Problem 29 - Algebra and Trigonometry

Question

Computing Work Find the work done by a force of 3 pounds acting in a direction of \(60^{\circ}\) to the horizontal in moving an object 6 feet from (0,0) to (6,0)

Step-by-Step Solution

Answer

9 foot-pounds

Step 1 - Identify the Components

Determine the components of the force. Given that the force is 3 pounds acting at an angle of 60 degrees to the horizontal, decompose the force into horizontal and vertical components. The horizontal component is calculated as: \( F_x = F \times \text{cos}(60^{\text{circ}}) \) and the vertical component is \( F_y = F \times \text{sin}(60^{\text{circ}}) \).

Step 2 - Calculate the Horizontal Component of the Force

Compute the horizontal component \( F_x \) of the force. Using \( F = 3 \text{ pounds} \) and \( \text{cos}(60^{\text{circ}}) = 0.5 \), we get: \( F_x = 3 \times 0.5 = 1.5 \text{ pounds} \).

Step 3 - Determine the Distance Moved

Recognize that the object is moved from (0,0) to (6,0), which means it is moved horizontally 6 feet.

Step 4 - Compute the Work Done

The work done by a force is given by the product of the force in the direction of motion and the distance moved. Since the force's horizontal component (1.5 pounds) acts entirely in the direction of motion: \( \text{Work} = F_x \times \text{distance} = 1.5 \text{ pounds} \times 6 \text{ feet} = 9 \text{ foot-pounds} \).

Key Concepts

Force Components

In physics, when studying forces, it's essential to understand how to break down a force into its components. The force applied at an angle can be divided into two parts: the horizontal component and the vertical component. These components help us analyze the motion more effectively and calculate other quantities like work.
For an applied force \( F \) acting at an angle \( \theta \) to the horizontal, the horizontal component and vertical component can be found using trigonometric functions. The horizontal component of the force, \( F_x \), is given by \( F \times \text{cos}(\theta) \), and the vertical component of the force, \( F_y \), is given by \( F \times \text{sin}(\theta) \).
Let's apply this to our example briefly:

• Horizontal force component: \( F_x = 3 \times \text{cos}(60^\text{circ}) = 3 \times 0.5 = 1.5 \) pounds
• Vertical force component: \( F_y = 3 \times \text{sin}(60^\text{circ}) = 3 \times 0.866 = 2.598 \) pounds

Changing a force from a single vector into its components makes it easier to work with, especially when calculating the work done by each component separately.

Trigonometry in Physics

Trigonometry is a branch of mathematics that deals with the relationships between side lengths and angles of triangles. It is highly useful in physics for resolving forces, analyzing waves, and understanding circular motions.
In our problem, we use basic trigonometric principles: sine \( \text{sin} \) and cosine \( \text{cos} \). These functions help us project the force onto the horizontal and vertical axes. For an angle \( \theta \):

• \( \text{cos}(\theta) \) gives the ratio of the adjacent side (horizontal component) to the hypotenuse (force magnitude).
• \( \text{sin}(\theta) \) gives the ratio of the opposite side (vertical component) to the hypotenuse.

In our example:

• In calculating the horizontal component: \( F_x = 3 \times \text{cos}(60^\text{circ}) = 1.5 \)
• In calculating the vertical component: \( F_y = 3 \times \text{sin}(60^\text{circ}) = 2.598 \)

Using trigonometry in this way allows you to break down forces for calculations and better understand their impacts on an object's motion.

Work Calculation

Work in physics is defined as the amount of energy transferred by a force acting through a distance. For a constant force making an angle with the direction of motion, the work done is calculated using the formula:
\[ \text{Work} = F \times d \times \text{cos}(\theta) \]
In our problem, the force applied is partially in the direction of motion; hence, we only consider the horizontal component of the force for calculating work. Work done by the vertical component is ignored because it does no work in the direction of the horizontal motion.

• Horizontal Force Component: 1.5 pounds
• Distance Moved: 6 feet
• Work: \( 1.5 \times 6 = 9 \text{ foot-pounds} \)

So, the force of 3 pounds acting at 60 degrees to the horizontal does 9 foot-pounds of work in moving an object 6 feet horizontally. Remember, only the force component in the direction of the object's displacement contributes to the work done.