Problem 70 - Trigonometry

Question

Force After they are finished swinging, Tyler and Kelly decide to rollerskate. They come to a hill that is inclined at \(8.5^{\circ}\). Tyler pushes Kelly halfway up the hill and then holds her there (Figure 14). If Kelly weighs \(58.0\) pounds, find the magnitude of the force Tyler must push with to keep Kelly from rolling down the hill. (We are assuming that the rollerskates make the hill into a frictionless surface so that the only force keeping Kelly from rolling backwards down the hill is the force Tyler is pushing with.)

Step-by-Step Solution

Answer

Tyler must push with a force of approximately 38.42 N to keep Kelly from rolling down the hill.

Step 1: Understand the Problem

The hill is inclined at an angle of \(8.5^{\circ}\), and Kelly weighs \(58.0\) pounds. To keep her from rolling down the hill, Tyler must apply a force to counteract the component of gravity pulling her back down the slope. We need to find this force.

Step 2: Calculate the Weight in Newtons

First, convert Kelly's weight from pounds to newtons, since the standard unit of force is the newton. We know that 1 pound is approximately 4.44822 newtons. Thus, Kelly's weight in newtons is \(58.0 \times 4.44822 = 258.0\) newtons.

Step 3: Determine Gravitational Force Component

Identify the component of the gravitational force that acts parallel to the inclined plane. This can be calculated using the formula \[ F_{\text{parallel}} = W \cdot \sin(\theta) \]where \(W\) is Kelly's weight in newtons, and \(\theta\) is the incline angle.

Step 4: Plug in the Values

Using Kelly's weight \(W = 258.0\) N and the incline angle \(\theta = 8.5^{\circ}\), the force component parallel to the incline is:\[ F_{\text{parallel}} = 258.0 \cdot \sin(8.5^{\circ}) \]Calculate the sine of the angle and then this force component.

Step 5: Perform the Calculation

Calculate \(\sin(8.5^{\circ})\) which is approximately \(0.1489\). Thus, the gravitational component is:\[ F_{\text{parallel}} = 258.0 \times 0.1489 \approx 38.42 \text{ N} \].

Step 6: Conclude with Tyler's Required Force

The magnitude of the force that Tyler must push with to keep Kelly from rolling down the hill is equal to the parallel gravitational force acting down the incline. Therefore, Tyler must exert a force of about \(38.42 \text{ N}\).

Key Concepts

Gravitational Force Component

When dealing with forces on an inclined plane, understanding the gravitational force component is key. Imagine a box on a hill: gravity pulls it directly down toward the Earth. However, since the box is on a slope, part of this gravitational force acts along the hill's surface, trying to pull it downwards. This is the component we need to consider when calculating the net force acting on an object on an incline.

To find this force component parallel to the incline, you use the equation:

• \( F_{\text{parallel}} = W \cdot \sin(\theta) \)

Here, \( W \) represents the weight of the object in newtons, and \( \theta \) is the angle of the incline. The function \( \sin(\theta) \) helps to determine the proportion of the weight that acts down the slope.

In simple terms, the steeper the hill (larger \( \theta \)), the greater the gravitational force trying to slide the object down. This concept is crucial as it defines how much effort is required to hold or move an object along an inclined plane.

Frictionless Surfaces

Frictionless surfaces might sound like a slippery fantasy, but they serve as essential models in physics. In reality, most surfaces have some friction, but by considering them as frictionless, we can simplify complex scenarios to understand the core forces at work.

On a frictionless inclined plane, the only forces acting on an object are its weight (or gravitational force) and any applied forces. Without friction, there’s nothing to slow down or resist the movement of an object once it's set in motion. This condition helps us clearly analyze how gravity alone impacts motion.

For example, in the problem with Kelly on rollerskates, we disregard friction to focus purely on the gravitational force component pulling her down the hill. Understanding this idealized situation allows us to precisely calculate the force needed just to counteract gravity, without the added complexity of calculating resistance factors that would complicate into real-world friction scenarios.

Weight Conversion to Newtons

Understanding how to convert weight into newtons is fundamental when working with physics problems involving force. Weight is the force exerted by gravity on an object, and to standardize measurements, we often need to convert it from units like pounds (common in the U.S.) to newtons, which are widely used in scientific calculations.

The conversion is straightforward: 1 pound is equivalent to approximately 4.44822 newtons. Therefore, to convert any weight in pounds to newtons, simply multiply by this conversion factor:

• For example, Kelly's weight of 58.0 pounds becomes \( 58.0 \times 4.44822 = 258.0 \) newtons.

This conversion allows us to use the weight in formulas that require metric units, helping calculate accurate force values. The process might seem like a small step, but without it, calculations for forces like those on inclined planes would be impossible or inaccurate.