Question

: The resultant of two forces has a magnitude of 20 N. If one of the forces is of, magnitude \[20\sqrt{3}N\] and makes an angle of \[{{30}^{0}}\] with the resultant then, the other forces must be of the magnitude
A- \[10\sqrt{3}N\]
B-20 N
C- \[20\sqrt{3}N\]
D-10 N

Answer 1

Hint: From parallelogram law, we are able to find the formula for the resultant of two forces. Since we’re given that the magnitude of the forces is equal, the calculation gets easier and we can generate a definite value as output.
\[R=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \varphi }\] , where A and B are two forces, R is their resultant and \[\varphi \] is the angle between them.

Complete step by step answer:
Given \[{{F}_{1}}=20\sqrt{3}N\]
And here we are not given\[\varphi \], the angle between the two original vectors but we are given the angle between the resultant and one of the given forces.
Let the magnitude of the other force be x and the angle it makes with resultant be θ.
Now, the components of the two vectors along the resultant should be of magnitude of the resultant and the component of these two vectors resolved in the direction perpendicular to resultant should be zero.
Writing it mathematically,
\[\begin{align}
& 20\sqrt{3}\times \dfrac{\sqrt{3}}{2}+x\cos \theta =20 \\
&\Rightarrow 30+x\cos \theta =20 \\
\end{align}\]
\[x\cos \theta =-10\] ------(1)
Also, resolving for the perpendicular direction,
\[10\sqrt{3}+x\sin \theta =0\]
\[x\sin \theta =-10\sqrt{3}\] ----------(2)
In order to solve for x we square and add eq (1) and (2)
\[\begin{align}
& {{(x\cos \theta )}^{2}}+{{(x\sin \theta )}^{2}}=100+300 \\
&\Rightarrow {{x}^{2}}=400 \\
&\therefore x=20 \\
\end{align}\]
So the value of x comes out to be 20 N, so the correct option is (B).

Note:We could have used parallelogram law where the two vectors acting at a point are shown based on magnitude and direction, by the two adjacent sides of a parallelogram generated from a particular point and the obtained resultant is represented by the diagonal of the parallelogram that is drawn from the same point. But we were not given the angle between the two vectors.