Question

2. The sum of magnitudes of two forces acting at
a point is 16 N and their resultant 8V3 N is at
900 with the force of smaller magnitude. The
two force (in N) are​

Answer 1

Answer:

$\large\rm {A + B = 16..........(1) }$

$\large\rm { tanα = \frac { B \ sin θ}{A + B \ cos θ} = tan 90}$

$\large\rm { Thus, \ A + B \ cos θ = 0 ⇒ cos θ = \frac {-A}{B} ......(2) }$

We also have : $\large\rm { 8 = \sqrt { A² + B² + 2AB \ cos θ} ........(3)}$

Solving these we get, A = 6N and B = 10N

Answer 2

Answer:

A+B=16..........(1)

\large\rm { tanα = \frac { B \ sin θ}{A + B \ cos θ} = tan 90}tanα=

A+B cosθ

B sinθ

=tan90

\large\rm { Thus, \ A + B \ cos θ = 0 ⇒ cos θ = \frac {-A}{B} ......(2) }Thus, A+B cosθ=0⇒cosθ=

B

−A

......(2)

We also have : \large\rm { 8 = \sqrt { A² + B² + 2AB \ cos θ} ........(3)}8=

A²+B²+2AB cosθ

........(3)

Solving these we get, A = 6N and B = 10N