The sum of magnitudes of two forces acting at a point is . . .

Question : The sum of magnitudes of two forces acting at a point is 16 and magnitude of their resultant is 8√3. If the resultant is at 90° with the force of smaller magnitude, their magnitudes are

a) 3, 13

b) 2, 14

c) 5, 11

d) 4, 12

Doubt by Avani

Solution : 

R=8√3 N

A+B=16 (Given) B=16-A — (1)

tanβ=Bsinθ/(A+Bcosθ) tan90°=Bsinθ/(A+Bcosθ) 1/0=Bsinθ/(A+Bcosθ) A+Bcoθ = 0 × Bsinθ

A+Bcosθ=0 Bcosθ=-A — (2)

R²=A²+B²+2ABcosθ R²=A²+(16-A)²+2A(-A) [Using Equation (1) and (2)] R²=A²+(16)²+A²-2(16)(A)-2A² (8√3)²=2A²+256-32A-2A² 192-256=-32A -64=-32A 64/32 = A 2=A A=2 N

Putting in equation (1)

B=16-A B=16-2 B=14 N

The magnitude of two forces are 2 N and 14 N. Hence b) 2, 14 would be the correct option.

Similar Question : The sum of the magnitudes of two forces acting at a point is 18 N and the magnitude of their resultant is 12 N. If the resultant makes an angle of 90° with the force of smaller magnitude, what are the magnitudes of the two forces?