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  4. The sum of magnitudes of two forces acting at a point is 16 N and their resultant 8√3 N is at 90° with the force of smaller magnitude. The two forces (in N) are

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Q. The sum of magnitudes of two forces acting at a point is $16\, N$ and their resultant $8\sqrt{3}$ N is at $90^\circ$ with the force of smaller magnitude. The two forces $(in\, N)$ are

KEAMKEAM 2012Motion in a Plane

Solution:

According to question,
$P+Q=16 \Rightarrow P=16-Q$
and resultant
$R =\sqrt{P^{2}+Q^{2}+2 P Q \cos \theta} $
$8 \sqrt{3} =\sqrt{P^{2}+Q^{2}+2 P Q \cos \theta} $
$\tan 90^{\circ} =\frac{\sin \theta}{P+Q \cos \theta} $
$P+Q \cos \theta =0 $
$\cos \theta =-\frac{P}{Q} $
$\therefore 8 \sqrt{3} =\sqrt{P^{2}+Q^{2}+2 P Q\left(-\frac{P}{Q}\right)} $
$=\sqrt{P^{2}+Q^{2}-2 P^{2}} $
$8 \sqrt{3} =\sqrt{Q^{2}-P^{2}} $
$Q^{2}-P^{2} =192 $
$Q^{2} -(Q-16)^{2}=192$
$Q^{2}-Q^{2}- 256+32 Q=192 $
$32 Q =192+256$
$Q =14$
$\therefore P =2$