Problem 48 - Contemporary Precalculus

Question

An object at the origin is acted upon by two forces, \(u\) and \(v,\) with direction angle \(\theta_{u}\) and \(\theta_{w}\) respectively. Find the direction and magnitude of the resultant force. $$\mathbf{u}=6 \text { pounds }, \theta_{\mathbf{u}}=45^{\circ} ; \mathbf{v}=6 \text { pounds }, \theta_{\mathbf{v}}=120^{\circ}$$

Step-by-Step Solution

Answer

Question: Determine the direction and the magnitude of the resultant force, given that the horizontal force is 6 pounds at 45 degrees and the vertical force is 6 pounds at an angle of 120 degrees. Answer: The direction and magnitude of the resultant force are approximately 75.26 degrees and 9.90 pounds.

Step 1: Write the forces in component form

Given the magnitude and direction of the forces, we can write the forces in component form as: $$\mathbf{u}= u_x\mathbf{\hat{i}} + u_y\mathbf{\hat{j}}$$ $$\mathbf{v}= v_x\mathbf{\hat{i}} + v_y\mathbf{\hat{j}}$$ We will find the components \(u_x, u_y, v_x\) and \(v_y\) next.

Step 2: Calculate the components of the forces

To find the components of the forces, multiply the magnitudes by the cosine of the angle for the x component and the sine of the angle for the y component. $$u_x = 6\cos(45^{\circ}) = 6\frac{\sqrt{2}}{2} = 3\sqrt{2}$$ $$u_y = 6\sin(45^{\circ}) = 6\frac{\sqrt{2}}{2} = 3\sqrt{2}$$ $$v_x = 6\cos(120^{\circ}) = -3$$ $$v_y = 6\sin(120^{\circ}) = 6\frac{\sqrt{3}}{2} = 3\sqrt{3}$$ Now we have: $$\mathbf{u} = 3\sqrt{2}\mathbf{\hat{i}} + 3\sqrt{2}\mathbf{\hat{j}}$$ $$\mathbf{v} = -3\mathbf{\hat{i}} + 3\sqrt{3}\mathbf{\hat{j}}$$

Step 3: Find the sum of the components

To find the sum of the components, we add the x components together and the y components together. $$R_x = u_x + v_x = 3\sqrt{2} - 3$$ $$R_y = u_y + v_y = 3\sqrt{2} + 3\sqrt{3}$$ So the resultant force is: $$\mathbf{R} = (3\sqrt{2} - 3)\mathbf{\hat{i}} + (3\sqrt{2} + 3\sqrt{3})\mathbf{\hat{j}}$$

Step 4: Calculate the magnitude of the resultant force

We can find the magnitude of the resultant force using the Pythagorean theorem: $$\|\mathbf{R}\| = \sqrt{R_x^2 + R_y^2} = \sqrt{(3\sqrt{2} - 3)^2 + (3\sqrt{2} + 3\sqrt{3})^2} \approx 9.90\:\text{pounds}$$

Step 5: Find the angle of the resultant force

We can find the angle of the resultant force using the arctangent function: $$\theta_{\mathbf{R}} = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{3\sqrt{2} + 3\sqrt{3}}{3\sqrt{2} - 3}\right) \approx 75.26^{\circ}$$ So, the direction and magnitude of the resultant force are approximately \(\boxed{75.26^{\circ}}\) and \(\boxed{9.90\:\text{pounds}}\).

Key Concepts

Resultant Force

When multiple forces act on an object, they can be combined into one single force known as the resultant force. This resultant force gives us an idea of the overall effect of the multiple individual forces. It's like the net influence those forces have on the object. To determine the resultant force, we use vector addition, which means summing up all the vector components of the forces involved.

The process to find the resultant force typically involves breaking each force into its components, adding these components together, and then using those sums to find the magnitude and direction of the resultant. By doing this, you can replace a complex system of forces with just one force that has the same effect.

Vector Components

Vectors break down into components, which are essential for computations in physics and engineering. Each force vector can be divided into a horizontal (x-component) and a vertical (y-component) based on its direction angle. This concept is crucial for vector addition, especially when forces don't act along the same line.

To calculate the components of a vector:

• The x-component is found by multiplying the vector's magnitude by the cosine of its angle, noted as: \( u_x = u \cdot \cos(\theta) \).
• The y-component is determined by multiplying the magnitude by the sine of its angle, noted as: \( u_y = u \cdot \sin(\theta) \).

This method allows you to transform vectors into a common frame of reference, making it easier to add or subtract them.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, play a pivotal role in determining vector components. They help in converting the magnitude and direction of forces into usable components for calculations.

When dealing with vectors:

• Cosine relates the adjacent side of an angle in a right triangle to the hypotenuse, perfect for finding the x-component.
• Sine relates the opposite side to the hypotenuse, ideal for determining the y-component.

These functions simplify the process of translating real-world forces into a mathematical form that can be easily manipulated to find the resultant force.

Direction Angles

Direction angles are angles that describe the orientation of a vector relative to a reference direction, usually the positive x-axis. They are essential to converting a force into its vector components. These angles directly influence the calculation of each component using trigonometric functions.

To find the resultant direction angle after calculating the resultant components:

• Use the arctangent function, \( \theta_{\mathbf{R}} = \tan^{-1}\left(\frac{R_y}{R_x}\right) \), to determine the angle of the resultant vector relative to the x-axis.

This angle helps in understanding the overall direction in which the forces will move the object. Understanding and accurately calculating direction angles is fundamental in predicting how an object will behave under the influence of combined forces.