Question

The resultant of two forces acting at an angle of \( 120^\circ \) is \( 10 \, \text{kg-W} \) and is perpendicular to one of the forces. That force is

• A. \( \frac{10}{\sqrt{3}} \, \text{kg-W} \)
• B. \( 10 \, \text{kg-W} \)
• C. \( 20\sqrt{3} \, \text{kg-W} \)
• D. \( 10\sqrt{3} \, \text{kg-W} \)

Hint:

Use the vector resultant formula and the condition of perpendicularity (dot product = 0) to solve for unknown forces.

Answer 1

The Correct Option is A

Let the two forces be \( F_1 \) and \( F_2 \), and suppose the resultant \( R \) is perpendicular to \( F_1 \).
Using the formula for resultant: \[ R^2 = F_1^2 + F_2^2 + 2F_1F_2\cos(120^\circ) \] But since \( R \perp F_1 \), we can also use: \[ F_1 \cdot R = 0 \Rightarrow \text{Dot product zero} \] Hence, \[ R^2 = F_1^2 + F_2^2 - F_1F_2 \] Set \( R = 10 \) and solve for \( F_1 \) or \( F_2 \). After simplification, we get: \[ F = \frac{10}{\sqrt{3}} \, \text{kg-W} \]