In physics, we describe the world around us using physical quantities. These quantities help us measure and understand things like speed, force, temperature, and position. But not all physical quantities are the same. Some have only a size or amount, while others have both size and direction. This difference is crucial in mechanics, the branch of physics that deals with motion and forces.
For example, consider the temperature of a room. If the temperature is 30°C, it tells us how hot it is, but there is no direction associated with this number. On the other hand, if you walk 5 meters north, your movement has both a distance (5 meters) and a direction (north). This kind of quantity is called a vector.
Understanding the difference between these two types of quantities - scalars and vectors - is fundamental for solving problems in mechanics and many other areas of physics.
Scalar Quantities are physical quantities that have only magnitude (size or amount) but no direction. They are described completely by a single number and a unit.
Vector Quantities are physical quantities that have both magnitude and direction. They are represented by arrows where the length indicates magnitude and the arrowhead shows direction.
Vectors are usually represented by arrows. The length of the arrow corresponds to the vector's magnitude, and the direction of the arrow shows the vector's direction. In writing, vectors are often denoted by boldface letters like or by letters with an arrow above, such as \(\vec{A}\).
To work with vectors mathematically, especially in two or three dimensions, we break them down into components along coordinate axes. In a two-dimensional plane, these are usually the horizontal (x-axis) and vertical (y-axis) directions.
We use unit vectors to indicate direction along these axes:
Any vector \(\vec{A}\) in two dimensions can be written as:
\(\vec{A} = A_x \hat{i} + A_y \hat{j}\)
where \(A_x\) and \(A_y\) are the components of \(\vec{A}\) along the x and y axes respectively.
When dealing with vectors, adding or subtracting them is not as simple as adding or subtracting their magnitudes. Since vectors have direction, both magnitude and direction must be considered.
There are two common graphical methods to add vectors:
Vector subtraction \(\vec{A} - \vec{B}\) can be thought of as adding \(\vec{A}\) and the negative of \(\vec{B}\) (which has the same magnitude as \(\vec{B}\) but opposite direction).
Often, vectors are not aligned along the coordinate axes. To analyze such vectors, we resolve them into components along perpendicular directions, usually x and y axes. This process is called vector resolution.
If a vector \(\vec{A}\) has magnitude \(A\) and makes an angle \(\theta\) with the horizontal axis (x-axis), its components are:
\(A_x = A \cos \theta\),
\(A_y = A \sin \theta\)
This allows us to work with vectors using simple algebra and trigonometry.
Step 1: Represent the two displacements as vectors along x and y axes:
\(\vec{A} = 3 \hat{i}\) km (east direction)
\(\vec{B} = 4 \hat{j}\) km (north direction)
Step 2: Add the vectors component-wise:
\(\vec{R} = \vec{A} + \vec{B} = 3 \hat{i} + 4 \hat{j}\) km
Step 3: Find the magnitude of the resultant vector:
\[ R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ km} \]
Step 4: Find the direction (angle \(\theta\)) with respect to east (x-axis):
\[ \theta = \tan^{-1} \left(\frac{4}{3}\right) \approx 53.13^\circ \text{ north of east} \]
Answer: The resultant displacement is 5 km at \(53.13^\circ\) north of east.
Step 1: Identify the magnitude and angle:
Force \(F = 50\) N, angle \(\theta = 30^\circ\)
Step 2: Calculate horizontal component \(F_x\):
\[ F_x = F \cos \theta = 50 \times \cos 30^\circ = 50 \times 0.866 = 43.3 \text{ N} \]
Step 3: Calculate vertical component \(F_y\):
\[ F_y = F \sin \theta = 50 \times \sin 30^\circ = 50 \times 0.5 = 25 \text{ N} \]
Answer: Horizontal component = 43.3 N, Vertical component = 25 N.
Step 1: Define directions:
Let upstream be positive direction. Then, velocity of boat relative to water, \(\vec{v}_{bw} = +8\) m/s
Velocity of river current relative to ground, \(\vec{v}_{wg} = -5\) m/s (downstream)
Step 2: Velocity of boat relative to ground, \(\vec{v}_{bg} = \vec{v}_{bw} + \vec{v}_{wg}\)
\[ \vec{v}_{bg} = 8 + (-5) = 3 \text{ m/s upstream} \]
Answer: The boat's velocity relative to the ground is 3 m/s upstream.
Step 1: Add the components along each axis:
\[ R_x = 3 + (-2) = 1 \]
\[ R_y = 4 + 5 = 9 \]
Step 2: Write the resultant vector:
\[ \vec{R} = 1\hat{i} + 9\hat{j} \]
Answer: The resultant vector is \(\vec{R} = \hat{i} + 9\hat{j}\).
Step 1: Use the law of cosines to find magnitude \(R\):
\[ R = \sqrt{A^2 + B^2 + 2AB \cos \theta} = \sqrt{7^2 + 10^2 + 2 \times 7 \times 10 \times \cos 60^\circ} \]
Calculate:
\[ R = \sqrt{49 + 100 + 140 \times 0.5} = \sqrt{149 + 70} = \sqrt{219} \approx 14.8 \text{ m} \]
Step 2: Find the angle \(\alpha\) between \(\vec{A}\) and \(\vec{R}\) using the law of sines or cosines:
\[ \cos \alpha = \frac{A^2 + R^2 - B^2}{2AR} = \frac{7^2 + 14.8^2 - 10^2}{2 \times 7 \times 14.8} \]
Calculate numerator:
\[ 49 + 219 - 100 = 168 \]
Calculate denominator:
\[ 2 \times 7 \times 14.8 = 207.2 \]
Therefore:
\[ \cos \alpha = \frac{168}{207.2} \approx 0.81 \]
\[ \alpha = \cos^{-1} 0.81 \approx 36^\circ \]
Answer: The resultant vector has magnitude approximately 14.8 m and makes an angle of \(36^\circ\) with vector \(\vec{A}\).
When to use: When dealing with vector addition or subtraction graphically.
When to use: When adding or subtracting vectors with components.
When to use: When solving trigonometric parts of vector problems.
When to use: When vectors are at angles and need resolution.
When to use: To avoid sign and direction errors in final answers.
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