The sum of two forces P and Q is R such that |R| = |P|. The angle theta (in degrees) that the result

Q: Why did we assume the magnitude of Q is not zero?

If the magnitude of Q were zero, the problem would be trivial. The resultant of P and Q would just be P, and the resultant of 2P and Q would just be 2P. There would be no meaningful angle to calculate relative to a zero vector.

Q: What does the equation $2|\mathbf{P}|\cos\beta = -|\mathbf{Q}|$ actually mean in real life?

It means that if you shine a light from above, the shadow (projection) of vector P onto the line of vector Q is exactly half the length of Q, and it points in the exact opposite direction of Q.

Q: Why does a zero in the denominator mean the angle is 90 degrees?

The formula for $\tan\theta$ is essentially (y-component) / (x-component). If the denominator (x-component) is zero, it means the vector has no horizontal length; it points straight up or straight down. The angle for a purely vertical vector relative to the horizontal x-axis is 90 degrees.

Q: Could the angle be 270 degrees instead of 90?

Mathematically, $\tan(270^\circ)$ is also undefined. However, the numerator $2|\mathbf{P}|\sin\beta$ determines the direction. Assuming $\beta$ is between 0 and 180 degrees (standard for vector angles), $\sin\beta$ is positive, so the resultant points in the positive y-direction, making it exactly 90 degrees.

Q: Is there a way to solve this purely by drawing?

Yes! If you draw a triangle with sides P, Q, and R where P=R, it forms an isosceles triangle. If you extend the side P to 2P, you can use geometry to prove that the median to the base of the new larger triangle is perpendicular to the base Q.

Question

The sum of two forces $\mathbf{P}$ and $\mathbf{Q}$ is $\mathbf{R}$ such that $|\mathbf{R}| = |\mathbf{P}|$. The angle $	heta$ (in degrees) that the resultant of $2\mathbf{P}$ and $\mathbf{Q}$ will make with $\mathbf{Q}$ is, .........

Sigma Percentile · Accepted Answer

Vector addition often hides beautiful geometric symmetries within its algebraic equations. This problem is a classic example of how a simple constraint on magnitudes can dictate a rigid spatial relationship between vectors.

### Analyzing the Initial Setup

We are given two force vectors, $\mathbf{P}$ and $\mathbf{Q}$, and their resultant $\mathbf{R} = \mathbf{P} + \mathbf{Q}$. Let the angle between $\mathbf{P}$ and $\mathbf{Q}$ be $\beta$. The fundamental law of vector addition tells us that the magnitude of the resultant is given by:

$$|\mathbf{R}|^2 = |\mathbf{P}|^2 + |\mathbf{Q}|^2 + 2|\mathbf{P}||\mathbf{Q}|\cos\beta$$

The problem introduces a fascinating constraint: the magnitude of the resultant $\mathbf{R}$ is exactly equal to the magnitude of the force $\mathbf{P}$, meaning $|\mathbf{R}| = |\mathbf{P}|$.

Let's substitute this condition into our master equation:

$$|\mathbf{P}|^2 = |\mathbf{P}|^2 + |\mathbf{Q}|^2 + 2|\mathbf{P}||\mathbf{Q}|\cos\beta$$

Notice how $|\mathbf{P}|^2$ beautifully cancels out from both sides. This leaves us with:

$$0 = |\mathbf{Q}|^2 + 2|\mathbf{P}||\mathbf{Q}|\cos\beta$$

We can factor out the magnitude of $\mathbf{Q}$:

$$|\mathbf{Q}|(|\mathbf{Q}| + 2|\mathbf{P}|\cos\beta) = 0$$

Since $\mathbf{Q}$ is a force vector, its magnitude $|\mathbf{Q}|$ is non-zero. Therefore, the term inside the parentheses must be zero. This yields a crucial geometric relationship:

$$|\mathbf{Q}| + 2|\mathbf{P}|\cos\beta = 0 \implies 2|\mathbf{P}|\cos\beta = -|\mathbf{Q}|$$

Physically, this means the projection of vector $\mathbf{P}$ onto vector $\mathbf{Q}$ is exactly half the magnitude of $\mathbf{Q}$, and it points in the opposite direction.

### The Second Scenario: Doubling P

Now, the problem asks us to consider a new resultant, let's call it $\mathbf{R}'$, which is the vector sum of $2\mathbf{P}$ and $\mathbf{Q}$. We need to find the angle $	heta$ that this new resultant makes with the vector $\mathbf{Q}$.

The standard formula for the direction of a resultant vector gives us:

$$	an	heta = \frac{|2\mathbf{P}|\sin\beta}{|\mathbf{Q}| + |2\mathbf{P}|\cos\beta}$$

This is where the magic happens. Look closely at the denominator: $|\mathbf{Q}| + 2|\mathbf{P}|\cos\beta$. Does this look familiar? It is exactly the expression we evaluated to zero in the first part of our analysis!

Let's substitute our finding ($2|\mathbf{P}|\cos\beta = -|\mathbf{Q}|$):

$$	an	heta = \frac{2|\mathbf{P}|\sin\beta}{|\mathbf{Q}| - |\mathbf{Q}|}$$

$$	an	heta = \frac{2|\mathbf{P}|\sin\beta}{0}$$

### The Final Conclusion

A denominator of zero in the tangent function implies that the value approaches infinity. In trigonometry, $	an	heta = \infty$ corresponds to an angle of $90^\circ$.

$$	heta = 90^\circ$$

Geometrically, by doubling the vector $\mathbf{P}$, its horizontal projection (which was $-|\mathbf{Q}|/2$) also doubles, becoming exactly $-|\mathbf{Q}|$. This perfectly cancels out the vector $\mathbf{Q}$ in the horizontal direction, leaving the new resultant purely vertical, hence making a $90^\circ$ angle with $\mathbf{Q}$.

Animated Solution for Physics - Kinematics: The sum of two forces $P$ and $Q$ is $R$ such that $∣ R ∣ = ∣ P ∣$ . The angle $θ$ (in degrees) that the resultant of $2 P$ and $Q$ will make with $Q$ is, .........

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Vectors}

\text{Magnitude of Resultant}

\text{Applying the Given Condition}

\text{Simplifying the Equation}

\text{Finding the Relation}

\text{The New Resultant } \mathbf{R}'

\text{Angle of the New Resultant}

\text{Calculating } \theta

The Sigma Insight: Vector Addition, Subtraction, and Resolution

Analyzing the Initial Setup

The Second Scenario

The Final Conclusion

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