Forces P and Q acting at a point O make an angle `150^(@)` between them. Their resultant acts at O, has magnitude 2 units and is perpendicular to P. Then in the same unit, the magnitudes of P and Q are

Options

A. `2 sqrt""3, 4`
B. `sqrt""(3//2), 2`
C. 3,4
D. 4,5

Answer

To solve the problem of finding the magnitudes of forces P and Q that act at point O, we will use vector resolution and the properties of triangles. Here's a step-by-step solution:

### Step 1: Understand the Given Information
We know:
- Forces P and Q make an angle of \(150^\circ\) between them.
- The resultant force \(R\) has a magnitude of \(2\) units and is perpendicular to force \(P\).

### Step 2: Draw the Diagram
Draw a vector diagram:
- Let \(P\) be directed along the positive x-axis.
- Force \(Q\) will then be at an angle of \(150^\circ\) to \(P\).
- The resultant \(R\) is perpendicular to \(P\), which means it makes an angle of \(90^\circ\) with \(P\).

### Step 3: Resolve Forces into Components
Since \(R\) is perpendicular to \(P\), we can express the components of \(Q\):
- The x-component of \(Q\) is \(Q \cos(150^\circ)\).
- The y-component of \(Q\) is \(Q \sin(150^\circ)\).

### Step 4: Use the Perpendicular Condition
Since \(R\) is perpendicular to \(P\), we can write:
\[ R_y = Q \sin(150^\circ) = 2 \]
\[ R_x = P + Q \cos(150^\circ) = 0 \]

### Step 5: Calculate the Components
Using the known values:
- \( \sin(150^\circ) = \frac{1}{2} \)
- \( \cos(150^\circ) = -\frac{\sqrt{3}}{2} \)

From \(R_y\):
\[ Q \cdot \frac{1}{2} = 2 \implies Q = 4 \text{ units} \]

From \(R_x\):
\[ P + Q \cdot \left(-\frac{\sqrt{3}}{2}\right) = 0 \]
Substituting \(Q = 4\):
\[ P - 4 \cdot \frac{\sqrt{3}}{2} = 0 \implies P = 2\sqrt{3} \text{ units} \]

### Step 6: Final Answer
Thus, the magnitudes of forces \(P\) and \(Q\) are:
- \(P = 2\sqrt{3} \text{ units}\)
- \(Q = 4 \text{ units}\)