Question

The resultant of two forces P & Q ( such that P > Q) acting along the same straight line but in opposite direction, is given by

• A. \(P + Q\)
• B. \(P - Q\)
• C. \(P \times Q\)
• D. \(\frac{P - Q}{2}\)

Hint:

For collinear forces:
- Same direction: Add magnitudes (\(P + Q\)).
- Opposite directions: Subtract smaller magnitude from larger (\(P - Q\)).

Answer 1

The Correct Option is B

Step 1: Set up the forces on a number line.
Since P and Q act along the same line but in opposite directions, treat one direction as positive and the other as negative. Let P act along the positive direction and Q act along the negative direction.
Step 2: Add the forces algebraically instead of using the general vector formula.
The resultant is \( R = P + (-Q) = P - Q \). Because it is given that \( P > Q \), this difference is positive, so the resultant acts in the direction of the larger force P, with a magnitude equal to the arithmetic difference of the two magnitudes.
Step 3: Cross check against the general two force formula.
For two collinear forces the general resultant formula \( R = \sqrt{P^2+Q^2+2PQ\cos\theta} \) reduces, at \( \theta = 180^\circ \) for opposite directions, to \( R = \sqrt{P^2+Q^2-2PQ} = \sqrt{(P-Q)^2} = P - Q \), confirming the simple subtraction.
\[ \boxed{P - Q} \]