Question

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

• A. $12N,6N$
• B. $13N,5N$
• C. $10N,8N$
• D. $16N,2N$

Answer 1

The correct option is B $13N,5N$

Let the smaller force be $B$ ,the larger force be $A$, resultant being $C$.
Given: $|A|+|B|=18,|C|=12$
A,B,C simply form a right angled triangle with base B vector,hypotenuse being A.
So, you have one equation with you: $A+B=18$

$A^2=B^2+C^2{A}^2-C^2=B^2(A+B)(A-B)=144(A-B)18=144A-B=\frac{144}{18}=8$
So simply solving both equations: A-B=8 and A+B=18
Adding both:
$2A=26$
$A=13$
$B=5$