Question

A firefighter who weighs 192 lb slides down an infinitely long fire pole that exerts a frictional resistive force with magnitude proportional to his speed, with \(k=2.5 \mathrm{lb}-\mathrm{s} / \mathrm{ft}\). Assuming that he starts from rest, find his velocity as a function of time and find his terminal velocity.

Short answer

The firefighter's velocity as a function of time is given by the formula: \(v(t) = \frac{192\ \mathrm{lb}}{2.5\ \mathrm{lb\cdot s/ft}} \Bigg[ 1 - e^{\frac{-2.5\ \mathrm{lb\cdot s/ft}}{5.96\ \mathrm{slug}}t} \Bigg]\). The terminal velocity, which is the limiting value of the velocity function as time approaches infinity, is approximately 76.8 ft/s.

Step-by-step solution

Determine the net force on the firefighter

In this situation, there are two forces acting on the firefighter: gravitational force (downward) and frictional force (upward). Let's denote them as \(F_g\) and \(F_f\), respectively. The gravitational force can be calculated as \(F_g = m \cdot g\), where \(m\) is the mass of the firefighter, and \(g\) is the acceleration due to gravity (\(32\ \mathrm{ft/s^2}\)). We are given the weight (force) of the firefighter (192 lb) which is equal to \(F_g\). The frictional force, \(F_f\), is given as proportional to the firefighter's speed, \(v\). Therefore, \(F_f = k \cdot v\) where \(k = 2.5\ \mathrm{lb\cdot s/ft}\). The net force can be found by subtracting the frictional force from the gravitational force: \(F_{net} = F_g - F_f\). Since we have \(F_{net} = F_g - F_f = m \cdot a\), where \(a\) is the acceleration, we can write the equation as: \(m \cdot a = 192\ \mathrm{lb} - 2.5\ \mathrm{lb\cdot s/ft} \cdot v\).

Convert weight (force) to mass

To calculate the acceleration and subsequently the velocity, we need the mass (m) in the equation. Using the conversion factor of 1 slug = 32.2 lb, we have: \(m = \frac{192\ \mathrm{lb}}{32.2\ \mathrm{lb/slug}} = 5.96\ \mathrm{slug}\). Now, we can write the equation as: \(5.96\ \mathrm{slug} \cdot a = 192\ \mathrm{lb} - 2.5\ \mathrm{lb\cdot s/ft} \cdot v\). Rewriting acceleration (a) as the time derivative of velocity \(a = \frac{dv}{dt}\), the equation becomes: \(5.96\ \mathrm{slug} \cdot \frac{dv}{dt} = 192\ \mathrm{lb} - 2.5\ \mathrm{lb\cdot s/ft} \cdot v\).

Solve the differential equation

We will separate variables and then integrate both sides to solve the differential equation: \(\frac{5.96\ \mathrm{slug}}{192\ \mathrm{lb} - 2.5\ \mathrm{lb\cdot s/ft} \cdot v} \cdot dv = dt\). Integrating both sides: \(\int \frac{5.96\ \mathrm{slug}}{192\ \mathrm{lb} - 2.5\ \mathrm{lb\cdot s/ft} \cdot v} \cdot dv = \int dt\). Let \(C\) be the constant of integration. Then we have: \(-\frac{5.96\ \mathrm{slug}}{2.5\ \mathrm{lb\cdot s/ft}} \ln{ \Big| 1 - \frac{2.5\ \mathrm{lb\cdot s/ft}}{192\ \mathrm{lb}} \cdot v} \Big| = t + C\). Since the firefighter starts from rest, initial velocity (\(v(0)\)) is 0, using this initial condition, we can find the constant \(C\): \(C = -\frac{5.96\ \mathrm{slug}}{2.5\ \mathrm{lb\cdot s/ft}} \ln{(1)}\). As \(\ln(1) = 0\), then \(C = 0\). Now we can write the velocity as a function of time: \(v(t) = \frac{192\ \mathrm{lb}}{2.5\ \mathrm{lb\cdot s/ft}} \Bigg[ 1 - e^{\frac{-2.5\ \mathrm{lb\cdot s/ft}}{5.96\ \mathrm{slug}}t} \Bigg]\).

Determine the terminal velocity

We find the terminal velocity by calculating the limiting value of the velocity function as time approaches infinity: \(\lim_{t\to\infty} v(t) = \lim_{t\to\infty} \frac{192\ \mathrm{lb}}{2.5\ \mathrm{lb\cdot s/ft}} \Bigg[ 1 - e^{\frac{-2.5\ \mathrm{lb\cdot s/ft}}{5.96\ \mathrm{slug}}t} \Bigg]\). As \(e^{\frac{-2.5\ \mathrm{lb\cdot s/ft}}{5.96\ \mathrm{slug}}t}\) approaches 0 as \(t\to\infty\), we have: Terminal velocity: \(v_{terminal} = \frac{192\ \mathrm{lb}}{2.5\ \mathrm{lb\cdot s/ft}} \approx 76.8\ \mathrm{ft/s}\). So, the firefighter's velocity as a function of time is: \(v(t) = \frac{192\ \mathrm{lb}}{2.5\ \mathrm{lb\cdot s/ft}} \Bigg[ 1 - e^{\frac{-2.5\ \mathrm{lb\cdot s/ft}}{5.96\ \mathrm{slug}}t} \Bigg]\). And his terminal velocity is approximately 76.8 ft/s.

Key concepts

Frictional Force

When an object moves through a medium or along a surface, it encounters resistance often referred to as frictional force. This resistance force is directly proportional to the speed of the moving object and opposes the direction of its motion. In the case of our firefighter sliding down the pole, this force is described by the equation \( F_f = k \times v \), where \( k \) is the frictional coefficient in units of \( \text{lb}\text{-s}/\text{ft} \) and \( v \) is the velocity. The frictional force plays a crucial role in determining the net force acting on the body and, consequently, its acceleration.

To fully understand the impact of the frictional force on the firefighter's descent, one might consider the fact that as his velocity increases, so does the magnitude of the frictional force working against him. This frictional force is a form of dissipative force that absorbs kinetic energy, which often leads to heat production. In practical applications, frictional forces are essential for braking systems, writing with a pen on paper, and even walking.

Terminal Velocity

Terminal velocity is a key concept when analyzing motions involving frictional forces. It's the constant speed that a freely falling object eventually reaches when the resistance of the medium through which it is falling prevents further acceleration. In our exercise, the firefighter achieves terminal velocity when the upward frictional force exactly balances the downward gravitational force, resulting in a net force of zero.

To conceptualize this further, imagine the firefighter accelerating due to gravity but as his velocity increases, the frictional force grows until the point where it equals the gravitational force. At this juncture, known as terminal velocity, the acceleration ceases, and the firefighter continues to move at this constant velocity. For the firefighter, we calculate this to be approximately 76.8 ft/s. This concept has wide real-world applications, including skydiving, where skydivers reach terminal velocity and experience the sensation of floating because of the balance between gravity and air resistance.

Solving Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They are integral in describing various phenomena in engineering, physics, economics, and beyond. In situations like our firefighter's descent, a differential equation models the relationship between velocity and acceleration with time as a variable. To solve these equations, one typically separates variables and integrates on both sides, often applying initial conditions to find the value for the constant of integration.

Solving the firefighter's equation gave us \( v(t) = \frac{192\ \text{lb}}{2.5\ \text{lb}\cdot \text{s/ft}} \Big[ 1 - e^{\frac{-2.5\ \text{lb}\cdot \text{s/ft}}{5.96\ \text{slug}}t} \Big] \) which shows how his velocity, \( v(t) \), changes over time. When you're faced with a differential equation, it's essential to understand the underlying physics or real-life context to set up the correct equation, separate variables appropriately, and identify initial conditions to integrate and find the specific solution.