Vectors in Mechanics: Displacement, Velocity, and Force

Problem Statement

In one-dimensional mechanics, we can denote direction using simple positive and negative signs. However, as soon as a particle moves in two or three dimensions, we need a more robust mathematical tool: vectors. Vectors natively encode both magnitude (how much/how fast) and direction (where).

In HSC Mechanics, vectors are heavily used to represent displacement ($\mathbf{r}$), velocity ($\mathbf{v}$), and acceleration ($\mathbf{a}$), as well as to resolve multiple forces acting on a single body. Vector calculus allows us to model complex trajectories, like projectile motion, by treating the horizontal ($\mathbf{i}$) and vertical ($\mathbf{j}$) components independently.

Consider a particle of mass $m=2$ kg subjected to three concurrent forces (in Newtons) acting in a 2D plane:

• $\mathbf{F_1} = 3\mathbf{i} + 4\mathbf{j}$
• $\mathbf{F_2} = -5\mathbf{i} + 2\mathbf{j}$
• $\mathbf{F_3} = 8\mathbf{i} - 10\mathbf{j}$

The particle starts from rest at the origin $O$ at time $t=0$.

(a) Find the resultant force $\mathbf{F_{net}}$ acting on the particle, and calculate its magnitude.

(b) Find the vector equation for the acceleration $\mathbf{a}$ of the particle.

(c) Using vector integration, find the position vector $\mathbf{r}(t)$ of the particle at time $t$.

(d) Find the distance of the particle from the origin after 5 seconds.

Hints

• Part (a): The resultant force is simply the vector sum of all individual forces. Add the $\mathbf{i}$ components together and the $\mathbf{j}$ components together. The magnitude of a vector $x\mathbf{i} + y\mathbf{j}$ is $\sqrt{x^2 + y^2}$.
• Part (b): Apply Newton's Second Law in vector form: $\mathbf{F_{net}} = m\mathbf{a}$. Divide your resultant force vector by the mass $m$.
• Part (c): Acceleration is the derivative of velocity ($\mathbf{a} = \frac{d\mathbf{v}}{dt}$), and velocity is the derivative of displacement ($\mathbf{v} = \frac{d\mathbf{r}}{dt}$). Integrate the acceleration vector twice with respect to time. Remember to include vector constants of integration ($\mathbf{C_1}, \mathbf{C_2}$) and use the initial conditions to find them.
• Part (d): Substitute $t=5$ into your position vector $\mathbf{r}(t)$. The distance from the origin is the magnitude of this position vector, $|\mathbf{r}(5)|$.

Solutions

Part (a): Resultant Force and Magnitude

1. The net force $\mathbf{F_{net}}$ is the sum of the three forces: $\mathbf{F_{net}} = \mathbf{F_1} + \mathbf{F_2} + \mathbf{F_3}$ $\mathbf{F_{net}} = (3\mathbf{i} + 4\mathbf{j}) + (-5\mathbf{i} + 2\mathbf{j}) + (8\mathbf{i} - 10\mathbf{j})$
2. Group the $\mathbf{i}$ and $\mathbf{j}$ components: $\mathbf{F_{net}} = (3 - 5 + 8)\mathbf{i} + (4 + 2 - 10)\mathbf{j}$ $\mathbf{F_{net}} = 6\mathbf{i} - 4\mathbf{j} \text{ Newtons}$
3. The magnitude of the resultant force is: $|\mathbf{F_{net}}| = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \text{ Newtons}$

Part (b): Acceleration Vector

1. Using Newton's Second Law in vector form: $\mathbf{F_{net}} = m\mathbf{a}$.
2. Substitute the known values ($m=2$ kg and $\mathbf{F_{net}} = 6\mathbf{i} - 4\mathbf{j}$): $6\mathbf{i} - 4\mathbf{j} = 2\mathbf{a}$
3. Divide by 2: $\mathbf{a} = 3\mathbf{i} - 2\mathbf{j} \text{ ms}^{-2}$

Part (c): Position Vector via Integration

1. We know that $\mathbf{a} = \frac{d\mathbf{v}}{dt}$. So, we integrate acceleration to find velocity: $\mathbf{v}(t) = \int (3\mathbf{i} - 2\mathbf{j}) \, dt$ $\mathbf{v}(t) = 3t\mathbf{i} - 2t\mathbf{j} + \mathbf{C_1}$
2. The particle starts from rest, so at $t=0$, $\mathbf{v} = 0\mathbf{i} + 0\mathbf{j}$. $0\mathbf{i} + 0\mathbf{j} = 0\mathbf{i} - 0\mathbf{j} + \mathbf{C_1} \implies \mathbf{C_1} = \mathbf{0}$ So, $\mathbf{v}(t) = 3t\mathbf{i} - 2t\mathbf{j}$.
3. We know that $\mathbf{v} = \frac{d\mathbf{r}}{dt}$. Integrate velocity to find displacement: $\mathbf{r}(t) = \int (3t\mathbf{i} - 2t\mathbf{j}) \, dt$ $\mathbf{r}(t) = \frac{3}{2}t^2\mathbf{i} - t^2\mathbf{j} + \mathbf{C_2}$
4. The particle starts at the origin, so at $t=0$, $\mathbf{r} = 0\mathbf{i} + 0\mathbf{j}$. $0\mathbf{i} + 0\mathbf{j} = 0\mathbf{i} - 0\mathbf{j} + \mathbf{C_2} \implies \mathbf{C_2} = \mathbf{0}$ Therefore, the position vector is: $\mathbf{r}(t) = \frac{3}{2}t^2\mathbf{i} - t^2\mathbf{j}$

Part (d): Distance after 5 Seconds

1. Substitute $t=5$ into the position vector: $\mathbf{r}(5) = \frac{3}{2}(5^2)\mathbf{i} - (5^2)\mathbf{j}$ $\mathbf{r}(5) = \frac{75}{2}\mathbf{i} - 25\mathbf{j}$ $\mathbf{r}(5) = 37.5\mathbf{i} - 25\mathbf{j}$
2. The distance from the origin is the magnitude of this vector: $|\mathbf{r}(5)| = \sqrt{(37.5)^2 + (-25)^2}$ $|\mathbf{r}(5)| = \sqrt{1406.25 + 625} = \sqrt{2031.25}$ $|\mathbf{r}(5)| \approx 45.07 \text{ metres}$

Takeaways

• Independence of Axes: The greatest strength of vector mechanics is that it allows us to treat horizontal (i) and vertical (j) motion completely independently when integrating or differentiating.
• Vector Calculus: The fundamental relationships $\mathbf{a} = \frac{d\mathbf{v}}{dt}$ and $\mathbf{v} = \frac{d\mathbf{r}}{dt}$ hold true in vector form. Always remember to include vector constants of integration.
• Resultant Forces: Adding multiple forces is vastly simplified using column vectors or $\mathbf{i}, \mathbf{j}$ notation compared to drawing complex trigonometric force polygons.

Further Readings

• HSC Vectors: https://vumaths.com/booklets/hsc-vectors/
• HSC Mechanics: https://vumaths.com/booklets/hsc-mechanics/
• HSC Integrals: https://vumaths.com/booklets/hsc-integrals/

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