The Resultant Vector: Finding the Single Combined Force
What Are Vectors and Their Components?
To understand the resultant vector, we must first understand what a vector is. A vector is a quantity that has both magnitude (size) and direction. This is different from a scalar, which only has magnitude. Speed is a scalar (e.g., 60 km/h), while velocity is a vector (e.g., 60 km/h north). Force, displacement, and acceleration are other common examples of vectors.
Vectors are often represented by arrows. The length of the arrow represents the magnitude, and the direction the arrow points indicates the direction of the vector.
A vector in a two-dimensional plane can be broken down into its horizontal (x) and vertical (y) components. If a vector $\vec{A}$ has a magnitude $A$ and makes an angle $\theta$ with the x-axis, its components are:
$A_x = A \cos \theta$
$A_y = A \sin \theta$
Methods for Finding the Resultant Vector
There are two primary ways to find the resultant of two or more vectors: the graphical method and the mathematical method. The graphical method is great for visualization, while the mathematical method gives a precise answer.
The Head-to-Tail Graphical Method
This is a simple way to find the resultant vector by drawing.
- Draw the first vector to scale.
- Place the tail of the second vector at the head of the first vector.
- Continue this process for all remaining vectors.
- The resultant vector is the arrow drawn from the tail of the first vector to the head of the last vector.
Imagine you walk 4 m east, then 3 m north. By drawing these vectors head-to-tail, the resultant vector is the straight line from your starting point to your ending point.
Mathematical Calculation Using Components
For a more accurate result, we use mathematics. The process involves these steps:
- Resolve all vectors into their x and y components.
- Sum all the x-components to find the total x-component of the resultant ($R_x$).
- Sum all the y-components to find the total y-component of the resultant ($R_y$).
- Find the magnitude of the resultant vector $R$ using the Pythagorean theorem: $R = \sqrt{R_x^2 + R_y^2}$.
- Find the direction of the resultant vector using trigonometry: $\theta = \tan^{-1}(R_y / R_x)$, where $\theta$ is the angle the resultant makes with the x-axis.
| Method | Description | Best Used For |
|---|---|---|
| Head-to-Tail | Drawing vectors sequentially and connecting the start to the end. | Visualization, quick estimates, understanding the concept. |
| Component Method | Breaking vectors into x and y parts and adding them mathematically. | Precise calculations, problems with many vectors, any angle. |
| Pythagorean Theorem | Calculating magnitude when vectors are at right angles. | Simple problems with two perpendicular vectors. |
Applying the Resultant Vector to Real-World Problems
The concept of a resultant vector is not just theoretical; it is used to solve real-world problems in physics and engineering. Let's look at two detailed examples.
Example 1: The Boat Crossing a River
Imagine a boat that can travel at 5 m/s in still water. It needs to cross a river that is flowing east at 3 m/s. If the captain points the boat directly north, what will its actual path and speed be?
- Vector 1: Boat's Velocity ($\vec{V_b}$) = 5 m/s North. Components: $V_{bx} = 0$, $V_{by} = 5$.
- Vector 2: River's Velocity ($\vec{V_r}$) = 3 m/s East. Components: $V_{rx} = 3$, $V_{ry} = 0$.
The resultant velocity ($\vec{R}$) is the sum of these two vectors.
- Find the components of the resultant: $R_x = V_{bx} + V_{rx} = 0 + 3 = 3$
$R_y = V_{by} + V_{ry} = 5 + 0 = 5$ - Find the magnitude (the boat's actual speed): $R = \sqrt{R_x^2 + R_y^2} = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} \approx 5.83 \text{ m/s}$
- Find the direction (the boat's actual path): $\theta = \tan^{-1}(R_y / R_x) = \tan^{-1}(5 / 3) \approx \tan^{-1}(1.667) \approx 59^\circ$
So, the boat will not reach the point directly opposite its starting point. Instead, it will move with a speed of about 5.83 m/s at an angle of 59° north of east.
Example 2: Tugboats Pulling a Ship
Two tugboats are pulling a large ship. Tugboat A pulls with a force of 3000 N at an angle 30° north of east. Tugboat B pulls with a force of 4000 N at an angle 45° south of east. What is the resultant force on the ship?
First, we find the x and y components of each force vector.
| Tugboat | Force (N) | X-Component (N) | Y-Component (N) |
|---|---|---|---|
| A | 3000 at 30° | $3000 \cos 30^\circ \approx 2598$ | $3000 \sin 30^\circ = 1500$ |
| B | 4000 at -45° | $4000 \cos 45^\circ \approx 2828$ | $4000 \sin (-45^\circ) \approx -2828$ |
| Resultant (R) | $R_x = 2598 + 2828 = 5426$ | $R_y = 1500 + (-2828) = -1328$ |
Now, we calculate the magnitude and direction of the resultant force.
- Magnitude: $R = \sqrt{(5426)^2 + (-1328)^2} \approx \sqrt{29441476 + 1763584} \approx \sqrt{31205060} \approx 5586 \text{ N}$
- Direction: $\theta = \tan^{-1}(R_y / R_x) = \tan^{-1}(-1328 / 5426) \approx \tan^{-1}(-0.245) \approx -13.8^\circ$
The resultant force on the ship is about 5586 N at an angle of 13.8° south of east. This single force would have the same effect on the ship's motion as the two individual tugboats pulling together.
Common Mistakes and Important Questions
Q: I added the magnitudes of two vectors, but my answer is wrong. Why?
Q: When using the component method, how do I know if my y-component is positive or negative?
A: You need to define a coordinate system. Conventionally:
- Right and Up are positive.
- Left and Down are negative.
So, a vector pointing north has a positive y-component, while a vector pointing south has a negative y-component. A vector pointing east has a positive x-component, while one pointing west has a negative x-component.
Q: Can the resultant vector ever be zero?
Footnote
[1] Equilibrium: A state where the net force on an object is zero, resulting in no change in its motion. The object is either at rest or moving with a constant velocity.