The resultant R of two forces acting on a particle is at right angles to one of them and its magnitu

Q: Wait, why is $\vec{F_1}$ the hypotenuse? Shouldn't the resultant be the longest side?

That is a fantastic question! In vector addition $\vec{R} = \vec{F_1} + \vec{F_2}$, the resultant $\vec{R}$ is the vector that closes the triangle. If $\vec{R}$ is perpendicular to $\vec{F_2}$, then $\vec{F_2}$ and $\vec{R}$ are the two legs of a right-angled triangle, and $\vec{F_1}$ must be the side connecting the start of $\vec{F_2}$ to the tip of $\vec{R}$. By the triangle law of vector addition, $\vec{F_1}$ is indeed the hypotenuse.

Q: Can I use the parallelogram law formula $R^2 = F_1^2 + F_2^2 + 2F_1F_2 \cos \theta$ instead?

You absolutely can! It is a perfectly valid approach. However, notice that the problem states $\vec{R} \perp \vec{F_2}$. This implies the angle between $\vec{R}$ and $\vec{F_2}$ is $90^\circ$. Using the triangle method is often faster and less prone to algebraic errors than expanding the full parallelogram law, but both paths lead to the same beautiful destination.

Q: How do I know for sure that $\vec{F_1}$ is the larger force?

Look at our right-angled triangle. In any right-angled triangle, the hypotenuse is always the longest side. Since $\vec{F_1}$ is the hypotenuse, its magnitude must be greater than the magnitude of the other leg, $\vec{F_2}$. Therefore, $\vec{F_1}$ is mathematically guaranteed to be the larger force.

Question

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

Options:

(A) 2 : 1

(B) 3 : $\sqrt{2}$

(C) 3 : 2

(D) 3 : $2\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a perfectly flat, frictionless surface, and a particle is being pulled by two distinct forces, $\vec{F_1}$ and $\vec{F_2}$. These forces are the invisible hands shaping the motion of the particle.

Our goal is to find the ratio of these forces. We are given that the resultant $\vec{R}$, defined as the vector sum $\vec{R} = \vec{F_1} + \vec{F_2}$, is perpendicular to $\vec{F_2}$.

### Visualizing the Triangle

When we talk about vector addition, the triangle law is our best tool. If we place $\vec{F_2}$ along the x-axis and draw $\vec{R}$ vertically along the y-axis, we complete the triangle by placing $\vec{F_1}$ such that it connects the origin to the tip of $\vec{R}$.

Suddenly, the chaos of vectors transforms into a clean, elegant right-angled triangle. In this triangle, $\vec{F_1}$ is the hypotenuse, while $\vec{F_2}$ and $\vec{R}$ are the legs.

### The Algebraic Dance

Now that we have our triangle, the physics becomes pure mathematics. We know from the Pythagorean theorem that the square of the hypotenuse is equal to the sum of the squares of the other two sides:

$$F_1^2 = F_2^2 + R^2$$

The problem provides a crucial constraint: the magnitude of the resultant $\vec{R}$ is exactly one-third of the magnitude of the larger force, $\vec{F_1}$. Mathematically, this is expressed as:

$$R = \frac{F_1}{3}$$

Substituting this into our Pythagorean equation, we must be careful to square both the numerator and the denominator:

$$F_1^2 = F_2^2 + \left( \frac{F_1}{3} \right)^2$$

Expanding this, we obtain:

$$F_1^2 = F_2^2 + \frac{F_1^2}{9}$$

Now, we perform an algebraic rearrangement to isolate the term involving $\vec{F_2}$ by subtracting $\frac{F_1^2}{9}$ from both sides:

$$F_2^2 = F_1^2 - \frac{F_1^2}{9}$$

By finding a common denominator of $9$, we simplify the right side:

$$F_2^2 = \frac{9F_1^2 - F_1^2}{9} = \frac{8F_1^2}{9}$$

### Final Calculation

To find the magnitude of $\vec{F_2}$, we take the square root of both sides:

$$F_2 = \sqrt{\frac{8}{9}} F_1 = \frac{2\sqrt{2}}{3} F_1$$

The question asks for the ratio of the larger force $\vec{F_1}$ to the smaller force $\vec{F_2}$. We set up the ratio as follows:

$$\frac{F_1}{F_2} = \frac{F_1}{\frac{2\sqrt{2}}{3}F_1}$$

The $F_1$ terms cancel out, leaving us with the final result:

$$\frac{F_1}{F_2} = \frac{3}{2\sqrt{2}}$$

Through the power of visualization and systematic algebra, we have decoded the relationship between these two forces. The final ratio is **$\frac{3}{2\sqrt{2}}$**.

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

Select Answer:

Visualized Solution

Visualizing the Forces

The Perpendicular Resultant

Forming the Vector Triangle

The Magnitude Condition

Applying Pythagoras Theorem

Substituting $R$

Expanding the Square

Isolating $F_{2}$

Simplifying the Expression

Taking the Square Root

Finding the Final Ratio

The Sigma Insight: Addition of Vectors

Analyzing the Setup

Visualizing the Triangle

The Algebraic Dance

Final Calculation

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