2.9 Resistive Forces

Motion with Resistive Force

Resistive Force Definition

Resistive forces act in opposition to an object's motion, effectively slowing it down over time. These forces increase in magnitude as the object's speed increases, creating a natural limiting effect on motion.

• Acts in the opposite direction of an object's velocity vector
• Magnitude typically increases with speed (faster objects experience stronger resistance)
• Examples include air resistance and fluid drag. In this topic, a resistive force means a velocity-dependent force opposite the object's velocity, such as $\vec{F}_{r}=-k \vec{v}$; do not confuse this with ordinary kinetic friction, which is typically modeled separately.

Air resistance is the most common example we encounter daily. It can be modeled mathematically as:

$\vec{F}_{r}=-k \vec{v}$

Where $k$ is a positive constant that depends on the object's shape, size, and the medium it's moving through, and $\vec{v}$ is the object's velocity vector. The negative sign indicates the force opposes motion.

Differential Equation for Velocity

When an object experiences a resistive force, its motion follows a differential equation derived from Newton's second law. This mathematical approach reveals important insights about the object's behavior over time.

Applying Newton's second law ($\vec{F}_{net} = m\vec{a}$) to an object experiencing both a constant force and a resistive force:

$m\frac{dv}{dt} = F_{constant} - kv$

This first-order differential equation is solved using separation of variables with proper limits of integration. Here is the explicit derivation:

Starting from $m\frac{dv}{dt} = F_{constant} - kv$, we separate variables by moving all $v$-terms to one side and $t$-terms to the other:

$\frac{dv}{F_{constant} - kv} = \frac{dt}{m}$

Now integrate both sides using limits. If the object has initial velocity $v_0$ at $t = 0$ and velocity $v$ at time $t$:

$\int_{v_0}^{v} \frac{dv'}{F_{constant} - kv'} = \int_{0}^{t} \frac{dt'}{m}$

Evaluating the left side with a $u$-substitution ($u = F_{constant} - kv'$, $du = -k\,dv'$):

$-\frac{1}{k}\ln\left(\frac{F_{constant} - kv}{F_{constant} - kv_0}\right) = \frac{t}{m}$

Solving for $v(t)$ by exponentiating both sides and rearranging:

$F_{constant} - kv = (F_{constant} - kv_0)\,e^{-kt/m}$

$v(t) = \frac{F_{constant}}{k} + \left(v_0 - \frac{F_{constant}}{k}\right)e^{-kt/m}$

This result shows how the initial condition $v_0$ enters the solution directly through the separation of variables process with limits of integration.

The time constant $\tau = \frac{m}{k}$ is a critical parameter that describes how quickly the object approaches its terminal velocity. This value represents the time it takes for the velocity to reach approximately 63% of its final value.

• Objects with larger mass take longer to reach terminal velocity
• Stronger resistive forces (larger $k$ values) cause objects to reach terminal velocity more quickly
• The approach to terminal velocity follows an exponential pattern

Time-Dependent Functions: Velocity, Acceleration, and Position

For a constant applied force $F_{constant}$ in the positive direction and resistive force $\vec{F}_{r} = -k\vec{v}$, we can write the complete set of time-dependent functions. Defining the terminal velocity as $v_{term} = \frac{F_{constant}}{k}$:

Velocity:

$v(t) = v_{term} + (v_0 - v_{term})\,e^{-kt/m}$

Acceleration (found by differentiating velocity):

$a(t) = \frac{dv}{dt} = -\frac{k}{m}(v_0 - v_{term})\,e^{-kt/m}$

Position (found by integrating velocity):

$x(t) = x_0 + v_{term}\,t + \frac{m}{k}(v_0 - v_{term})(1 - e^{-kt/m})$

Each function depends on the initial conditions $x_0$ and $v_0$. The velocity and acceleration are exponential functions of time, while the position is a combination of a linear term and an exponential term. Their long-term behavior is governed by asymptotes:

• As $t \to \infty$, $v(t) \to v_{term}$ (velocity approaches terminal velocity asymptotically)
• As $t \to \infty$, $a(t) \to 0$ (acceleration decays exponentially to zero)
• As $t \to \infty$, $x(t)$ grows approximately linearly at the rate $v_{term}$, since the exponential correction term vanishes

Terminal Velocity

Terminal velocity is the maximum speed reached when a constant force and a resistive force act in opposite directions. It occurs when the net force is zero, so acceleration is zero. In general, if the constant force has magnitude $F_{constant}$ and the resistive force is $kv$ opposite the motion, then at terminal speed $F_{constant} - kv_{terminal} = 0$, so:

$v_{terminal} = \frac{F_{constant}}{k}$

For a falling object, $F_{constant} = mg$, giving:

$v_{terminal} = \frac{mg}{k}$

This equilibrium explains many everyday phenomena:

• Skydivers control their descent rate by changing body position, effectively altering their $k$ value
• Raindrops fall at different terminal velocities depending on their size
• Heavier objects generally have higher terminal velocities in the same medium

The time required to reach terminal velocity depends on how far the initial velocity is from terminal velocity and the time constant $\tau$. Generally, an object reaches approximately 95% of its terminal velocity after 3 time constants.

Practice Problem 1: Terminal Velocity

Solution

First, we need to find the terminal velocity using the equation: $v_{terminal} = \frac{mg}{k}$

Given:

• Mass (m) = 75 kg
• Gravitational acceleration (g) = 9.8 m/s²
• Resistance constant (k) = 15 kg/s

$v_{terminal} = \frac{75 \times 9.8}{15} = \frac{735}{15} = 49$ m/s

To find the time to reach 90% of terminal velocity, we use the exponential approach equation: $v(t) = v_{terminal}(1 - e^{-t/\tau})$

Where $\tau = \frac{m}{k} = \frac{75}{15} = 5$ seconds

We need to find t when $v(t) = 0.9 \times v_{terminal}$: $0.9 = 1 - e^{-t/5}$

$e^{-t/5} = 0.1$ $-t/5 = \ln(0.1)$ $t = -5 \times \ln(0.1) = -5 \times (-2.3) = 11.5$ seconds

Therefore, the skydiver will reach 90% of terminal velocity (44.1 m/s) after approximately 11.5 seconds.

Practice Problem 2: Motion with Resistive Force

Solution

This problem requires us to solve the differential equation of motion with both gravity and resistive force.

Given:

• Mass (m) = 2 kg
• Initial velocity (v₀) = 20 m/s
• Resistance constant (k) = 0.4 kg/s
• Gravitational acceleration (g) = 9.8 m/s²

The differential equation is: $m\frac{dv}{dt} = -mg - kv$

The velocity as a function of time is: $v(t) = (v_0 + \frac{mg}{k})e^{-kt/m} - \frac{mg}{k}$

To find the maximum height, we need to find when v = 0 and then calculate the displacement.

Setting v(t) = 0: $(20 + \frac{2 \times 9.8}{0.4})e^{-0.4t/2} - \frac{2 \times 9.8}{0.4} = 0$

$(20 + 49)e^{-0.2t} = 49$ $69e^{-0.2t} = 49$ $e^{-0.2t} = \frac{49}{69}$ $-0.2t = \ln(\frac{49}{69})$ $t = -\frac{1}{0.2}\ln(\frac{49}{69}) = -5\ln(\frac{49}{69}) \approx 1.713$ seconds

The height can be found by integrating the velocity function from t = 0 to t = 1.713: $h = \int_0^{1.713} v(t) \, dt$

Using the position formula for this type of motion: $x(t) = x_0 + \frac{m}{k}(v_0 + \frac{mg}{k})(1-e^{-kt/m}) - \frac{mg}{k}t$

Substituting our values with x₀ = 0: $h = \frac{2}{0.4}(20 + \frac{2 \times 9.8}{0.4})(1-e^{-0.4 \times 1.713/2}) - \frac{2 \times 9.8}{0.4} \times 1.713$

$h = 5(20 + 49)(1-e^{-0.3426}) - 49 \times 1.713$

Using $e^{-0.3426} \approx 0.7101$, we have $1 - e^{-0.3426} \approx 0.2899$.

$h = 5 \times 69 \times 0.2899 - 49 \times 1.713$

$h \approx 99.97 - 83.94 = 16.03$ meters

Therefore, the object reaches a maximum height of approximately 16.0 meters.