Question

35.
Two forces 9 N and 15 N act at 120°. The third force required to keep the body in equilibrium is
(A) 2-17 N
(B) 3/19 N
(C) 2/19 N
(D) 3 17 N​

Answer 1

Answer:

Answer is 3/19 N is the answer of above question

Answer 2

Given:

Two forces 9 N and 15 N act at 120°

To find:

The third force required to keep the body in equilibrium

Solution:

Let "A" & "B" be the two forces and "R" be the resultant force of the two forces A and B.

here A = 9 N and B = 15 N

and

The angle between A & B, θ = 120°

We will first find the magnitude of the resultant of the two forces which is given by the following formula:

$\boxed{\bold{R\:=\:\sqrt{A^2\:+\:B^2\:+\:2AB\:cos\theta} }}$

By substituting the values of A, B & θ, we get

$R\:=\:\sqrt{9^2\:+\:15^2\:+\:2\times9\times15\times\cos120}$

⇒ $R\:=\:\sqrt{81\:+\:225\:+\:2\times9\times15\times\frac{-1}{2} }$

⇒ $R\:=\:\sqrt{81\:+\:225\:-\:9\times15}$

⇒ $R\:=\:\sqrt{81\:+\:225\:-135 }$

⇒ $R\:=\:\sqrt{306\:-135 }$

⇒ $R\:=\:\sqrt{171 }$

⇒ $R\:=\:\sqrt{3\times3\times19 }$

⇒ $R\:=\:3\sqrt{19}\:N$

Now, we know that the third force which is required to keep the body in equilibrium will have the same magnitude and just in the opposite direction as that of the resultant force.

∴ The third force = $\bold{- 3\sqrt{19}\:N }$

Thus, the third force required to keep the body in equilibrium is $\underline{-3\sqrt{19}\:N}$.

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