Problem 7
Question
Consider two displacements, one of magnitude \(3 \mathrm{~m}\) and another of magnitude \(4 \mathrm{~m}\). Show how the displacement vectors may be combined to get a resultant displacement of magnitude (a) \(7 \mathrm{~m}\), (b) \(1 \mathrm{~m}\), and (c) \(5 \mathrm{~m}\).
Step-by-Step Solution
Verified Answer
(a) Same direction, 7 m; (b) Opposite direction, 1 m; (c) Perpendicular, 5 m.
1Step 1: Understanding Vector Addition
Vectors are quantities that have both magnitude and direction. When combining vectors, the resultant vector is often the sum of the individual vectors. The magnitude of two vectors can result from either adding them directly or subtracting one from the other, depending on their directions.
2Step 2: When Vectors are in the Same Direction
To get a resultant displacement of 7 m, both vectors (3 m and 4 m) must point in the same direction. The resultant vector magnitude is calculated by vector addition: \[ 3 \text{ m} + 4 \text{ m} = 7 \text{ m}. \]
3Step 3: When Vectors are in Opposite Directions
To achieve a resultant displacement of 1 m, the two vectors should be pointing in opposite directions. The resultant vector magnitude can be found by subtracting the smaller vector from the larger one: \[ 4 \text{ m} - 3 \text{ m} = 1 \text{ m}. \]
4Step 4: When Vectors Create a Right Triangle
To obtain a resultant vector of 5 m, the two vectors (3 m and 4 m) should form a right-angle triangle. By Pythagoras' theorem, the resultant magnitude is the hypotenuse: \[ \sqrt{(3 \text{ m})^2 + (4 \text{ m})^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ m}. \]
Key Concepts
Resultant DisplacementPythagorean TheoremVector Magnitude
Resultant Displacement
The resultant displacement is the overall effect when combining two or more displacement vectors. It represents the shortest straight-line distance between the starting and ending points after all considered movements. This can be visualized as follows:
- If the displacements are in the same direction, they simply add up to give a greater resultant displacement.
- If they oppose each other, the resultant is smaller and is the difference between them.
- If they are neither equal nor opposite, they follow a more complex interaction, often using geometric methods such as forming triangles.
Pythagorean Theorem
The Pythagorean Theorem is a fundamental concept in geometry that relates the sides of a right-angled triangle. Specifically, it states that for a right triangle:
\[ c^2 = a^2 + b^2 \]
where \( c \) is the hypotenuse, and \( a \) and \( b \) are the other two sides. This theorem is particularly useful in vector mechanics when two vectors form a right angle. By applying this rule, the resultant displacement vector can be calculated easily.
For example, in the exercise, where displacements of 3 m and 4 m form a right triangle, applying the Pythagorean Theorem gives:
\[ \sqrt{(3\, \mathrm{m})^2 + (4\, \mathrm{m})^2} = 5\, \mathrm{m} \]
This shows how vector components can resolve spatial relationships when vectors are not aligned linearly. Understanding and applying the Pythagorean Theorem is critical for solving such vector problems.
\[ c^2 = a^2 + b^2 \]
where \( c \) is the hypotenuse, and \( a \) and \( b \) are the other two sides. This theorem is particularly useful in vector mechanics when two vectors form a right angle. By applying this rule, the resultant displacement vector can be calculated easily.
For example, in the exercise, where displacements of 3 m and 4 m form a right triangle, applying the Pythagorean Theorem gives:
\[ \sqrt{(3\, \mathrm{m})^2 + (4\, \mathrm{m})^2} = 5\, \mathrm{m} \]
This shows how vector components can resolve spatial relationships when vectors are not aligned linearly. Understanding and applying the Pythagorean Theorem is critical for solving such vector problems.
Vector Magnitude
Vector magnitude is the measure of how long a vector is, regardless of its direction. It is calculated by taking the square root of the sum of the squares of its components. Mathematically, for a vector \( \mathbf{v} = (x, y) \), the magnitude \(|\mathbf{v}|\) is:
\[ |\mathbf{v}| = \sqrt{x^2 + y^2} \]
This measure helps quantify vectors in both physical and spatial contexts, assisting in understanding how different vectors combine in physics.
In the provided exercise, consider vectors of magnitudes 3 m and 4 m. These vectors can create different resultant magnitudes depending on their orientation. When aligned perfectly to form a 90-degree angle, as illustrated in the solution, these separate magnitudes create a resultant of 5 m. Calculating the resultant magnitude correctly is crucial for accurate vector analysis and is foundational to mastering physics problems involving vector quantities.
\[ |\mathbf{v}| = \sqrt{x^2 + y^2} \]
This measure helps quantify vectors in both physical and spatial contexts, assisting in understanding how different vectors combine in physics.
In the provided exercise, consider vectors of magnitudes 3 m and 4 m. These vectors can create different resultant magnitudes depending on their orientation. When aligned perfectly to form a 90-degree angle, as illustrated in the solution, these separate magnitudes create a resultant of 5 m. Calculating the resultant magnitude correctly is crucial for accurate vector analysis and is foundational to mastering physics problems involving vector quantities.
Other exercises in this chapter
Problem 4
Express the following angles in radians: (a) \(20.0^{\circ}\), (b) \(50.0^{\circ}\), (c) \(100^{\circ}\). Convert the following angles to degrees: (d) \(0.330\)
View solution Problem 5
A ship sets out to sail to a point \(120 \mathrm{~km}\) due north. An unexpected storm blows the ship to a point \(100 \mathrm{~km}\) due east of its starting p
View solution Problem 8
A person walks in the following pattern: \(3.1 \mathrm{~km}\) north, then \(2.4 \mathrm{~km}\) west, and finally \(5.2 \mathrm{~km}\) south. (a) Sketch the vect
View solution Problem 9
Two vectors are given by and $$\begin{aligned}&\vec{a}=(4.0 \mathrm{~m}) \hat{\mathrm{i}}-(3.0 \mathrm{~m}) \hat{\mathrm{j}}+(1.0 \mathrm{~m}) \hat{\mathrm{k}}
View solution