Animated Solution for Physics - Kinematics: Two forces are such that the sum of their magnitudes is $18\text{ N}$ and their resultant which has magnitude $12\text{ N}$, is perpendicular to the smaller force. Then, the magnitudes of the forces are

Welcome to a classic problem in vector mechanics! This question beautifully intertwines the geometric intuition of vectors with the algebraic elegance of trigonometric identities. Let's embark on a journey to decode the mystery of the perpendicular resultant.

Imagine you are standing in a field, and two ropes are pulling you in different directions. These are our two forces, let's call them $\vec{A}$ and $\vec{B}$. We are given a fascinating constraint: the resultant force $\vec{R}$ (the net effect of both ropes) is perfectly perpendicular to the smaller force, which we will assume is $\vec{A}$.

We are also armed with two numerical facts: 1. The sum of their magnitudes is $18\text{ N}$, meaning $A+B=18$. 2. The magnitude of the resultant is $12\text{ N}$, meaning $R=12$.

To solve this, we need to translate our geometric reality into mathematical equations. The direction of a resultant vector $\vec{R}$ relative to vector $\vec{A}$ is given by the formula:

Here, $\alpha$ is the angle between $\vec{R}$ and $\vec{A}$, and $\theta$ is the angle between the two forces $\vec{A}$ and $\vec{B}$. Since the resultant is perpendicular to $\vec{A}$, we know that $\alpha =90^{\circ }$.

What happens when we plug this in? We know that $\tan 90^{\circ }$ approaches infinity (it is undefined). For a fraction to be undefined, its denominator must be exactly zero. This gives us a powerful relationship:

Rearranging this, we find the cosine of the angle between the forces:

Now, let's bring in the magnitude of the resultant. The universal law of vector addition states:

We know $R=12$, and we just discovered that $\cos \theta =-\frac{A}{B}$. Let's substitute these in:

Watch the magic happen as the $B$ in the numerator and denominator cancel out:

We have arrived at a beautiful difference of squares. Using the classic algebraic identity $x^2-y^2=(x-y)(x+y)$, we can expand our equation:

Remember our very first clue? The sum of the magnitudes is $18$, so $B+A=18$. Substituting this in:

Dividing both sides by $18$, we find the difference between the forces:

Now we have a simple system of linear equations: 1. $B+A=18$ 2. $B-A=8$

Adding the two equations together yields $2B=26$, which means $B=13\text{ N}$. Substituting this back into the first equation gives $A=5\text{ N}$.

Vector Triangles

Could we have solved this without the heavy trigonometry? Absolutely! Let's look at the problem geometrically.

By definition, vector addition follows the triangle law: $\vec{A}+\vec{B}=\vec{R}$. If we rearrange this, we get $\vec{B}=\vec{R}-\vec{A}$.

Since we are told that $\vec{R}$ is perpendicular to $\vec{A}$, the vectors $\vec{A}$ and $\vec{R}$ form the two perpendicular legs of a right-angled triangle, and $\vec{B}$ acts as the hypotenuse!

Applying the Pythagorean theorem directly to this vector triangle:

Rearranging this gives $B^2-A^2=R^2$, which immediately leads to $B^2-A^2=144$. From here, you just apply the difference of squares as we did before. This geometric shortcut bypasses the $\cos \theta$ substitution entirely and showcases the true beauty of physics!