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

Question

Two forces are such that the sum of their magnitudes is 18 N and their resultant which has magnitude 12 N, is perpendicular to the smaller force. Then the magnitudes of the forces are : (A) 12 N , 6 N (B) 13 N , 5 N (C) 10 N , 8 N (D) 16 N , 2 N

Tardigrade Admin · Accepted Answer

A+B=18...(1) 12 =√A2+B2+2 A B cos θ...(2) tan α =(B sin θ/A+B cos 0) ⇒ tan 90° =(B sin θ/A+B cos θ) ⇒ cos θ=(-A/B)...(3) Solving Eqs. (1), (2) and (3), A=5 N, B=13 N

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

$12 N, 6 N$

$13 N, 5 N$

$10 N, 8 N$

$16 N, 2 N$

$A + B = 18$ ...(1)
$12 = A^{2} + B^{2} + 2 A B cos θ$ ...(2)
$tan α = \frac{B s i n θ}{A + B c o s 0}$
$\Rightarrow tan 9 0^{\circ} = \frac{B s i n θ}{A + B c o s θ}$
$\Rightarrow cos θ = \frac{- A}{B}$ ...(3)
Solving Eqs. (1), (2) and (3), $A = 5 N, B = 13 N$

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

12N,6N

13N,5N

10N,8N

16N,2N