Problem 49 - Trigonometry

Question

Work An automobile is pushed down a level street by exerting a force of 85 pounds at an angle of \(15^{\circ}\) with the horizontal (Figure 6). How much work is done in pushing the car 110 fect?

Step-by-Step Solution

Answer

The work done is approximately 9025.565 foot-pounds.

Step 1: Understand the Concept of Work

Work done is calculated by the formula \( W = Fd \cos(\theta) \), 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 Key Components

From the problem, we are given: - Force \( F = 85 \) pounds- Distance \( d = 110 \) feet- Angle \( \theta = 15^{\circ} \).

Step 3: Calculate Cosine of the Angle

Determine \( \cos(\theta) \) using trigonometry. In this case, \( \cos(15^{\circ}) \) is needed. Using a calculator, \( \cos(15^{\circ}) \approx 0.9659 \).

Step 4: Substitute Values into Work Formula

Plug the values into the work formula: \[ W = 85 \times 110 \times \cos(15^{\circ}) \] \[ W = 85 \times 110 \times 0.9659 \].

Step 5: Perform the Multiplication

Calculate the multiplication step-by-step: - First multiply the force and the distance: \( 85 \times 110 = 9350 \)- Then multiply by the cosine: \( 9350 \times 0.9659 \approx 9025.565 \).

Step 6: Result for Work Done

The work done in pushing the car is approximately 9025.565 foot-pounds.

Key Concepts

Trigonometry

Trigonometry is a branch of mathematics dealing with the relationships between the angles and sides of triangles. It plays a crucial role in physics, especially when forces are involved at angles. In this exercise, we see how trigonometry helps determine the effectiveness of a force applied at an angle.

Consider when you exert force, not directly, but at some angle to the direction of motion. Only the component of the force that acts along the direction of motion contributes to the work done. Trigonometry provides us with a tool to find this component. The cosine function is particularly useful here. For an angle \( \theta \) between the force direction and the direction of motion, the cosine of \( \theta \) represents the fraction of the force acting along the direction of motion.

In the problem, the angle is \( 15^{\circ} \). Using a calculator, we find \( \cos(15^{\circ}) \) which tells us how much of that 85-pound force effectively moves the car forward. The cosine value at \( 15^{\circ} \) is approximately 0.9659, meaning most of the force is indeed contributing to moving the car. Trigonometry thus helps isolate only the useful part of the force for our calculations.

Force and Motion

Force and motion are fundamental concepts in physics that describe why and how objects move. In this context, it's all about understanding how a force applied to an object can cause it to change its motion, i.e., to speed up, slow down, or change direction.

When you apply a force to an object, like pushing a car, the force can be broken down into components—thanks to vectors—to better understand its effect. These components determine how much of the force is actually involved in moving the object in a given direction. For example, pushing down at an angle involves one component that moves the object horizontally and another that might push it downwards.

Here, the force of 85 pounds is at an angle, and drawing a vector diagram can help visualize this. Vectors are tools that have direction and magnitude, which we use in physics to visualize forces. Therefore, the applied force at an angle means that not all of it moves the car forward; some of the force may be wasted if it goes into the ground, especially without a lot of friction.

Work Formula

The work formula is a straightforward way to calculate the work done when moving an object with a force. Given by \( W = F d \cos(\theta) \), it combines the effects of the force's magnitude, the distance moved, and the angle between the force direction and motion.

Let's break it down:

• \( F \) stands for the force applied, in this case, 85 pounds.
• \( d \) is the distance over which the force moves the object, given as 110 feet in this problem.
• \( \cos(\theta) \) is the cosine of the angle, which adjusts the force depending on its direction.

The angle \( \theta \) is crucial because it accounts for the direction of force. If the force acts directly along the direction of motion, \( \theta = 0 \) and \( \cos(\theta) = 1 \), implying the entire force contributes to work done. However, any deviation means some force does not contribute to motion, as seen here with \( \theta = 15^{\circ} \) and its cosine is approximately 0.9659.

Apply these values to the work formula: \[ W = 85 \times 110 \times 0.9659 \] Results in about 9025.565 foot-pounds. This value tells us how much work is done against friction to move the car that specific distance.