With two forces acting at point, the maximum affect is obtained when their resultant is 4N. If they

Q: Why is the maximum resultant simply $P + Q$?

The general formula for the resultant of two vectors $\vec{P}$ and $\vec{Q}$ is $R = \sqrt{P^2 + Q^2 + 2PQ \cos(\theta)}$. When the forces act in the same direction, the angle $\theta$ is $0^\circ$. Since $\cos(0^\circ) = 1$, the formula becomes $R = \sqrt{P^2 + Q^2 + 2PQ}$, which is $\sqrt{(P+Q)^2}$, simplifying to $P+Q$. It is the physical limit of reinforcement where the forces work in perfect harmony.

Q: Why use the identity method instead of substituting $Q = 4 - P$?

While substitution is a valid algebraic path, it forces you to solve a quadratic equation $P^2 + (4-P)^2 = 9$. The identity method is a powerful tool that keeps the algebra symmetric and elegant. It reduces the chance of calculation errors and is a technique you will use repeatedly in higher-level physics problems involving energy and momentum conservation.

Q: What if the angle between the forces was not $90^\circ$ but some other angle $\alpha$?

If the angle were $\alpha$, the second equation would be $R^2 = P^2 + Q^2 + 2PQ \cos(\alpha)$. You would still have the sum $P+Q=4$ and the product $PQ$ from the first condition. You would then solve for $P$ and $Q$ using the Law of Cosines. The methodology remains the same: translate the physical geometry into algebraic constraints.

Question

With two forces acting at point, the maximum affect is obtained when their resultant is 4N. If they act at right angles, then their resultant is 3N. Then the forces are

Options:

(A) $(2 + \frac{1}{2}\sqrt{3}) N$ and $(2 - \frac{1}{2}\sqrt{3}) N$

(B) $(2 + \sqrt{3}) N$ and $(2 - \sqrt{3}) N$

(C) $(2 + \frac{1}{2}\sqrt{2}) N$ and $(2 - \frac{1}{2}\sqrt{2}) N$

(D) $(2 + \sqrt{2}) N$ and $(2 - \sqrt{2}) N$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, open field, and you have two invisible forces, $P$ and $Q$, acting on a single point. Our goal is to uncover their true magnitudes by translating these physical conditions into mathematical constraints.

The problem states that the maximum resultant force is $4 	ext{ N}$. This occurs when the two forces act in the same direction, meaning the angle between them is $	heta = 0^\circ$.

In this configuration, the magnitudes simply add up. Thus, our first equation is:
$$P + Q = 4$$

### The Orthogonal Condition

Next, we consider the scenario where these same forces act at right angles to each other. In this case, the resultant $R$ is the diagonal of a rectangle formed by vectors $P$ and $Q$.

According to the Pythagorean theorem, the magnitude of the resultant is given by $R = \sqrt{P^2 + Q^2}$. We are given that this resultant is $3 	ext{ N}$, leading to:
$$\sqrt{P^2 + Q^2} = 3$$

Squaring both sides of this equation, we obtain our second foundation:
$$P^2 + Q^2 = 9$$

### Solving the System

We now have a system of two equations: $P + Q = 4$ and $P^2 + Q^2 = 9$. Rather than using substitution, we utilize algebraic identities for a more elegant solution.

Recall the expansion:
$$(P + Q)^2 = P^2 + Q^2 + 2PQ$$

Substituting the known values $P + Q = 4$ and $P^2 + Q^2 = 9$ into this identity:
$$4^2 = 9 + 2PQ$$
$$16 = 9 + 2PQ$$
$$2PQ = 7$$

### Final Calculation

To find the individual values of $P$ and $Q$, we determine the difference $(P - Q)$ using the identity:
$$(P - Q)^2 = P^2 + Q^2 - 2PQ$$

Substituting our known values:
$$(P - Q)^2 = 9 - 7 = 2$$
$$P - Q = \sqrt{2}$$

We now have a simple linear system:
1) $P + Q = 4$
2) $P - Q = \sqrt{2}$

Adding these equations yields $2P = 4 + \sqrt{2}$, which simplifies to:
$$P = 2 + \frac{\sqrt{2}}{2} 	ext{ N}$$

Subtracting the equations yields $2Q = 4 - \sqrt{2}$, which simplifies to:
$$Q = 2 - \frac{\sqrt{2}}{2} 	ext{ N}$$

The magnitudes of the two forces are **$P = (2 + \frac{\sqrt{2}}{2}) 	ext{ N}$** and **$Q = (2 - \frac{\sqrt{2}}{2}) 	ext{ N}$**.

Animated Solution for Mathematics - Vector Algebra: With two forces acting at point, the maximum affect is obtained when their resultant is 4N. If they act at right angles, then their resultant is 3N. Then the forces are

Select Answer:

Visualized Solution

Defining the Forces $P$ and $Q$

Condition 1: Maximum Resultant

Formulating Equation 1

Condition 2: Forces at Right Angles

Resultant at $9 0^{\circ}$

Formulating Equation 2

Using Algebraic Identities

Finding the Product $2 P Q$

Finding the Difference $(P - Q)$

Calculating $(P - Q)$

Solving for $P$

Solving for $Q$

Final Answer

The Sigma Insight: Addition of Vectors

Analyzing the Setup

The Orthogonal Condition

Solving the System

Final Calculation

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