First slide

Scalars and Vectors

Question

Two forces of magnitudes P and Q are inclined at an angle ( $θ$ ) the magnitude of their resultant is 3Q. When the inclination is changed to (180 - $θ$ ) the magnitude of the
resultant force becomes Q. The ratio of the forces $(\frac{P}{Q})$ is

Moderate

A

$\frac{4}{1}$
B

$\frac{2}{1}$
C

$\frac{1}{4}$
D

$\frac{1}{2}$

Solution

$\begin{array}{l} Case 1 : (3 Q)^{2} = P^{2} + Q^{2} + 2 P Q \cos θ . . . . . . . . . . . . . (i) \\ Case 2 : Q^{2} = P^{2} + Q^{2} + 2 P Q \cos (180 - θ) . . . . . . . . . . . . . (i i) \\ 9 Q^{2} = P^{2} + Q^{2} + 2 P Q \cos θ \\ Q^{2} = P^{2} + Q^{2} - 2 P Q \cos θ \\ Adding 10 Q^{2} = 2 P^{2} + 2 Q^{2} \\ or 8 Q^{2} = 2 P^{2} \\ or \frac{P^{2}}{Q^{2}} = \frac{8}{2} = \frac{4}{1} or \frac{P}{Q} = 2 \end{array}$

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