Concept:Vector addition using right-triangle properties when the resultant is perpendicular to one of the component forces.
Formula:$$F_1 + F_2 = 16\text{ N}$$
$$F_1^2 + R^2 = F_2^2$$
Solution:Let the smaller force be \(F_1\) and the larger force be \(F_2\). Thus, \(F_2 = 16 - F_1\).
Since the resultant \(R = 8\text{ N}\) is perpendicular to \(F_1\), these vectors form a right-angled triangle where \(F_2\) is the hypotenuse:
$$F_1^2 + 8^2 = (16 - F_1)^2$$
$$F_1^2 + 64 = 256 - 32F_1 + F_1^2$$
$$32F_1 = 256 - 64$$
$$32F_1 = 192 \implies F_1 = 6\text{ N}$$
Calculating the larger force:
$$F_2 = 16 - 6 = 10\text{ N}$$
Why other options are incorrect:- 8 N and 8 N is incorrect because if both are 8 N, their sum is 16 N but they cannot form a right triangle with a perpendicular resultant of 8 N.
- 4 N and 12 N, and 2 N and 14 N are incorrect because these pairs do not satisfy the Pythagorean relation with a third side of 8 N. For example, \(4^2 + 8^2 = 80 \neq 12^2\).
Quality & Fidelity Assurance:
Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.