Problem 23 - College Trigonometry

Question

Suppose Taylor fills her wagon with rocks and must exert a force of 13 pounds to pull her wagon across the yard. If she maintains a \(15^{\circ}\) angle between the handle of the wagon and the horizontal, compute how much work Taylor does pulling her wagon 25 feet. Round your answer to two decimal places.

Step-by-Step Solution

Answer

Taylor does approximately 314.64 pounds-feet of work.

Step 1: Identify the Work Formula

Recall that work done, which is the dot product between force and displacement vectors, is expressed as: \\[W = F \, d \, \cos(\theta)\] where \(W\) is the work, \(F\) is the force, \(d\) is the displacement, and \(\theta\) is the angle between the force and the direction of motion.

Step 2: Plug in Known Values

Insert the given values into the formula: \- Force \(F = 13\) pounds, \- Displacement \(d = 25\) feet, \- Angle \(\theta = 15^{\circ}\). \The formula becomes: \\[W = 13 \, \times \, 25 \, \times \, \cos(15^{\circ})\]

Step 3: Calculate the Cosine of the Angle

Use a calculator to find the cosine of \(15^{\circ}\). \\[\cos(15^{\circ}) \approx 0.9659\]

Step 4: Calculate the Work Done

Now substitute \(\cos(15^{\circ}) \approx 0.9659\) back into the equation to calculate the work: \\[W = 13 \, \times \, 25 \, \times \, 0.9659\] \\[W \approx 314.63575\] pounds-feet.

Step 5: Round the Answer

Round the answer to two decimal places: \\[W \approx 314.64\] pounds-feet.

Key Concepts

Understanding the Dot Product in Physics

Work in physics is not just any physical labor, it's a mathematical operation that often involves something known as the dot product. The dot product, sometimes called the scalar product, is vital when we need to calculate the work done by a force. In simpler terms, when we talk about work in physics, we actually refer to the product of two vectors:

• The force applied
• The displacement or the distance over which the force is applied

The beauty of the dot product lies in its ability to combine these two vectors into a single scalar value, which represents the work done.
The formula for calculating work using the dot product is \(W = F \, d \, \cos(\theta)\), where \(F\) is the force, \(d\) is the displacement, and \(\theta\) is the angle between the force vector and the displacement vector. This calculation gives us one number, a scalar, which tells us the amount of work performed.

Trigonometric Functions and their Role in Work Calculations

Trigonometry often comes into play when dealing with angles, especially when calculating work. The trigonometric functions, which include sine, cosine, and tangent, help us understand the components of forces and displacements. In the formula for work, \(W = F \, d \, \cos(\theta)\), the cosine function is crucial. It allows us to find the component of the force that acts in the direction of the displacement.
Consider this: if you pull a wagon at a certain angle to the ground, not all your pulling force contributes to moving the wagon horizontally. It's the horizontal component of the force, found using \(\cos(\theta)\), that actually makes the wagon move. Thus, cosine helps bridge the angle between force and direction, simplifying the calculation of work done in situations where force isn't applied straight along the path of motion.

The Importance of Angle Measurement in Physics

Understanding and accurately measuring angles is critical in physics, particularly in work calculations. The angle \(\theta\) in the work formula \(W = F \, d \, \cos(\theta)\) determines how much of the applied force effectively contributes to moving an object.

• When \(\theta = 0^{\circ}\), the force is entirely in the direction of the displacement, and it's fully effective in doing the work.
• If \(\theta = 90^{\circ}\), the force is perpendicular to the direction of displacement and does no work on the object.

In real-world scenarios, such as pulling a wagon at an angle, knowing this angle is essential to accurately compute the work done. Angles can be found using instruments like protractors or more advanced tools, ensuring that the calculations reflect the actual physical situation and don't just rely on assumptions.