Problem 47 - 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 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 Work Formula

Work done is calculated using the formula \( W = F \cdot d \cdot \cos(\theta) \), where \( W \) is the work done, \( F \) is the force applied, \( d \) is the distance over which the force is applied, and \( \theta \) is the angle between the force and the direction of motion.

Step 2: Identify Given Values

From the problem statement, we have - Force \( F = 50 \) lb - Distance \( d = 200 \) ft - Angle \( \theta = 30^{\circ} \).

Step 3: Convert Angle to Radians if Needed

Since the work formula uses the cosine function, ensure your calculator is set to degrees, or convert the angle to radians if necessary. For this scenario, leave the angle in degrees because \( \cos(30^{\circ}) \) is a standard trigonometric value.

Step 4: Calculate the Cosine of the Angle

Find \( \cos(30^{\circ}) \). We know from trigonometry that \( \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \approx 0.866 \).

Step 5: Substitute Values into the Work Formula

Substitute the given values and the cosine value into the work formula: \[ W = 50 \times 200 \times \frac{\sqrt{3}}{2} = 50 \times 200 \times 0.866 \].

Step 6: Perform the Calculation

Now perform the multiplication: \[ W = 50 \times 200 \times 0.866 \approx 8660 \] ft-lb.

Key Concepts

Trigonometric Functions

Trigonometric functions are essential in physics and engineering, as they relate the angles and lengths in right-angled triangles. They help in analyzing different components of motion, especially when forces act at various angles. In this example, the force applied to the lawnmower forms a right triangle with the ground.

• The handle of the lawnmower is at an angle of 30° to the horizontal.
• In trigonometry, the cosine of an angle is the ratio of the adjacent side to the hypotenuse in a right triangle.
• Known trigonometric values provide quick solutions in specific cases. For example, \(\cos(30^{\circ})\) equals \(\frac{\sqrt{3}}{2}\) or approximately 0.866.

Using trigonometric functions, we can resolve the force exerted at an angle into components, making it possible to calculate work, which involves forces at angles other than perpendicular or parallel to the direction of motion.

Force and Motion

In physics, understanding force and motion is crucial for analyzing various physical phenomena. Force is an interaction that changes the motion of an object. When force is applied over a distance, it results in work being done.

• In the given problem, a force of 50 lb is applied to push a lawnmower across a 200 ft path.
• Because the force is applied at an angle, it's necessary to consider the component of the force that acts in the direction of the mower's movement.
• This is where the cosine function (\(\cos(30^{\circ})\)) comes into play, determining how much of the force contributes to the forward motion.

Without understanding how forces interact with angles, calculating the true effectiveness of the applied force would be challenging. This concept is fundamental in mechanics, dictating how objects respond to varying forces and angles.

Physics Calculations

Physics calculations often involve applying various formulas and understanding the underlying principles behind the numbers.

• The concept of "work" in physics involves multiplying force by the distance over which the force acts and the cosine of the angle between the force and direction of motion.
• In this case, the formula \( W = F \cdot d \cdot \cos(\theta) \) is used to determine the total work done.
• By substituting in the known values (\( F = 50 \) lb, \( d = 200 \) ft, and \( \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \)), the resulting calculation simplifies to \( 8660 \) ft-lb of work being performed.

Understanding this formula and how to employ it in different scenarios is invaluable for accurately assessing work done, energy used, and overall efficiency in real-life applications. Knowing how to substitute values and calculate the result empowers students to tackle similar problems with confidence.