Question

The sum of magnitude of two forces acting at a point is 16N. If their resultant is normal to the smaller force and has a magnitude of 8N, the forces are?​

Answer 1

Let:-

There be two forces P and Q.

And as the given resultant angle is normal to the smaller force so, a = 90°

And as the angle with the resultant,

$tan \: a \: = \frac{Qsin \: a}{p + Qcos \: a}$

$tan \: 90 = \frac{Qsin \: a}{p + Qcos \: a}$

$p + Qcos \: a = 0........eq1 \\ cos \: a = \frac{ - p}{Q}$

And as the resultant of the two forces given is 8N,

Therefore:-

${p}^{2} +{ Q}^{2} + 2pQcos \: a = {8}^{2}$

Putting the value of eq 1, we get

${p}^{2} +{ Q}^{2} + 2pQ( \frac{ - p}{Q}) = 64$

=> P² + Q² - 2P² = 64

=> Q²b- P² = 64

=> (Q + P)(Q - P) = 64

And as the sum of two forces is 16N.

=> 16(Q - P) = 64

=> (Q - P) = 4

Now,

=> (P+Q) =16

=> (P-Q)=4

Adding both, we get

=> 2P = 20

=> P =10N

=> Q = 6N

Therefore:-

The two forces are 10N and 6N.

Answer 2

REFER TO THE BELOW ATTACHMENT.....