2.9 Resistive Forces

Motion with Resistive Force

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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, fluid drag, and friction in mechanical systems

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 can be solved using separation of variables and integration with appropriate limits. The solution reveals that:

• Velocity approaches a limiting value asymptotically rather than increasing indefinitely
• Acceleration decreases exponentially over time, approaching zero
• Position function follows a more complex pattern that depends on initial conditions

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

Terminal Velocity

Terminal velocity represents the equilibrium state where forces balance perfectly, resulting in zero acceleration. At this point, the object continues moving at a constant speed.

When an object falls through a fluid like air, it eventually reaches terminal velocity when the upward resistive force exactly balances the downward gravitational force. At this point:

$F_{resistive} = F_{gravitational}$

$kv_{terminal} = mg$

Solving for terminal velocity:

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

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

$(20 + 49)e^{-0.2t} = 49$ $69e^{-0.2t} = 49$ $e^{-0.2t} = \frac{49}{69}$ $-0.2t = \ln(\frac{49}{69})$ $t = -\frac{1}{0.2}\ln(\frac{49}{69}) = 1.72$ seconds

The height can be found by integrating the velocity function from t = 0 to t = 1.72: $h = \int_0^{1.72} v(t) dt$

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: $h = \frac{2}{0.4}(20 + \frac{2 \times 9.8}{0.4})(1-e^{-0.4 \times 1.72/2}) - \frac{2 \times 9.8}{0.4} \times 1.72$

$h = 5(20 + 49)(1-e^{-0.344}) - 49 \times 1.72$

$h = 5 \times 69 \times 0.291 - 84.28$

$h = 100.4 - 84.28 = 16.1$ meters

Therefore, the object reaches a maximum height of approximately 16.1 meters.