Problem 41 - Algebra and Trigonometry

Question

Work A lawn mower is pushed a distance of 200 \(\mathrm{ft}\) along a horizontal path by a constant force of 50 \(\mathrm{lb}\) . The handle of the lawn mower is held at an angle of \(30^{\circ}\) from the horizontal (see the figure). Find the work done.

Step-by-Step Solution

Answer

The work done is approximately 8660 ft·lb.

Step 1: Understand the Concept of Work

Work is calculated as the product of the force applied and the distance over which the force is applied. However, the force must be in the direction of the displacement. If the force is at an angle, only the component of the force in the direction of the displacement is used to calculate work.

Step 2: Find the Horizontal Component of the Force

The force applied by the handle is at an angle, so we need to find its horizontal component using the cosine function. The horizontal component of the force is given by \( F_x = F \cdot \cos(\theta) \), where \( F = 50 \ \text{lb} \) and \( \theta = 30^{\circ} \). Thus: \[ F_x = 50 \times \cos(30^{\circ}) \] \[ F_x = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \ \text{lb} \]

Step 3: Calculate the Work Done

The work done \( W \) is the product of the horizontal component of the force and the distance the lawn mower is pushed. Therefore: \[ W = F_x \times d \] where \( d = 200 \ \text{ft} \). Thus: \[ W = 25\sqrt{3} \times 200 \] \[ W = 5000\sqrt{3} \ \text{ft}\cdot\text{lb} \]

Step 4: Express the Work in Decimal Form

Calculate the final value of the work by approximating \( \sqrt{3} \) as 1.732. Therefore: \[ W \approx 5000 \times 1.732 = 8660 \ \text{ft}\cdot\text{lb} \].

Key Concepts

Force and Displacement

In physics, the concept of work hinges on understanding the relationship between force and displacement. Work is only done when a force causes an object to move. Consider a situation where a lawnmower is pushed across a lawn. The force applied must actually move the mower, covering a specific distance.
For work to be calculated correctly, the force must be in the same direction as the displacement. This means if you push something forward, it should ideally move forward in the direction of that push.
In any case where the force isn't directly in the line of motion—like when pushing with a handle that is inclined—it becomes essential to find the component of the force that acts in the desired direction. This ensures that only the effective force contributing to the movement is used in our calculations.

Trigonometric Functions

Trigonometric functions are invaluable tools in physics, especially when dealing with forces at angles. They allow us to dissect forces into components that align with our desired coordinate system.
When the handle of a lawnmower is pushed at an angle to the horizontal, we can't use the whole force to calculate work. Instead, we use the cosine function to find the horizontal component of this force. The cosine of an angle in a right triangle relates the adjacent side to the hypotenuse.
In our scenario, the force applied is divided into two components: one parallel and one perpendicular to the horizontal path. By using the equation \( F_x = F \cdot \cos(\theta) \), the horizontal component is extracted. This component is the effective force that contributes to moving the mower along its path.

Vector Components

Understanding vector components is fundamental in physics when forces don't align with our axes, much like our inclined lawnmower handle scenario. Force vectors can point in any direction, but our calculations often need them aligned along horizontal and vertical axes.

Breaking down vectors into components helps us focus on individual influences. In the case of the lawnmower, the force is broken down into horizontal and vertical components. For calculating work, we focus on the horizontal force component because vertical components typically do not contribute to work along a horizontal path.

This decomposition is done using trigonometric functions—using cosine for horizontal and sine for vertical components. This technique highlights how vector components streamline the process of problem-solving, making it easier to apply concepts to real-world physics scenarios.