Two forces are such that the sum of their magnitudes is $18 N$ and their resultant is perpendicular to the smaller force and magnitude of resultant is $12 N$ . Then the magnitude of the forces are

12 N, 6 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

13 N, 5 N

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

10 N, 8 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

16 N, 2 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $13 N, 5 N$
Let the two forces be $F_{1}$ and $F_{2}$ such that ( $F_{1} < F_{2}$ )
Let $R$ be the resultant of these two forces.

According to question:
$F_{1} + F_{2} = 18 N$ ...(i)
Angle between $F_{1}$ and resultant ( $R$ ) is $90^{\circ}$
$∴ tan 90^{\circ} = \frac{F_{2} sin θ}{F_{2} cos θ + F_{1}} = \infty$
$⟹ F_{1} + F_{2} cos θ = 0$
$⟹ cos θ = - \frac{F_{1}}{F_{2}}$ ....(ii)
$R^{2} = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos θ$
$⟹ 144 = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \times \frac{- F_{1}}{F_{2}}$
$⟹ 144 = F_{2}^{2} - F_{1}^{2}$ ...(iii)
Using (i) and (iii), we get
$F_{1} = 5 N$ and $F_{2} = 13 N$

Suggest Corrections

Similar questions

Two forces are such that the sum of their magnitude is 18N and their resultant which has magnitude 12N, is perpendicular to the smaller force. Then the magnitudes of the forces are

Two forces are such that the sum of their magnitudes is 18 N and their resultant is perpendicular to the smaller force and magnitude of resultant is 12 N. Then the magnitudes of the forces are

Q. Two forces are such that the sum of their magnitude is