Topic: Engineering Mechanics
The resultant of two forces acting at $60^{\circ}$ is 7 N and when acting at right angles is $\sqrt{34}N$ The forces are
- A. 25 N and 9 N
- B. 3 N and $\sqrt{5}N$
- C. 3 N and 5 N
- D. $\sqrt{3}N$ and $\sqrt{5}N$
Answer: C. 3 N and 5 N
Explanation: By Parallelogram Method, $R=\sqrt{P^{2}+Q^{2}+2PQ~\cos~\theta}$. For $60^{\circ}$: $7=\sqrt{P^{2}+Q^{2}+2PQ\cos60^{\circ}} = \sqrt{P^{2}+Q^{2}+PQ} \Rightarrow 49=P^{2}+Q^{2}+PQ$ (Eq 1). When acting at right angles ($90^{\circ}$): $\sqrt{34}=\sqrt{P^{2}+Q^{2}} \Rightarrow 34=P^{2}+Q^{2}$ (Eq 2). Substituting (2) in (1): $49=34+PQ \Rightarrow PQ=15$. Solving $P^{2}+Q^{2}=34$ and $PQ=15$ yields $P=3N$ and $Q=5N$.