Vectors in physics notes

In physics, vectors play an essential role in representing quantities that have both magnitude and direction. Many fundamental physical quantities, such as displacement, velocity, force, and momentum, are vectors, and their correct handling is crucial for solving problems in mechanics, electromagnetism, and other areas of physics.

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1. What Are Vectors?

A vector is a mathematical object used to represent quantities that require both a magnitude (how much) and a direction (where). It is typically drawn as an arrow, where:

• Length of the arrow represents the magnitude of the vector.
• Direction of the arrow shows the direction of the vector.

Common examples of vectors in physics include:

• Displacement: How far and in what direction an object moves.
• Velocity: How fast and in what direction an object moves.
• Force: The push or pull exerted on an object, with direction being important for the resulting motion.

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2. Representation of Vectors

Vectors can be represented in multiple ways, each having its significance depending on the context:

a. Graphical Representation

Vectors are commonly represented as arrows in two- or three-dimensional space, with their length and orientation indicating magnitude and direction, respectively.

For example, a displacement vector of 5 meters to the east would be drawn as an arrow pointing right (east) with a length corresponding to 5 units.

b. Component Form

In mathematics, vectors are often broken down into their components along the coordinate axes (e.g., $x$-axis, $y$-axis). A vector in two-dimensional space can be written as:

$$\mathbf{A} = A_x \hat{i} + A_y \hat{j}$$

Where:

• $A_x$ and $A_y$ are the components of the vector along the $x$- and $y$-axes.
• $\hat{i}$ and $\hat{j}$ are unit vectors in the direction of the $x$- and $y$-axes.

In three dimensions, a vector might look like:

$$\mathbf{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$$

Where $\hat{k}$ is the unit vector in the direction of the $z$-axis.

c. Magnitude and Direction Form

Vectors can also be expressed using magnitude and direction (often given as an angle). For a vector $\mathbf{A}$ in two-dimensional space, the magnitude $|\mathbf{A}|$ and the angle $\theta$ can be used to describe it:

$$|\mathbf{A}| = \sqrt{A_x^2 + A_y^2}$$

And the angle $\theta$ (measured counterclockwise from the positive $x$-axis) is given by:

$$\theta = \tan^{-1} \left(\frac{A_y}{A_x}\right)$$

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3. Operations on Vectors

Vectors are essential in physics because they obey specific mathematical rules that make it easy to manipulate them. Some of the primary operations include:

a. Vector Addition

To add two vectors graphically, you can use the tip-to-tail method:

• Place the tail of the second vector at the tip of the first vector.
• The resultant vector (sum) is drawn from the tail of the first vector to the tip of the second vector.

In component form, the sum of two vectors $\mathbf{A}$ and $\mathbf{B}$ is:

$$\mathbf{R} = \mathbf{A} + \mathbf{B} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j}$$

b. Vector Subtraction

Subtracting vectors is similar to addition but involves adding the negative of the vector. Graphically, you reverse the direction of the vector being subtracted.

In component form:

$$\mathbf{R}=\mathbf{A}-\mathbf{B}=(A_x-B_x)\widehat{i}+(A_y-B_y)\hat{j}$$

c. Scalar Multiplication

Multiplying a vector by a scalar changes its magnitude but not its direction. For instance, if $\mathbf{A}$ is a vector and $k$ is a scalar:

$$\mathbf{B} = k \mathbf{A}$$

This operation stretches or shrinks the vector depending on the value of $k$.

d. Dot Product (Scalar Product)

The dot product is a way to multiply two vectors to produce a scalar. It is defined as:

$$\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta$$

Where $\theta$ is the angle between the two vectors. The dot product is useful in calculating quantities like work in physics.

e. Cross Product (Vector Product)

The cross product produces a new vector perpendicular to the plane formed by two vectors. It is given by:

$$\mathbf{A} \times \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \sin \theta \hat{n}$$

Where $\hat{n}$ is a unit vector perpendicular to both $\mathbf{A}$ and $\mathbf{B}$, and $\theta$ is the angle between them. The cross product is commonly used in rotational dynamics and electromagnetism.

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Resolution of Vectors: A Detailed Explanation

In physics, resolution of vectors refers to the process of breaking a single vector into two or more components, typically along perpendicular directions such as the $x$-axis and $y$-axis in two-dimensional space. This process is particularly useful in simplifying vector-related problems by dealing with each component separately.

Understanding Vector Resolution

When resolving a vector, we express it in terms of its horizontal (along the $x$-axis) and vertical (along the $y$-axis) components. This process allows us to work with these components independently, especially in problems involving motion, forces, or fields.

For instance, consider a vector $\mathbf{A}$ with magnitude $A$ and angle $\theta$ relative to the horizontal axis. The vector $\mathbf{A}$ can be resolved into:

• Horizontal component ($A_x$): This is the projection of the vector along the $x$-axis.
• Vertical component ($A_y$): This is the projection of the vector along the $y$-axis.

The components can be found using basic trigonometry:

$$A_x = A \cos \theta$$
$$A_y = A \sin \theta$$

Where:

• $A_x$ is the horizontal component.
• $A_y$ is the vertical component.
• $A$ is the magnitude of the vector.
• $\theta$ is the angle between the vector and the positive $x$-axis.

The original vector can then be written as:

$$\mathbf{A} = A_x \hat{i} + A_y \hat{j}$$

Where $\hat{i}$ and $\hat{j}$ are unit vectors in the directions of the $x$-axis and $y$-axis, respectively.

Applications of Vector Resolution

a. Force Problems

In mechanics, the resolution of forces into horizontal and vertical components is a common technique. For example, if a force is applied at an angle to a surface, the force can be resolved into a component acting along the surface (horizontal) and a component acting perpendicular to the surface (vertical).

Example: A force $\mathbf{F}$ of magnitude 50 N is applied at an angle of 30° to the horizontal. The horizontal and vertical components are:

$$F_x = 50 \cos 30^\circ = 43.3 \, \text{N}$$
$$F_y = 50 \sin 30^\circ = 25 \, \text{N}$$

Thus, the force can be resolved into a horizontal force of 43.3 N and a vertical force of 25 N.

b. Projectile Motion

In projectile motion, the initial velocity of the object is often given at an angle to the horizontal. To analyze the motion, the velocity is resolved into horizontal and vertical components:

• The horizontal component determines the motion along the $x$-axis, typically with constant velocity (ignoring air resistance).
• The vertical component determines the motion along the $y$-axis, where the object is subject to acceleration due to gravity.

c. Electric and Magnetic Fields

When dealing with electric and magnetic fields, field vectors are often resolved into components to simplify calculations. For example, an electric field vector at an angle can be split into perpendicular components that can be analyzed independently.

How to Resolve Vectors

Here’s a step-by-step guide to resolving vectors:

1. Identify the vector and the angle: Begin by identifying the vector's magnitude and the angle it makes with a reference axis (usually the $x$-axis or horizontal).

2. Use trigonometric functions: To resolve the vector, apply trigonometric functions (sine and cosine) to calculate the components along each axis:

   • The cosine of the angle gives the component along the adjacent side (horizontal or $x$-axis).
   • The sine of the angle gives the component along the opposite side (vertical or $y$-axis).

3. Write the vector in component form: Once the components are calculated, express the vector in terms of its horizontal and vertical components.

4. Check the direction: Ensure that the signs of the components match the vector's direction in the coordinate system (positive or negative $x$, positive or negative $y$).

Example Problem: Resolving a Force Vector

Problem 1: A force of 100 N is applied at an angle of 45° to the horizontal. Resolve the force into its horizontal and vertical components.

Solution:

1. The magnitude of the force is $F=100\text{N.}$
2. The angle with the horizontal is $\theta = 45^\circ$.

Using trigonometric functions:

$$F_x = 100 \cos 45^\circ = 100 \times 0.707 = 70.7 \, \text{N}$$
$$F_y = 100 \sin 45^\circ = 100 \times 0.707 = 70.7 \, \text{N}$$

Thus, the force has a horizontal component of 70.7 N and a vertical component of 70.7 N.

Problem 2: A force of 100 N is applied at an angle of 30° to the horizontal. We can calculate the components as follows:

$$F_x = 100 \cos 30^\circ = 86.6 \, \text{N}$$
$$F_y = 100 \sin 30^\circ = 50 \, \text{N}$$

The diagram below illustrates the resolution of this force:

Problem 3: A force of $50 \, \text{N}$ is acting along a certain direction, with a vertical component of $32 \, \text{N}$ directed due north.

Calculate:

(a) The angle of inclination of the force with respect to the horizontal.

(b) The horizontal component of the force.

Solution

Given:

• Magnitude of the force ($F$) = $50\text{N}$
• Vertical component ($F_y$) = $32\text{N}$

To find:

1. The angle of inclination ($\theta$)
2. The horizontal component ($F_x$)

(a) Calculate the angle of inclination ($\theta$)

Using the relationship between the vertical component, the total force, and the angle:

$$F_y = F \sin \theta$$

Substituting the known values:

$$32 \, \text{N} = 50 \, \text{N} \cdot \sin \theta$$

To find $\sin \theta$:

$$\sin \theta = \frac{32}{50} = 0.64$$

Now, calculate $\theta$:

$$\theta = \sin^{-1}(0.64)$$

Calculating $\theta$:

$$\theta \approx 39.81^\circ$$

(b) Calculate the horizontal component ($F_{}$$x_{}$​)

Using the relationship between the horizontal component and the angle:

$$F_x = F \cos \theta$$

First, we need to calculate $\cos \theta$:

$$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - (0.64)^2}$$
$$\cos \theta = \sqrt{1 - 0.4096} = \sqrt{0.5904} \approx 0.768$$

Now, calculate $F_x$:

$$F_x = 50 \, \text{N} \cdot 0.768 \approx 38.4 \, \text{N}$$

Final Answers

1. (a) The angle of inclination ($\theta$) is approximately $39.81^\circ$.
2. (b) The horizontal component of the force ($F_x$) is approximately $38.4 \, \text{N}$.

Problem 4: An impulse of $25 \, \text{Ns}$ is exerted at an angle of $20^\circ$ with the horizontal.

Calculate:

(a) The horizontal component of the impulse.
(b) The vertical component of the impulse.

Solution

Given:

• Magnitude of the impulse ($I$) = $25 \, \text{Ns}$
  
• Angle ($\theta$) = $20^\circ$
  

To find:

1. The horizontal component of the impulse ($I_x$)
2. The vertical component of the impulse ($I_y$)

(a) Calculate the horizontal component ($I_{}$$x_{}$)

Using the cosine function for the horizontal component:

$$I_x = I \cos \theta$$

Substituting the known values:

$$I_x = 25 \, \text{Ns} \cdot \cos(20^\circ)$$

Calculating $I_x$:

$$I_x \approx 25 \, \text{Ns} \cdot 0.9397 \approx 23.49 \, \text{Ns}$$

(b) Calculate the vertical component ($I_{}$$y_{}$​)

Using the sine function for the vertical component:

$$I_y = I \sin \theta$$

Substituting the known values:

$$I_y = 25 \, \text{Ns} \cdot \sin(20^\circ)$$

Calculating $I_y$:

$$I_y \approx 25 \, \text{Ns} \cdot 0.3420 \approx 8.55 \, \text{Ns}$$

Final Answers

(a) The horizontal component of the impulse ($I_x$) is approximately $23.49\text{Ns}$.
(b) The vertical component of the impulse ($I_y$) is approximately $8.55\text{Ns}$.

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Addition of Vectors

There are two primary methods for adding vectors: graphical addition and analytical addition.

a. Graphical Addition of Vectors

Graphical addition involves visually combining vectors using a diagram. The most common methods are the tip-to-tail method and the parallelogram method.

Tip-to-Tail Method:

1. Draw the First Vector: Start by drawing the first vector, $\mathbf{A}$, as an arrow.
2. Place the Second Vector: Take the second vector, $\mathbf{B}$, and position its tail at the tip of the first vector.
3. Draw the Resultant Vector: The resultant vector $\mathbf{R}$ is drawn from the tail of the first vector to the tip of the second vector.

Example:

Suppose we have two vectors:

• $\mathbf{A} = 5$ units to the east
• $\mathbf{B} = 3$ units to the north

To add these vectors graphically:

1. Draw vector $\mathbf{A}$ pointing right (east).
2. Place vector $\mathbf{B}$ starting at the tip of vector $\mathbf{A}$ and point it upwards (north).
3. The resultant vector $\mathbf{R}$ is drawn from the starting point of $\mathbf{A}$ to the tip of $\mathbf{B}$.

Parallelogram Method:

1. Draw the Vectors: Draw both vectors $\mathbf{A}$ and $\mathbf{B}$ starting from the same point.
2. Complete the Parallelogram: Draw a line parallel to $\mathbf{A}$ from the tip of $\mathbf{B}$ and a line parallel to $\mathbf{B}$ from the tip of $\mathbf{A}$ to form a parallelogram.
3. Draw the Resultant Vector: The diagonal of the parallelogram that starts from the common point represents the resultant vector $\mathbf{R}$.

Example of the Parallelogram Method

Let’s consider two vectors:

• Vector A: 5 units to the east
• Vector B: 3 units to the north

Steps:

1. Draw Vector A: Start at point O and draw a horizontal arrow 5 units to the right (east).

2. Draw Vector B: From point O, draw a vertical arrow 3 units up (north).

3. Complete the Parallelogram:

   • From the tip of vector A, draw a line parallel to vector B (upwards).
   • From the tip of vector B, draw a line parallel to vector A (to the right).

4. Draw the Resultant Vector:

   • Draw a diagonal from point O (the tail of both vectors) to the point where the two parallel lines intersect. This diagonal is your resultant vector $\mathbf{R}$.

5. Label the Resultant: The resultant vector $\mathbf{R}$ can be computed as the combination of both vectors using trigonometric functions if needed.

Mathematical Representation

If the vectors $\mathbf{A}$ and $\mathbf{B}$ have an angle $\theta$ between them, the magnitude of the resultant vector $\mathbf{R}$ can be calculated using the law of cosines:

$$|\mathbf{R}| = \sqrt{|\mathbf{A}|^2 + |\mathbf{B}|^2 + 2 |\mathbf{A}| |\mathbf{B}| \cos(\theta)}$$

b. Analytical Addition of Vectors

Analytical addition involves breaking down vectors into their components and using algebra to combine them.

1. Identify the Components: A vector in two dimensions can be represented in terms of its components along the $x$- and $y$-axes:

   • $\mathbf{A} = A_x \hat{i} + A_y \hat{j}$
   • $\mathbf{B} = B_x \hat{i} + B_y \hat{j}$

2. Add the Components: The resultant vector $\mathbf{R}$ can be found by adding the corresponding components:

   $$\mathbf{R} = \mathbf{A} + \mathbf{B} = (A_x + B_x) \hat{i} + (A_y + B_y) \hat{j}$$

3. Magnitude and Direction: The magnitude of the resultant vector can be calculated using the Pythagorean theorem:

   $$|\mathbf{R}| = \sqrt{(A_x + B_x)^2 + (A_y + B_y)^2}$$
   

   The direction can be found using the arctangent function:

   $$\theta ={\tan }^{-1}\left(\frac{R_y}{R_x}\right)$$

Example:

Given:

• $\mathbf{A} = 4 \hat{i} + 3 \hat{j}$ (4 units right, 3 units up)
• $\mathbf{B} = 2 \hat{i} - 5 \hat{j}$ (2 units right, 5 units down)

To add them analytically:

1. Find the components:

   $$R_x = 4 + 2 = 6$$
   $$R_y=3-5=-2$$

2. The resultant vector is:

   $$\mathbf{R} = 6 \hat{i} - 2 \hat{j}$$

3. Magnitude:

   $$|\mathbf{R}| = \sqrt{(6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} \approx 6.32$$
   

4. Direction:

   $$\theta = \tan^{-1} \left(\frac{-2}{6}\right) \approx -18.43^\circ \text{ (below the x-axis)}$$
   

Properties of Vector Addition

• Commutative Property: The order in which vectors are added does not matter:

  $$\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}$$
  

• Associative Property: The grouping of vectors does not affect the result:

  $$\mathbf{A} + (\mathbf{B} + \mathbf{C}) = (\mathbf{A} + \mathbf{B}) + \mathbf{C}$$
  

• Zero Vector: Adding a zero vector to any vector does not change the original vector:

  $$\mathbf{A} + \mathbf{0} = \mathbf{A}$$

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4. Applications of Vectors in Physics

Vectors are used extensively in various fields of physics to represent quantities that have both magnitude and direction:

a. Displacement, Velocity, and Acceleration

In kinematics, displacement ($\mathbf{d}$), velocity ($\mathbf{v}$), and acceleration ($\mathbf{a}$) are all vector quantities. For example, an object's motion is described by the vector $\mathbf{v}$, which tells us both how fast it is moving and in what direction.

b. Force and Momentum

In mechanics, both force ($\mathbf{F}$) and momentum ($\mathbf{p}$) are vectors. Newton’s Second Law, for instance, is often written in vector form:

$$\mathbf{F} = m \mathbf{a}$$

This tells us that the direction of the acceleration vector $\mathbf{a}$ will always be the same as the force vector $\mathbf{F}$.

c. Electric and Magnetic Fields

In electromagnetism, the electric field ($\mathbf{E}$) and the magnetic field ($\mathbf{B}$) are vectors that describe the forces exerted by these fields on charges or currents.

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5. Vector Decomposition

In many physics problems, vectors need to be broken down into their components along mutually perpendicular axes. This process is called vector decomposition.

For example, a force vector acting at an angle can be broken down into its horizontal ($F_x$) and vertical ($F_y$) components:

$$F_x = F \cos \theta$$
$$F_y = F \sin \theta$$

Where $\theta$ is the angle between the vector and the horizontal axis. This decomposition makes it easier to apply equations to each direction separately.

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6. Conclusion

Vectors are fundamental in physics because they provide a complete description of quantities that have both magnitude and direction. Understanding how to manipulate and operate with vectors is essential for solving a wide range of problems in mechanics, electromagnetism, and many other areas of physics.

For further reading and practice, you may visit resources like:

• Physics Classroom - Vectors and Scalars
• Khan Academy - Introduction to Vectors

By mastering vectors, you can significantly enhance your ability to tackle complex physical problems in a systematic way.