Question

The resultant of two forces acting on a particle is at right angles to one of them and its magnitude is one-third of the magnitude of the other. The ratio of the larger force to the smaller is:

• A. $3:2\sqrt{2}$
• B. $3\sqrt{3}:2$
• C. $3:2$
• D. $4:3$

Answer 1

The correct option is A

$3:2\sqrt{2}$

Step 1: Given Data

Let $\overrightarrow{\mathrm{F}}$and $\overrightarrow{\mathrm{F}}\text{'}$be two forces and their resultant $\overrightarrow{\mathrm{R}}$ is at right angles to one of the forces.

Let the resultant vector have a magnitude of $\raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ and one of the forces have the magnitude of $F$.

Step 2: Formula used:

The resultant of the two forces is given with the help of the parallelogram law of vector addition.

A parallelogram can be formed, and as one of the angles is already a right angle, Pythagoras's theorem can be used to find the side of the right triangle formed.

Step 3. Calculating the magnitude

Now, considering the figure above.

In the figure, the resultant lies in the middle and other vectors are $F$ and $F\text{'}$.

$F\text{'}$ is the one we need to find.

We can place the force vector $F$ such that it starts at the head of the vector $F\text{'}$ and ends at the head of the resultant vector.

In vectors, direction and magnitude is important so we are preserving that in doing so.

Here, the parallelogram gives the magnitude of the remaining force.

$R=\sqrt{F^2+{\left(F^{\text{'}}\right)}^2+2F{F}^{\text{'}}\cos \theta }$, where $R$ is the resultant vector.

Also $R=\frac{F}{3}$ (given in question)

The remaining force is given as:

$$\begin{gathered} F^{\text{'}}=\sqrt{F^2-{\left(\frac{F}{3}\right)}^2} \\ {F}^{\text{'}}=\sqrt{F^2-\frac{F^2}{9}} \\ {F}^{\text{'}}=\sqrt{\frac{8F^2}{9}} \\ {F}^{\text{'}}=\frac{2\sqrt{2}F}{3} \end{gathered}$$

Therefore now the two forces in the question have a magnitude of ratio $3:2\sqrt{2}$.

Thus, option A is the correct option.