In a right angle ΔABC, ∠A = 90° and sides a, b, c are respectively, 5 cm, 4 cm and 3 cm. If a force

Q: Wait, why is the moment about A zero? Does that mean the force is zero?

Not at all! The moment of a force about a point is defined as the product of the force magnitude and the perpendicular distance from the point to the line of action. If the moment is zero, it simply means the perpendicular distance is zero. Physically, this means the line of action of the force passes directly through point A. Think of it like pushing a door right at the hinge—it won't rotate, no matter how hard you push!

Q: Why do we use absolute values in the distance formula?

Great question. The distance $d$ is a physical length, which must always be positive. When we use the formula $\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$, the absolute value ensures that we get a positive magnitude for the distance, regardless of which side of the line the point lies on.

Q: Can we use vector cross products instead of the distance formula?

Absolutely! You could define $\vec{F} = F_x \hat{i} + F_y \hat{j}$ and use $\vec{M} = \vec{r} \times \vec{F}$. However, in this specific problem, coordinate geometry is much more elegant and less prone to algebraic errors. The distance formula approach allows us to isolate the slope $m$ and the magnitude $|\vec{F}|$ very cleanly.

Question

In a right angle $	riangle ABC$, $\angle A = 90^\circ$ and sides a, b, c are respectively, 5 cm, 4 cm and 3 cm. If a force $\vec{F}$ has moments 0, 9 and 16 in N cm. units respectively about vertices A, B and C, then magnitude of $\vec{F}$ is

Options:

(A) 9

(B) 4

(C) 5

(D) 3

Sigma Percentile · Accepted Answer

### Analyzing the Geometry

Imagine you are standing in front of a coordinate plane, looking at a right-angled triangle $	riangle ABC$. We are given that $\angle A = 90^{\circ}$, with side lengths $a=5$, $b=4$, and $c=3$.

The most strategic move here is to place point $A$ at the origin $(0,0)$. By aligning $AB$ along the x-axis and $AC$ along the y-axis, we define our vertices as $A(0,0)$, $B(3,0)$, and $C(0,4)$.

This setup transforms an abstract geometry problem into a playground of coordinates.

### The Physics of the Moment

The problem provides a crucial piece of information: the moment of force $\vec{F}$ about vertex $A$ is zero. In the language of physics, this is the golden key.

If the moment about a point is zero, the line of action of the force must pass through that point. Since our pivot $A$ is at the origin, the line of action of $\vec{F}$ must be a line passing through $(0,0)$.

We can represent this line as $y = mx$, or in its general form, $mx - y = 0$. This is the path along which our force $\vec{F}$ is acting.

### The Algebraic Symphony

We recall the definition of the moment of a force: $M = |\vec{F}| \cdot d$, where $d$ is the perpendicular distance from the pivot to the line of action. We are given that the moment about $B(3,0)$ is $9$, and the moment about $C(0,4)$ is $16$.

Using the perpendicular distance formula $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$, we calculate the distances from $B$ and $C$ to our line $mx - y = 0$.

For point $B(3,0)$:
$$d_B = \frac{|m(3) - (0)|}{\sqrt{m^2 + (-1)^2}} = \frac{3|m|}{\sqrt{m^2 + 1}}$$

For point $C(0,4)$:
$$d_C = \frac{|m(0) - (4)|}{\sqrt{m^2 + 1}} = \frac{4}{\sqrt{m^2 + 1}}$$

### Solving the System

Now, we set up our two equations for the moments:

1) $|\vec{F}| \cdot \frac{3|m|}{\sqrt{m^2 + 1}} = 9 \Rightarrow \frac{|\vec{F}| \cdot |m|}{\sqrt{m^2 + 1}} = 3$

2) $|\vec{F}| \cdot \frac{4}{\sqrt{m^2 + 1}} = 16 \Rightarrow \frac{|\vec{F}|}{\sqrt{m^2 + 1}} = 4$

If we divide the first equation by the second, the term $\frac{|\vec{F}|}{\sqrt{m^2 + 1}}$ cancels out entirely. We are left with $|m| = \frac{3}{4}$.

### Final Calculation

Finally, we substitute $|m| = \frac{3}{4}$ back into our second equation:

$$\frac{|\vec{F}|}{\sqrt{(\frac{3}{4})^2 + 1}} = 4$$

$$\frac{|\vec{F}|}{\sqrt{\frac{9}{16} + 1}} = 4 \Rightarrow \frac{|\vec{F}|}{\frac{5}{4}} = 4$$

$$|\vec{F}| = 5$$

The magnitude of the force is **$5$ N**. This is a clean result that rewards a structured, logical approach to the problem.

Animated Solution for Mathematics - Straight Lines: In a right angle $△ A B C$ , $∠ A = 9 0^{\circ}$ and sides a, b, c are respectively, 5 cm, 4 cm and 3 cm. If a force $F$ has moments 0, 9 and 16 in N cm. units respectively about vertices A, B and C, then magnitude of $F$ is

Select Answer:

Visualized Solution

Coordinate Setup for $△ A B C$

Line of Action of $F$

Equation of Line of Action

Moment Formula

Moment about $B (3, 0)$

Equation for Moment at $B$

Moment about $C (0, 4)$

Equation for Moment at $C$

Calculating Slope $∣ m ∣$

Substituting $∣ m ∣$

Calculating Magnitude $∣ F ∣$

Final Result

The Sigma Insight: Distance of a Point from a Line

Analyzing the Geometry

The Physics of the Moment

The Algebraic Symphony

Solving the System

Final Calculation

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Visualized Solution

Coordinate Setup for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △ABC

Line of Action of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F