Question

Resultant of two forces \( F_1 \) and \( F_2 \) is of magnitude \( P \). If \( F_2 \) is reversed, the resultant of two forces is of magnitude \( Q \). The value of \( (P^2 + Q^2) \) is

• A. \( \frac{F_1^2 + F_2^2}{4} \)
• B. \( \frac{F_1^2 - F_2^2}{2} \)
• C. \( \frac{F_1^2 + F_2^2}{2} \)
• D. \( 2 (F_1^2 + F_2^2) \)

Hint:

The magnitude of the resultant of two forces changes based on the angle between them. Reversing one of the forces results in a different magnitude of the resultant.

Answer 1

The Correct Option is D

Step 1: Use the formula for the magnitude of the resultant force.
When two forces \( F_1 \) and \( F_2 \) are applied, the magnitude of the resultant force \( P \) is given by: \[ P = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos(\theta)} \] where \( \theta \) is the angle between the forces. When \( F_2 \) is reversed, the new resultant \( Q \) is: \[ Q = \sqrt{F_1^2 + F_2^2 - 2 F_1 F_2 \cos(\theta)} \] Step 2: Calculate \( (P^2 + Q^2) \).
Now, we calculate \( P^2 + Q^2 \): \[ P^2 = F_1^2 + F_2^2 + 2 F_1 F_2 \cos(\theta) \] \[ Q^2 = F_1^2 + F_2^2 - 2 F_1 F_2 \cos(\theta) \] Adding these two equations: \[ P^2 + Q^2 = 2 F_1^2 + 2 F_2^2 \] Thus, we get: \[ P^2 + Q^2 = 2 (F_1^2 + F_2^2) \] Step 3: Conclusion.
Thus, the value of \( (P^2 + Q^2) \) is \( 2 (F_1^2 + F_2^2) \), which corresponds to option (D).