Animated Solution for Physics - Kinematics: Two forces $P$ and $Q$ of magnitude $2F$ and $3F$, respectively, are at an angle $\theta$ with each other. If the force $Q$ is doubled, then their resultant also gets doubled. Then, the angle $\theta$ is

Imagine you are pulling a heavy block with two ropes. The forces you exert on the ropes are represented by vectors $\vec{P}$ and $\vec{Q}$. In our problem, the magnitudes of these forces are given as $2F$ and $3F$, respectively. They act at a certain angle $\theta$ to each other.

According to the parallelogram law of vector addition, the resultant force $\vec{R}$ is the diagonal of the parallelogram formed by these two vectors. The magnitude of this resultant is a crucial piece of the puzzle, and it depends heavily on the angle $\theta$ between the forces.

The parallelogram law gives us a direct formula to calculate the square of the resultant's magnitude:

Let's substitute our initial forces into this equation. We have $A=2F$ and $B=3F$:

Expanding the squares and multiplying the terms, we get:

This is our foundational equation, linking the unknown resultant $R$ to the unknown angle $\theta$.

Now, the problem introduces a twist: the force $\vec{Q}$ is doubled. Its new magnitude becomes $6F$. What happens to the resultant? The problem states that the resultant is also doubled, becoming $2R$.

Let's apply the parallelogram law to this new, scaled-up scenario:

Squaring the terms gives us:

We now have a system of two equations with two unknowns ($R$ and $\theta$).

To find $\theta$, we need to eliminate $R$. A clever way to do this is to multiply our first equation by 4, so the left-hand side matches the second equation:

Now, we can equate the right-hand sides of equation (2) and equation (3):

Notice that every term contains $F^2$. Since $F$ is a non-zero magnitude, we can safely divide the entire equation by $F^2$:

Now, it's a simple linear equation in terms of $\cos \theta$. Let's group the cosine terms on one side and the constants on the other:

The cosine of the angle is negative, which immediately tells us the angle is obtuse (between $90^{\circ }$ and $180^{\circ }$). The specific angle whose cosine is $-\frac{1}{2}$ is:

And there we have it! The forces must be acting at an angle of $120^{\circ }$ for this specific scaling relationship to hold true.