Problem 50 - Trigonometry

Question

Pulling a Crate A large crate is pulled across the ice with two ropes. A force of 47 pounds is applied to the first rope in the direction \(80^{\circ}\), and a force of 55 pounds is applied to the second rope in the direction \(105^{\circ}\). What are the magnitude and direction of the resultant force acting on the crate?

Step-by-Step Solution

Answer

Resultant force magnitude is 99.568 pounds at \(93.653^{\circ}\).

Step 1: Resolve Each Force

First, we need to resolve each force into its horizontal (x) and vertical (y) components. Use trigonometric functions for this. For the first force of 47 pounds at an angle of \(80^{\circ}\):\[ F_{1x} = 47 \cos(80^{\circ}) \]\[ F_{1y} = 47 \sin(80^{\circ}) \]For the second force of 55 pounds at an angle of \(105^{\circ}\):\[ F_{2x} = 55 \cos(105^{\circ}) \]\[ F_{2y} = 55 \sin(105^{\circ}) \]

Step 2: Calculate Component Values

Now calculate the values of each component using the trigonometric values for cosine and sine:\[ F_{1x} = 47 \times 0.1736 \approx 8.1592 \]\[ F_{1y} = 47 \times 0.9848 \approx 46.256 \]\[ F_{2x} = 55 \times (-0.2588) \approx -14.234 \]\[ F_{2y} = 55 \times 0.9659 \approx 53.125 \]

Step 3: Sum the Components

Add the x-components and the y-components separately to find the total force components:\[ R_x = F_{1x} + F_{2x} = 8.1592 - 14.234 \approx -6.0748 \]\[ R_y = F_{1y} + F_{2y} = 46.256 + 53.125 \approx 99.381 \]

Step 4: Calculate the Resultant Force Magnitude

Use the Pythagorean theorem to find the magnitude of the resultant force:\[ R = \sqrt{R_x^2 + R_y^2} = \sqrt{(-6.0748)^2 + (99.381)^2} \approx \sqrt{0.3690 + 9876.792} \approx 99.568 \text{ pounds} \]

Step 5: Determine the Resultant Force Direction

Determine the angle of the resultant force using the arctangent function:\[ \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{99.381}{-6.0748}\right) \approx \tan^{-1}(-16.361) \approx -86.347^{\circ} \]Since the components are in the third quadrant, add \(180^{\circ}\) to find the angle in standard position:\[ \text{Adjusted } \theta = 180 - 86.347 \approx 93.653^{\circ} \]

Key Concepts

Understanding Trigonometric Functions

Trigonometric functions are essential tools used to resolve forces and understand their components. They are primarily the sine, cosine, and tangent functions, which relate angles of a triangle to the lengths of its sides. When working with forces applied at an angle, we use these functions to break down the force into horizontal and vertical components.
For this exercise, when a force is applied at an angle,

• the cosine function helps us find the horizontal component (adjacent side),
• and the sine function gives us the vertical component (opposite side).

For example, to resolve the first force of 47 pounds at an angle of \(80^\circ\), we use
\[ F_{1x} = F \cos(\theta) \quad \text{and} \quad F_{1y} = F \sin(\theta) \] to find its components on the x and y-axis respectively. These functions allow us to precisely calculate where and how these forces will impact the object.

Breaking Down Force Components

Force components represent how an angled force is distributed in two perpendicular directions, typically the horizontal (x-axis) and vertical (y-axis) directions. The initial step in calculating a resultant force involves resolving each force into these components.
By identifying these components:

• We gain a clearer understanding of how much force is directed straight ahead or upward.
• This breakdown aids in evaluating the net effect of multiple forces on an object.

For instance, with 47 pounds at \(80^\circ\), we calculate

• horizontal force, \(F_{1x} = 8.1592\) pounds,
• and vertical force, \(F_{1y} = 46.256\) pounds.

Similarly, the second force of 55 pounds at \(105^\circ\) results in

• \(F_{2x} = -14.234\) pounds,
• and \(F_{2y} = 53.125\) pounds.

These values are decisive in calculating the overall impact of the forces on the crate.

Applying the Pythagorean Theorem

The Pythagorean theorem is a fundamental principle used to find the magnitude of the resultant force. It is expressed as \(a^2 + b^2 = c^2\) for right triangles, where \(c\) is the hypotenuse, or here, the resultant force.
To apply this theorem:

• Firstly, we sum up the square of the x-components \((R_x)\) and y-components \((R_y)\).
• Then, by taking the square root of this sum, we determine the magnitude of the resultant force.

So for this exercise, the calculation shows \[ R = \sqrt{(-6.0748)^2 + (99.381)^2} \approx 99.568 \text{ pounds} \] This value signifies the total force impacting the crate, combining all force vectors applied. It's crucial in predicting how the object will move on the ice.

Finding the Direction Using the Arctangent Function

The arctangent function, \(\tan^{-1}\), plays a critical role in determining the direction of the resultant force. Once the force components are determined, we utilize \(\tan^{-1}\) to find the angle of the resultant force with respect to the positive x-axis.
Here's how it works:

• The angle \(\theta\) is calculated using \(\tan^{-1}(\frac{R_y}{R_x})\).
• This gives the angle in the context of the coordinate system based on quadrant rules.

From our exercise, \[ \theta = \tan^{-1}\left(\frac{99.381}{-6.0748}\right) \approx -86.347^\circ \], but since it's in the third quadrant, we adjust by adding \(180^\circ\).
This adjustment results in an overall angle of approximately \(93.653^\circ\), indicating the precise direction of the effective force across the coordinate plane. This direction is crucial for understanding the precise movement path of the crate when the forces are applied.