Three forces 3P, 7P and 5P act along the sides AB, BC and CA of an equilateral triangle ABC. Find their resultant. [6]

Solution

Step 1: Understanding the Problem

We are given an equilateral triangle $ABC$ with forces acting along its sides:

• Force $3P$ along $AB$,
• Force $7P$ along $BC$,
• Force $5P$ along $CA$.

Since the triangle is equilateral, all sides are equal, and all angles are $60^{\circ }$. We need to find the resultant of these three forces.

---

Step 2: Representing Forces as Vectors

Forces are vector quantities, so we represent them using coordinate geometry. Let’s place the triangle in a coordinate system for easier calculation.

Assume:

• Side length of the equilateral triangle $ABC=a$.
• Place point $A$ at the origin $(0,0)$.
• Place point $B$ along the x-axis at $(a,0)$.
• Point $C$ will then be at $\left(\frac{a}{2},\frac{a\sqrt{3}}{2}\right)$ because the height of an equilateral triangle is $\frac{a\sqrt{3}}{2}$.

---

Step 3: Direction of Forces

The forces act along the sides of the triangle. We need to determine the direction of each force in terms of unit vectors.

1. Force along $AB$ (3P):

   • Direction: From $A$ to $B$, i.e., along the positive x-axis.
   • Unit vector: $\hat{i}$.
   • Force vector: ${\vec{F}}_{AB}=3P\hat{i}$.

2. Force along $BC$ (7P):

   • Direction: From $B$ to $C$.
   • Coordinates of $B=(a,0)$ and $C=\left(\frac{a}{2},\frac{a\sqrt{3}}{2}\right)$.
   • Displacement vector $BC=\left(\frac{a}{2}-a,\frac{a\sqrt{3}}{2}-0\right)=\left(-\frac{a}{2},\frac{a\sqrt{3}}{2}\right)$.
   • Unit vector: $\frac{BC}{|BC|}=\frac{1}{a}\left(-\frac{a}{2},\frac{a\sqrt{3}}{2}\right)=\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)$.
   • Force vector: ${\vec{F}}_{BC}=7P\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)=\left(-\frac{7P}{2},\frac{7P\sqrt{3}}{2}\right)$.

3. Force along $CA$ (5P):

   • Direction: From $C$ to $A$.
   • Coordinates of $C=\left(\frac{a}{2},\frac{a\sqrt{3}}{2}\right)$ and $A=(0,0)$.
   • Displacement vector $CA=\left(0-\frac{a}{2},0-\frac{a\sqrt{3}}{2}\right)=\left(-\frac{a}{2},-\frac{a\sqrt{3}}{2}\right)$.
   • Unit vector: $\frac{CA}{|CA|}=\frac{1}{a}\left(-\frac{a}{2},-\frac{a\sqrt{3}}{2}\right)=\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$.
   • Force vector: ${\vec{F}}_{CA}=5P\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)=\left(-\frac{5P}{2},-\frac{5P\sqrt{3}}{2}\right)$.

---

Step 4: Summing the Forces

The resultant force $\vec{R}$ is the vector sum of ${\vec{F}}_{AB}$, ${\vec{F}}_{BC}$, and ${\vec{F}}_{CA}$:

$$\vec{R}={\vec{F}}_{AB}+{\vec{F}}_{BC}+{\vec{F}}_{CA}$$

Substitute the components:

$$\vec{R}=\left(3P\hat{i}\right)+\left(-\frac{7P}{2}\hat{i}+\frac{7P\sqrt{3}}{2}\hat{j}\right)+\left(-\frac{5P}{2}\hat{i}-\frac{5P\sqrt{3}}{2}\hat{j}\right)$$

Combine like terms:

$$\vec{R}=\left(3P-\frac{7P}{2}-\frac{5P}{2}\right)\hat{i}+\left(\frac{7P\sqrt{3}}{2}-\frac{5P\sqrt{3}}{2}\right)\hat{j}$$

Simplify:

$$\vec{R}=\left(3P-\frac{12P}{2}\right)\hat{i}+\left(\frac{2P\sqrt{3}}{2}\right)\hat{j}$$

$$\vec{R}=\left(3P-6P\right)\hat{i}+\left(P\sqrt{3}\right)\hat{j}$$

$$\vec{R}=-3P\hat{i}+P\sqrt{3}\hat{j}$$

---

Step 5: Magnitude of the Resultant Force

The magnitude of $\vec{R}$ is given by:

$$|\vec{R}|=\sqrt{(-3P{)}^2+(P\sqrt{3}{)}^2}$$

$$|\vec{R}|=\sqrt{9P^2+3P^2}$$

$$|\vec{R}|=\sqrt{12P^2}$$

$$|\vec{R}|=2\sqrt{3}P$$

---

Step 6: Direction of the Resultant Force

The direction of $\vec{R}$ can be found using the angle $\theta$ it makes with the negative x-axis:

$$\tan \theta =\frac{\text{Vertical component}}{\text{Horizontal component}}=\frac{P\sqrt{3}}{-3P}=-\frac{\sqrt{3}}{3}$$

$$\theta ={\tan }^{-1}\left(-\frac{\sqrt{3}}{3}\right)=-30^{\circ }$$

This means the resultant force makes an angle of $30^{\circ }$ with the negative x-axis (or equivalently, $150^{\circ }$ with the positive x-axis).

---

Final Answer

The resultant of the three forces is a force of magnitude $2\sqrt{3}P$ acting at an angle of $150^{\circ }$ with the positive x-axis (or equivalently, $30^{\circ }$ with the negative x-axis).

---