Problem 69 - Physics for Scientists and Engineers with Modern Physics

Question

(III) A bicyclist can coast down a \(7.0^{\circ}\) hill at a steady \(9.5 \mathrm{~km} / \mathrm{h} .\) If the drag force is proportional to the square of the speed \(v,\) so that \(F_{\mathrm{D}}=-c v^{2},\) calculate \((a)\) the value of the constant \(c\) and \((b)\) the average force that must be applied in order to descend the hill at \(25 \mathrm{~km} / \mathrm{h}\). The mass of the cyclist plus bicycle is \(80.0 \mathrm{~kg} .\) Ignore other types of friction.

Step-by-Step Solution

Answer

(a) The constant \(c\) is approximately 14.54 kg/m. (b) An average applied force of 121.5 N is needed.

Step 1: Analysis of Forces on the Hill

When the cyclist coasts down the hill, the gravitational force component down the hill is balanced by the drag force. The gravitational component along the hill is given by \(mg \sin \theta\), where \(m\) is the mass (80 kg) and \(\theta = 7^{\circ}\). The drag force \(F_D\) is \(-c v^2\) where \(v = 9.5 \text{ km/h} = 2.64 \text{ m/s}\). At steady speed, the forces balance thus \(mg \sin \theta = c v^2\).

Step 2: Calculate the Drag Force Constant c

Using the equation from Step 1: \[ mg \sin \theta = c v^2 \]Substitute the known values:- \(m = 80 \text{ kg}\)- \(g = 9.81 \text{ m/s}^2\)- \(\theta = 7^{\circ}\)- \(v = 2.64 \text{ m/s}\)Thus, \[ 80 \times 9.81 \times \sin(7^{\circ}) = c \times (2.64)^2 \]Solve for \(c\):\[ c = \frac{80 \times 9.81 \times \sin(7^{\circ})}{(2.64)^2} \approx 14.54 \text{ kg/m} \]

Step 3: Force Required to Descend at 25 km/h

To descend at 25 km/h, convert speed to m/s: \(v = 25 \text{ km/h} = 6.94 \text{ m/s}\). The forces need to rebalance, where the applying force \(F_A\) plus the drag force \(F_D\) counteracts the gravity component:\[ F_A + c v^2 = mg \sin \theta \]Substitute \(c\) calculated previously and solve for \(F_A\):\[ F_A = mg \sin \theta - c v^2 = 80 \times 9.81 \times \sin(7^{\circ}) - 14.54 \times (6.94)^2 \]\[ F_A \approx 577.3 - 698.8 = -121.5 \text{ N} \]* The negative sign indicates the gravity component exceeds the resistance, so the same force magnitude is needed uphill.

Key Concepts

Drag force calculation

Calculating drag force is important in understanding how objects like bicycles or cars interact with air as they move. The drag force opposes the motion of an object and is often proportional to the square of the velocity for high-speed cases. This relationship is represented by the equation \( F_D = -c v^2 \), where

• \( F_D \) is the drag force,
• \( c \) is a proportionality constant, and
• \( v \) is the velocity.

In the given exercise, the cyclist coasts at a constant speed, suggesting that the drag force perfectly balances the gravitational pull downhill. To find the drag constant \( c \), the balance between the gravitational component and drag force allows setting up a direct equation using the known speed and the incline angle. This balance is given by \( mg \sin \theta = c v^2 \). From this equation, \( c \) can be derived using the specific measurements provided.

Gravitational force component

When calculating motion on an incline, it's important to break down gravitational force into components. Gravitational force acts directly down towards the center of the Earth, but when on an incline, it has two components:

• One perpendicular to the incline, which usually does not affect motion along the incline, and
• One parallel to the incline, which affects the object's motion.

The component of the gravitational force that acts along the hill is calculated using \( mg \sin \theta \), where

• \( m \) is the mass of the object,
• \( g \) is the acceleration due to gravity (approximately \( 9.81 \text{ m/s}^2 \)), and
• \( \theta \) is the angle of inclination.

For the cyclist, this parallel component drives the motion down the slope and needs to be balanced by the drag force when coasting steadily.

Physics of motion on an incline

Understanding the physics of an object moving on an incline involves analyzing how forward and resisting forces interact. When a cyclist goes down a hill, two primary forces are at play: the gravitational force component driving the cyclist down and the drag force resisting this motion. These forces can be defined as:

• The gravitational force component is calculated by \( mg \sin \theta \).
• The drag force opposing motion is influenced by the velocity, expressed as \( -c v^2 \).

When these forces are equal, the cyclist travels at a steady velocity. If we want to change this velocity, such as increasing speed to 25 km/h, an additional applied force \( F_A \) is necessary to adjust the motion dynamics to maintain a new balance, taking into account the faster velocity's increased drag.

Drag force proportionality

The concept of drag force proportionality is crucial in understanding the resistance experienced by objects moving through a fluid like air. In many situations, drag force becomes significant as velocity increases, especially at higher speeds. Here, the drag force follows a quadratic relationship to the speed, represented by \( F_D = -c v^2 \).This quadratic relationship implies:

• As speed doubles, the drag force increases by a factor of four.
• This force is highly dependent on the coefficient \( c \), which is determined by factors like the object's shape and surface texture.

In our exercise, changing the cyclist's speed from coasting to increasing speed alters the drag, impacting how much additional force is required to overcome this resistance. By understanding this relationship, it's possible to predict how different speeds impact drag force, helping predict necessary forces for desired velocities.