Problem 15 - Applied Mathematics

Question

A plane lamina has perpendicular axes \(O x\) and \(O y\) marked on it, and is acted upon by the following forces: \(5 P\) in the direction \(O y\), \(4 P\) in the direction \(O x\), \(6 P\) in the direction OA where \(A\) is the point \((3 a, 4 a\) ), \(8 P\) in the direction \(\mathrm{AB}\) where \(\mathrm{B}\) is the point \((-a, a)\). Express each force in the form \(p \mathbf{i}+q \mathbf{j}\) and hence calculate the magnitude and direction of the resultant of these forces.

Step-by-Step Solution

Answer

The resultant force vector is \( (7.6P - 4\sqrt{2}P) \mathbf{i} + (9.8P + 4\sqrt{2}P) \mathbf{j} \).

Step 1: Set Up Forces in Vector Form

Express each force in terms of its vector components: 1. Force of magnitude 5P in direction Oy: Vector form: \[5P \, \mathbf{j}\] 2. Force of magnitude 4P in direction Ox: Vector form: \[4P \, \mathbf{i}\]

Step 2: Determine the Vector for Force 6P (OA)

First, find the unit vector in the direction of OA where A is (3a, 4a). Calculate the magnitude of OA: \[\| \mathbf{OA} \| = \sqrt{(3a)^2 + (4a)^2} = 5a\]Unit vector in the direction of OA: \[\hat{u}_{OA} = \left( \frac{3a}{5a}, \frac{4a}{5a} \right) = (0.6, 0.8)\]Force of magnitude 6P in direction OA: Vector form: \[6P \cdot (0.6 \mathbf{i} + 0.8 \mathbf{j}) = 3.6P \mathbf{i} + 4.8P \mathbf{j}\]

Step 3: Determine the Vector for Force 8P (AB)

First, find the unit vector in the direction of AB where B is (-a, a) and A is (0, 0). Calculate the magnitude of AB: \[\| \mathbf{AB} \| = \sqrt{(-a - 0)^2 + (a - 0)^2} = \sqrt{2a^2} = a \sqrt{2}\]Unit vector in the direction of AB: \[\hat{u}_{AB} = \left( \frac{-a}{a\sqrt{2}}, \frac{a}{a\sqrt{2}} \right) = \left( \frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)\]Force of magnitude 8P in the direction AB: Vector form: \[8P \cdot \left( \frac{-1}{\sqrt{2}} \mathbf{i} + \frac{1}{\sqrt{2}} \mathbf{j} \right) = -4\sqrt{2}P \mathbf{i} + 4\sqrt{2}P \mathbf{j} \]

Step 4: Sum the Vectors to Find Resultant Force

Combine all the forces in vector form: \[\mathbf{R} = (4P \mathbf{i} + 3.6P \mathbf{i} - 4\sqrt{2} P \mathbf{i}) + (5P \mathbf{j} + 4.8P \mathbf{j} + 4\sqrt{2}P \mathbf{j}) \]Combine like terms:\[\mathbf{R} = (7.6P - 4\sqrt{2}P) \mathbf{i} + (9.8P + 4\sqrt{2}P) \mathbf{j} \]

Step 5: Calculate the Magnitude of the Resultant Force

Find the magnitude of the resultant force vector: e\[\| \mathbf{R} \| = \sqrt{(7.6P - 4\sqrt{2}P)^2 + (9.8P + 4\sqrt{2}P)^2}\] Simplify the magnitudes inside the square root.

Step 6: Find the Direction of the Resultant Force

Determine the angle θ that the resultant force makes with the positive x-axis: \[\theta = \tan^{-1} \left( \frac{R_y}{R_x} \right)\] Substitute the values found in the previous step to calculate θ.

Key Concepts

vector addition

In physics, vector addition is a fundamental way to combine forces. Each force is represented as a vector, which has both magnitude and direction. When we add vectors, we combine their corresponding components to find a resultant vector. For instance, if we have two forces, represented as vectors \( \textbf{A} = a_x \textbf{i} + a_y \textbf{j} \) and \( \textbf{B} = b_x \textbf{i} + b_y \textbf{j} \), their sum is \( \textbf{A} + \textbf{B} = (a_x + b_x) \textbf{i} + (a_y + b_y) \textbf{j} \). This simplification enables us to handle multiple forces acting in different directions easily.

resultant force

The resultant force is the single vector that has the same effect as all the individual forces acting together. To find the resultant force, we add all the vector forces component-wise. For example, if we sum the forces from the problem: \( \textbf{R} = (4P \textbf{i} + 3.6P \textbf{i} - 4\textbackslash{}sqrt{2} P \textbf{i}) + (5P \textbf{j} + 4.8P \textbf{j} + 4\textbackslash{}sqrt{2}P \textbf{j}) \), we get \( \textbf{R} = (7.6P - 4\textbackslash{}sqrt{2}P) \textbf{i} + (9.8P + 4\textbackslash{}sqrt{2}P) \textbf{j} \). This resultant vector can then be used to find the overall effect, including its magnitude and direction.

unit vector

A unit vector has a magnitude of 1 and indicates direction. To find a unit vector from any given vector, we divide the vector by its magnitude. For example, for OA in the problem, the unit vector is given by \( \textbackslash{}hat{u}\textsubscript{OA} = \textbackslash{}left( \frac{3a}{5a}, \frac{4a}{5a} \textbackslash{}right) = (0.6, 0.8) \). This unit vector shows the direction of OA without changing its original orientation, allowing us to apply scalars to it and find vectors in the same direction.

magnitude and direction of a vector

To fully understand a vector, we need to know its magnitude and direction. The magnitude of a vector \( \textbf{R} = (R_x \textbf{i} + R_y \textbf{j}) \) is calculated using the formula \( \textbackslash{}left| \textbf{R} \textbackslash{}right| = \textbackslash{}sqrt{R_x^2 + R_y^2} \). Direction is found using trigonometry. The angle \(\textbackslash{}theta\) that the vector makes with the x-axis can be found using \( \textbackslash{}theta = \tan^{-1} \textbackslash{}left( \frac{R_y}{R_x} \textbackslash{}right) \). These metrics give us a clear picture of the vector's effect.

trigonometry in physics

Trigonometry plays a vital role in vector analysis in physics. For instance, calculating the components of a force vector often involves trigonometric functions. In our problem, to find the direction of the force OA, we use sine and cosine. Let \( \theta \) be the angle the force makes with the x-axis, then \( F_x = F \textbackslash{}cos(\theta), \) and \( F_y = F \textbackslash{}sin(\theta) \). Trigonometry helps convert between component form and magnitude-direction form of vectors, making it a powerful tool in solving physics problems.