product Calculation:**
\[ \vec{R} \cdot \vec{A} = (\vec{A} + \vec{B}) \cdot \vec{A} = \vec{A} \cdot \vec{A} + \vec{B} \cdot \vec{A} \]
Since \(\vec{A} \cdot \vec{A} = A^2\) and \(\vec{B} \cdot \vec{A} = B A \cos \theta\), we get:
\[ A^2 + B A \cos \theta = 0 \]

4. **Substitute \(B = 2A\):**
\[ A^2 + 2A \cdot A \cos \theta = 0 \]
\[ A^2 + 2A^2 \cos \theta = 0 \]
\[ 1 + 2 \cos \theta = 0 \]
\[ \cos \theta = -\frac{1}{2} \]

5. **Determine the Angle \(\theta\):**
The angle \(\theta\) for which \(\cos \theta = -\frac{1}{2}\) is:
\[ \theta = 120^\circ \]

### Final Answer:
The angle between the two forces is \(120^\circ\).