Chapter 6 Readings: 6.1 Solving Problems with Newton’s Laws

Problem-Solving Strategy for Newton’s Laws

Applying Newton’s laws of motion systematically requires a structured approach to translate physical situations into solvable mathematical models. Success in mechanics depends on identifying the physical principles involved, separating the system from its surroundings, and checking the validity of the final results.

Core Learning Objectives

• Apply problem-solving techniques to determine unknown quantities in complex force systems.

• Integrate kinematics with Newton’s laws of motion to solve multi-step dynamics problems.

• Solve equilibrium problems involving multiple forces acting at arbitrary angles.

• Formulate equations of motion for accelerating single-body and multi-body systems.

• Utilize differential and integral calculus to solve advanced dynamics problems involving non-constant forces.

Five-Step Problem-Solving Procedure

1. Identify Physical Principles and Variables: Determine the physical principles governing the situation. Explicitly list all known quantities (givens) and identify the specific quantities to be calculated.

2. Sketch the Situation: Draw a clear visual representation of the physical scenario. Use directed line segments (arrows) to represent all acting forces, matching arrow directions and relative lengths to the forces where possible.

3. Isolate the System of Interest and Construct a Free-Body Diagram (FBD): Define the boundaries of the specific object or combination of objects being analyzed. The free-body diagram must show only the external forces acting directly on the system of interest. All internal forces (forces between parts of the chosen system) and forces exerted by the system on external bodies must be excluded. Only force vectors are shown; never include velocity or acceleration vectors in a free-body diagram.

4. Apply Newton’s Second Law: Formulate the vector equation of motion, projecting it along convenient coordinate axes: $\sum \mathbf{\vec{F}} = m \mathbf{\vec{a}}$ For one-dimensional motion where all forces are collinear, equations can be solved algebraically. For two-dimensional motion, decompose all vector quantities into orthogonal scalar components along chosen axes: $\sum F_x = m a_x$ $\sum F_y = m a_y$ When an object is in equilibrium or accelerates along only one axis (such as purely horizontal acceleration), set the acceleration along the inactive axis to zero: $a_y = 0$ If required, incorporate kinematic equations for straight-line motion to connect displacement, velocity, acceleration, and elapsed time.

5. Evaluate and Check the Solution: Assess whether the calculated magnitudes, directions, and units are physically reasonable. Dimensional analysis provides a primary check (e.g., verifying that a force evaluates to Newtons rather than units of speed). Physical intuition provides a second check (e.g., confirming that frictional resistance slows an object on an incline rather than speeding it up).

Defining the System of Interest: Piano Hoisting Analysis

Consider hoisting a grand piano toward a second-story apartment window:

• Forces Identified in the Physical Sketch: The upward tension from the cable supporting the rigging is $\mathbf{\vec{T}}$, the downward gravitational force (weight) acting on the piano is $\mathbf{\vec{w}}$, and the downward pull that the piano exerts on the supporting rope is $\mathbf{\vec{F}}_{\text{T}}$. Minor ambient forces (such as a light breeze) are assumed to be negligible.

• Exclusion of Extraneous Forces: When defining the piano as the isolated system of interest, $\mathbf{\vec{F}}_{\text{T}}$ must be omitted from the free-body diagram. By Newton’s third law, $\mathbf{\vec{F}}_{\text{T}}$ represents the action force that the piano exerts on the external rigging, not a force acting on the piano itself. Including both $\mathbf{\vec{T}}$ and $\mathbf{\vec{F}}_{\text{T}}$ on the same object would produce an erroneous double-counting of force interactions.

• Equilibrium Condition: When the piano remains stationary, the net external force is zero, requiring the upward tension to balance the downward weight exactly: $\mathbf{\vec{T}} = -\mathbf{\vec{w}}$

Particle Equilibrium

A particle is in equilibrium when the vector sum of all external forces acting upon it equals zero: $\sum \mathbf{\vec{F}} = 0$

• Static Equilibrium: Applies to systems that remain permanently at rest relative to an inertial reference frame.

• Dynamic Equilibrium: Applies to systems moving with a constant velocity vector ($\mathbf{\vec{v}} = \text{constant}$, $\mathbf{\vec{a}} = 0$).

• Relativity of Equilibrium: The distinction between static and dynamic equilibrium depends strictly on the observer's inertial frame of reference. An object observed to be static in one frame is in dynamic equilibrium when viewed from another frame moving at a constant relative velocity.

Example 6.1: Different Tensions at Different Angles

A traffic light of mass $m = 15.0\,\text{kg}$ is suspended in static equilibrium from two light cables connected to support points overhead. Cable 1 pulls upward and to the left at an angle of $30^\circ$ below the horizontal overhead support (or elevated at $30^\circ$ above the horizontal axis at the junction). Cable 2 pulls upward and to the right at an angle of $45^\circ$ above the horizontal axis. Neglect the mass of the cables.

Coordinate Resolution and Free-Body Diagram

Define the junction point containing the traffic light as the system of interest. The forces acting on the point are:

• Tension $\mathbf{\vec{T}}_1$ acting at an angle of $30^\circ$ above the negative $x$-axis.

• Tension $\mathbf{\vec{T}}_2$ acting at an angle of $45^\circ$ above the positive $x$-axis.

• Gravitational weight $\mathbf{\vec{w}}$ acting downward along the negative $y$-axis.

Analytical Solution

Evaluate horizontal force equilibrium ($x$-axis): $F_{\text{net } x} = T_{2x} - T_{1x} = 0$ $|T_{1x}| = |T_{2x}|$ $T_1 \cos(30^\circ) = T_2 \cos(45^\circ)$ $T_2 = \frac{\cos(30^\circ)}{\cos(45^\circ)} T_1 = \frac{0.8660}{0.7071} T_1 = 1.225 T_1$

Evaluate vertical force equilibrium ($y$-axis): $F_{\text{net } y} = T_{1y} + T_{2y} - w = 0$ $T_{1y} + T_{2y} = w$ $T_1 \sin(30^\circ) + T_2 \sin(45^\circ) = w$

Substitute the expression for $T_2$ in terms of $T_1$: $T_1 (0.500) + (1.225 T_1)(0.707) = m g$ $0.500 T_1 + 0.866 T_1 = (15.0\,\text{kg})(9.80\,\text{m/s}^2)$ $1.366 T_1 = 147\,\text{N}$ $T_1 = \frac{147\,\text{N}}{1.366} = 108\,\text{N}$

Calculate the magnitude of $T_2$: $T_2 = 1.225 (108\,\text{N}) = 132\,\text{N}$

Physical Significance

Because Cable 2 is oriented more vertically ($45^\circ$ relative to the horizontal) than Cable 1 ($30^\circ$), Cable 2 supports a greater share of the vertical load, resulting in $T_2 > T_1$. The tensions in both cables would increase substantially if the cables were stretched closer to a purely horizontal orientation. The tensions are equal if and only if the angles on both sides are identical.

Particle Acceleration

When the net external force acting on an object is non-zero, the object experiences an acceleration proportional to the net force and inversely proportional to its mass, according to Newton's second law: $\sum \mathbf{\vec{F}} = m \mathbf{\vec{a}}$

Example 6.2: Drag Force on a Barge

Two tugboats push a flat-bottomed barge through water. Tugboat 1 exerts a force of $F_1 = 2.7 \times 10^5\,\text{N}$ in the positive $x$-direction. Tugboat 2 exerts a force of $F_2 = 3.6 \times 10^5\,\text{N}$ in the positive $y$-direction. The mass of the barge is $m = 5.0 \times 10^6\,\text{kg}$. The barge accelerates at $a = 7.5 \times 10^{-2}\,\text{m/s}^2$ in the direction of the net applied tugboat force. Determine the magnitude and direction of the water's drag force $\mathbf{\vec{F}}_{\text{D}}$ resisting the motion.

Free-Body Analysis

• Vertical Plane: The downward gravitational weight of the barge and the upward buoyant force exerted by the water cancel each other completely and are omitted from planar horizontal calculations.

• Horizontal Plane: The applied forces $\mathbf{\vec{F}}_1$ and $\mathbf{\vec{F}}_2$ act along perpendicular axes. The total applied force is: $\mathbf{\vec{F}}_{\text{app}} = \mathbf{\vec{F}}_1 + \mathbf{\vec{F}}_2$

• Fluid drag $\mathbf{\vec{F}}_{\text{D}}$ acts directly opposite to the velocity and acceleration vectors.

Analytical Solution

Compute the resultant applied force magnitude using the Pythagorean theorem: $F_{\text{app}} = \sqrt{F_1^2 + F_2^2} = \sqrt{(2.7 \times 10^5\,\text{N})^2 + (3.6 \times 10^5\,\text{N})^2} = \sqrt{7.29 \times 10^{10} + 1.296 \times 10^{11}}\,\text{N} = 4.5 \times 10^5\,\text{N}$

Determine the angle $\theta$ of the applied force relative to the positive $x$-axis: $\theta = \tan^{-1}\left(\frac{F_2}{F_1}\right) = \tan^{-1}\left(\frac{3.6 \times 10^5\,\text{N}}{2.7 \times 10^5\,\text{N}}\right) = \tan^{-1}(1.333) = 53.1^\circ$

Because the barge accelerates along this line of action, the problem simplifies to a single dimension along the line of motion: $F_{\text{net}} = F_{\text{app}} - F_{\text{D}}$

Apply Newton’s second law: $F_{\text{app}} - F_{\text{D}} = m a$ $F_{\text{D}} = F_{\text{app}} - m a$ $F_{\text{D}} = (4.5 \times 10^5\,\text{N}) - (5.0 \times 10^6\,\text{kg})(7.5 \times 10^{-2}\,\text{m/s}^2)$ $F_{\text{D}} = (4.5 \times 10^5\,\text{N}) - (3.75 \times 10^5\,\text{N}) = 7.5 \times 10^4\,\text{N}$

The direction of the drag force is oriented opposite to the applied force vector, pointing at an angle of $53^\circ$ south of west (or $180^\circ + 53.1^\circ = 233.1^\circ$).

Physical Significance

The calculated drag force ($7.5 \times 10^4\,\text{N}$) is less than $1/600\text{th}$ of the gravitational weight of the barge ($w = m g \approx 4.9 \times 10^7\,\text{N}$). This relatively low resistance is consistent with a well-designed hull moving at low operational speeds to prevent harbor collisions.

Normal Force and Scale Readings in an Accelerating Elevator

The normal force is a contact force exerted perpendicular to a supporting surface. A bathroom scale registers the magnitude of the normal force ($F_{\text{s}}$) that it exerts upward to support an object resting on its plate. By Newton's third law, this equals the downward contact force ($F_{\text{p}}$) exerted by the person on the scale plate.

Example 6.3: What Does the Bathroom Scale Read in an Elevator?

A person with mass $m = 75.0\,\text{kg}$ (weight $w = m g = (75.0\,\text{kg})(9.80\,\text{m/s}^2) = 735\,\text{N}$, approximately $165\,\text{lb}$) stands on a bathroom scale inside an elevator. Determine the scale reading under two operational conditions:

1. The elevator accelerates upward at $a = 1.20\,\text{m/s}^2$.

2. The elevator moves upward at a constant velocity of $v = 1\,\text{m/s}$.

Formulation of Equations

Isolate the person as the system of interest. The two vertical forces acting on the person are:

• Upward normal contact force from the scale plate: $\mathbf{\vec{F}}_{\text{s}}$

• Downward gravitational force: $\mathbf{\vec{w}}$

Applying Newton's second law along the vertical axis: $F_{\text{net}} = F_{\text{s}} - w = m a$ $F_{\text{s}} = m a + w = m a + m g = m(g + a)$

Calculations

Condition (a): Upward Acceleration $F_{\text{s}} = (75.0\,\text{kg})(9.80\,\text{m/s}^2) + (75.0\,\text{kg})(1.20\,\text{m/s}^2)$ $F_{\text{s}} = 735\,\text{N} + 90.0\,\text{N} = 825\,\text{N}$ An upward force of $825\,\text{N}$ corresponds to an apparent weight of approximately $185\,\text{lb}$.

Condition (b): Constant Velocity When velocity is constant, the rate of change of velocity is zero: $a = \frac{\Delta v}{\Delta t} = 0$ $F_{\text{s}} = m(0) + m g = (75.0\,\text{kg})(9.80\,\text{m/s}^2) = 735\,\text{N}$ The scale registers the true static weight ($735\,\text{N}$ or $165\,\text{lb}$).

Check Your Understanding 6.1: Downward Acceleration

When the elevator accelerates downward at $a = 1.20\,\text{m/s}^2$, the acceleration vector is negative relative to the positive upward coordinate axis: $F_{\text{s}} - w = -m a$ $F_{\text{s}} = m g - m a = m(g - a)$ $F_{\text{s}} = (75.0\,\text{kg})(9.80\,\text{m/s}^2 - 1.20\,\text{m/s}^2) = (75.0\,\text{kg})(8.60\,\text{m/s}^2) = 645\,\text{N}$

• Free-Fall Limit: If the elevator cables break and the cab enters free fall ($a = g$ downward): $F_{\text{s}} = m(g - g) = 0\,\text{N}$ The scale reads zero, representing apparent weightlessness.

Multi-Body Coupled Systems

When multiple objects are mechanically connected by light, inextensible cables running over frictionless, massless pulleys, the tension is uniform throughout the connecting cable, and the magnitudes of the linear accelerations of the connected bodies are kinematically linked.

Example 6.4: Two Attached Blocks

Block 1 of mass $m_1$ rests on a frictionless horizontal plane and is attached to a light, inextensible string that passes over a frictionless, massless pulley to a suspended Block 2 of mass $m_2$.

Component Force Equations

• Block 1 (Horizontal Motion): $\sum F_x = T = m_1 a_{1x}$ $\sum F_y = N - m_1 g = 0$

• Block 2 (Vertical Motion): $\sum F_y = T - m_2 g = m_2 a_{2y}$

Kinematic Constraint

Because the string has a fixed length, the displacement of Block 1 to the right equals the downward displacement of Block 2. Defining $a$ as the common acceleration magnitude: $a_{1x} = a$ $a_{2y} = -a$

Simultaneous Solution

Substitute the constraint into the equations of motion: $T = m_1 a$ $T - m_2 g = -m_2 a$

Subtract the first equation from the second (or substitute for $T$): $m_1 a - m_2 g = -m_2 a$ $(m_1 + m_2) a = m_2 g$ $a = \left(\frac{m_2}{m_1 + m_2}\right) g$

Substitute $a$ back to solve for cable tension $T$: $T = \left(\frac{m_1 m_2}{m_1 + m_2}\right) g$

Physical Significance

The tension in the string is strictly less than the gravitational weight of the hanging block ($T < m_2 g$). Setting $T = m_2 g$ is a common conceptual error that is valid only when the hanging mass is in static equilibrium ($a = 0$). An accelerating suspended mass requires an unbalanced net force ($m_2 g - T > 0$).

Check Your Understanding 6.2

Calculate the acceleration and tension when $m_1 = 5.00\,\text{kg}$ and $m_2 = 3.00\,\text{kg}$: $a = \left(\frac{3.00\,\text{kg}}{5.00\,\text{kg} + 3.00\,\text{kg}}\right)(9.80\,\text{m/s}^2) = \left(\frac{3.00}{8.00}\right)(9.80\,\text{m/s}^2) = 3.68\,\text{m/s}^2$ $T = \left(\frac{(5.00\,\text{kg})(3.00\,\text{kg})}{5.00\,\text{kg} + 3.00\,\text{kg}}\right)(9.80\,\text{m/s}^2) = \left(\frac{15.00}{8.00}\right)(9.80\,\text{N}) = 18.4\,\text{N}$

Example 6.5: The Atwood Machine

An Atwood machine consists of two unequal masses, $m_1$ and $m_2$, suspended vertically on opposite sides of a frictionless, massless pulley by an inextensible string. Let $m_1 = 2.00\,\text{kg}$ and $m_2 = 4.00\,\text{kg}$.

Formulation

Assuming $m_2 > m_1$, Block 2 accelerates downward ($a_2 = -a$) while Block 1 accelerates upward ($a_1 = +a$). The dynamic equations for the isolated masses are:

• For Block 1: $\sum F_y = T - m_1 g = m_1 a$

• For Block 2: $\sum F_y = T - m_2 g = -m_2 a$

Analytical Solution

Subtract the equation for Block 2 from the equation for Block 1: $(T - m_1 g) - (T - m_2 g) = m_1 a - (-m_2 a)$ $(m_2 - m_1) g = (m_1 + m_2) a$ $a = \left(\frac{m_2 - m_1}{m_1 + m_2}\right) g$

Substitute the numerical parameters: $a = \left(\frac{4.00\,\text{kg} - 2.00\,\text{kg}}{4.00\,\text{kg} + 2.00\,\text{kg}}\right)(9.80\,\text{m/s}^2) = \left(\frac{2.00\,\text{kg}}{6.00\,\text{kg}}\right)(9.80\,\text{m/s}^2) = 3.27\,\text{m/s}^2$

Calculate the string tension $T$ using the equation for Block 1: $T = m_1(g + a) = (2.00\,\text{kg})(9.80\,\text{m/s}^2 + 3.27\,\text{m/s}^2) = (2.00\,\text{kg})(13.07\,\text{m/s}^2) = 26.1\,\text{N}$

Significance and Gravity Measurement

The acceleration represents the ratio of the net unbalanced gravitational force on the system, $(m_2 - m_1)g$, to the total inertia of the system, $(m_1 + m_2)$. Historically, the Atwood machine provided an accurate laboratory method to dilute gravitational acceleration, allowing precise measurements of the local gravitational field strength $g$.

Check Your Understanding 6.3

Derive a general formula for the string tension $T$ in an Atwood machine in terms of $m_1$, $m_2$, and $g$: $T = m_1(g + a) = m_1\left[g + \left(\frac{m_2 - m_1}{m_1 + m_2}\right)g\right] = m_1 g \left[1 + \frac{m_2 - m_1}{m_1 + m_2}\right]$ $T = m_1 g \left[\frac{(m_1 + m_2) + (m_2 - m_1)}{m_1 + m_2}\right] = \left(\frac{2 m_1 m_2}{m_1 + m_2}\right) g$

Integrating Newton’s Laws with Kinematics

Newton’s second law provides the link between forces and motion by determining acceleration, which can then be used in kinematic equations.

Example 6.6: Horizontal Force of a Sprinter

A soccer player of mass $m = 70.0\,\text{kg}$ starts from rest ($v_0 = 0$) and accelerates forward in a straight line, reaching a top speed of $v = 8.00\,\text{m/s}$ in $\Delta t = 2.50\,\text{s}$. Assume air resistance is negligible.

Kinematic Acceleration

$a = \frac{\Delta v}{\Delta t} = \frac{8.00\,\text{m/s} - 0\,\text{m/s}}{2.50\,\text{s}} = 3.20\,\text{m/s}^2$

Force Calculation

By Newton’s third law, the forward force driving the runner is the reaction force exerted forward by the ground against the player's feet as she pushes backward against the ground. Applying Newton’s second law: $F_{\text{net}} = m a = (70.0\,\text{kg})(3.20\,\text{m/s}^2) = 224\,\text{N}$ An average forward force of $224\,\text{N}$ is approximately $50\,\text{lb}$, which is typical for a trained athlete.

Check Your Understanding 6.4

The player comes to a complete stop, then spots a ball located $d = 2.00\,\text{m}$ away. If she generates a net forward force of $F = 126\,\text{N}$, find the time required to reach the ball from rest: $a = \frac{F}{m} = \frac{126\,\text{N}}{70.0\,\text{kg}} = 1.80\,\text{m/s}^2$ $d = v_0 t + \frac{1}{2} a t^2 = 0 + \frac{1}{2} a t^2$ $t = \sqrt{\frac{2 d}{a}} = \sqrt{\frac{2(2.00\,\text{m})}{1.80\,\text{m/s}^2}} = \sqrt{\frac{4.00}{1.80}} = \sqrt{2.222} = 1.49\,\text{s}$

Example 6.7: Two-Dimensional Vector Dynamics of a Model Helicopter

A $1.50\,\text{kg}$ model helicopter has an initial velocity of $\mathbf{\vec{v}}_0 = 5.00\,\mathbf{\hat{j}}\,\text{m/s}$ at $t = 0$. It experiences constant acceleration over a duration of $\Delta t = 2.00\,\text{s}$, reaching a final velocity of $\mathbf{\vec{v}} = (6.00\,\mathbf{\hat{i}} + 12.00\,\mathbf{\hat{j}})\,\text{m/s}$. Determine the magnitude of the net resultant force.

Acceleration Vector

$\Delta \mathbf{\vec{v}} = \mathbf{\vec{v}} - \mathbf{\vec{v}}_0 = (6.00\,\mathbf{\hat{i}} + 12.00\,\mathbf{\hat{j}})\,\text{m/s} - 5.00\,\mathbf{\hat{j}}\,\text{m/s} = (6.00\,\mathbf{\hat{i}} + 7.00\,\mathbf{\hat{j}})\,\text{m/s}$ $\mathbf{\vec{a}} = \frac{\Delta \mathbf{\vec{v}}}{\Delta t} = \frac{6.00\,\mathbf{\hat{i}} + 7.00\,\mathbf{\hat{j}}\,\text{m/s}}{2.00\,\text{s}} = (3.00\,\mathbf{\hat{i}} + 3.50\,\mathbf{\hat{j}})\,\text{m/s}^2$

Force Vector and Magnitude

$\sum \mathbf{\vec{F}} = m \mathbf{\vec{a}} = (1.50\,\text{kg})(3.00\,\mathbf{\hat{i}} + 3.50\,\mathbf{\hat{j}}\,\text{m/s}^2) = (4.50\,\mathbf{\hat{i}} + 5.25\,\mathbf{\hat{j}})\,\text{N}$ $F = \sqrt{(4.50\,\text{N})^2 + (5.25\,\text{N})^2} = \sqrt{20.25 + 27.5625}\,\text{N} = \sqrt{47.8125}\,\text{N} = 6.91\,\text{N}$

Check Your Understanding 6.5

Calculate the direction angle $\theta$ of the resultant force relative to the positive horizontal axis ($\mathbf{\hat{i}}$): $\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right) = \tan^{-1}\left(\frac{5.25\,\text{N}}{4.50\,\text{N}}\right) = \tan^{-1}(1.1667) = 49.4^\circ$

Advanced Dynamics with Calculus

When forces vary with time, position, or velocity, acceleration is not constant, and standard kinematic equations cannot be used. Instead, calculus must be applied directly to Newton's second law.

Time-Dependent Forces

When net force is a function of time, $F(t) = m a(t)$, acceleration is determined by differentiation or integration: $a(t) = \frac{dv}{dt}$ $v(t) = v_0 + \int_{t_0}^t a(t')\,dt'$

Example 6.8: Baggage Tractor and Luggage Carts

A baggage tractor of mass $m_{\text{tractor}} = 650.0\,\text{kg}$ pulls two carts along a straight tarmac: Cart A with mass $m_{\text{A}} = 250.0\,\text{kg}$ and Cart B with mass $m_{\text{B}} = 150.0\,\text{kg}$. The driving tractive force increases linearly with time according to $F(t) = (820.0 t)\,\text{N}$. The system starts from rest ($v_0 = 0$ at $t = 0$).

Part (a): Speed at $t = 3.00\,\text{s}$

Analyze the entire train as a unified system of interest: $m_{\text{system}} = m_{\text{tractor}} + m_{\text{A}} + m_{\text{B}} = 650.0\,\text{kg} + 250.0\,\text{kg} + 150.0\,\text{kg} = 1050.0\,\text{kg}$ $\sum F_x = 820.0 t = m_{\text{system}} a(t)$ $820.0 t = 1050.0 a(t)$ $a(t) = \frac{820.0}{1050.0} t = 0.7809 t\,\text{m/s}^2$

Integrate acceleration to determine velocity: $dv = a(t)\,dt$ $\int_0^v dv = \int_0^{3.00} 0.7809 t\,dt$ $v = \left[0.3905 t^2\right]_0^{3.00} = 0.3905 (3.00)^2 = 0.3905 (9.00) = 3.51\,\text{m/s}$

Part (b): Coupling Tension Between Tractor and Cart A at $t = 3.00\,\text{s}$

Isolate the tractor as an individual subsystem to reveal the internal coupling force $T$ exerted on it by Cart A: $\sum F_x = F(t) - T = m_{\text{tractor}} a(t)$ $820.0 t - T = (650.0)(0.7805 t)$

Evaluate at $t = 3.00\,\text{s}$: $(820.0)(3.00) - T = (650.0)(0.7805)(3.00)$ $2460.0 - T = 1521.975$ $T = 2460.0\,\text{N} - 1522.0\,\text{N} = 938\,\text{N}$

Position- and Velocity-Dependent Forces

When acceleration depends on displacement or velocity, eliminating the time parameter $dt$ is necessary. By the chain rule of calculus: $v = \frac{ds}{dt} \implies dt = \frac{ds}{v}$ $a = \frac{dv}{dt} \implies dt = \frac{dv}{a}$ $\frac{ds}{v} = \frac{dv}{a}$ $a\,ds = v\,dv$

Example 6.9: Projectile Fired Vertically with Quadratic Air Drag

A mortar shell of mass $m = 10.0\,\text{kg}$ is launched vertically upward from ground level ($y_0 = 0$) with an initial velocity of $v_0 = 50.0\,\text{m/s}$. During its upward flight, atmospheric drag opposes the motion according to: $F_{\text{D}} = (0.0100 v^2)\,\text{N}$ Calculate the maximum altitude $h$ reached by the projectile (where $v = 0$).

Equation of Motion

Isolate the mortar shell during its ascent. Both gravity and air resistance act downward: $\sum F_y = -F_{\text{D}} - w = m a_y$ $-0.0100 v^2 - m g = m a$ $-0.0100 v^2 - (10.0\,\text{kg})(9.80\,\text{m/s}^2) = 10.0 a$ $-0.0100 v^2 - 98.0 = 10.0 a$ $a = -0.00100 v^2 - 9.80$

Integration for Maximum Height

Substitute the expression for acceleration into the differential relation $a\,dy = v\,dv$: $(-0.00100 v^2 - 9.80)\,dy = v\,dv$ $dy = \frac{v\,dv}{-0.00100 v^2 - 9.80} = -\frac{v\,dv}{0.00100 v^2 + 9.80}$

Integrate from the launch position ($y = 0$, $v = 50.0\,\text{m/s}$) to the peak altitude ($y = h$, $v = 0$): $\int_0^h dy = -\int_{50.0}^0 \frac{v\,dv}{0.00100 v^2 + 9.80} = \int_0^{50.0} \frac{v\,dv}{0.00100 v^2 + 9.80}$

Use the standard logarithmic integral form $\int \frac{u}{a u^2 + b}\,du = \frac{1}{2a} \ln(a u^2 + b)$: $\int_0^h dy = \left[\frac{1}{2(0.00100)} \ln(0.00100 v^2 + 9.80)\right]_0^{50.0}$ $h = 500 \left[\ln(0.00100(50.0)^2 + 9.80) - \ln(0.00100(0)^2 + 9.80)\right]$ $h = 500 \left[\ln(0.00100(2500) + 9.80) - \ln(9.80)\right]$ $h = 500 \left[\ln(2.50 + 9.80) - \ln(9.80)\right]$ $h = 500 [\ln(12.30) - \ln(9.80)]$ $h = 500 [2.5096 - 2.2824] = 500 (0.2272) = 113.6\,\text{m} \approx 114\,\text{m}$

Physical Significance

In the absence of air drag, a projectile launched at $50.0\,\text{m/s}$ would reach a theoretical vacuum height of: $h_{\text{vacuum}} = \frac{v_0^2}{2g} = \frac{(50.0\,\text{m/s})^2}{2(9.80\,\text{m/s}^2)} = \frac{2500}{19.6} = 127.6\,\text{m}$ Atmospheric drag reduces the peak altitude to $114\,\text{m}$. Because fluid resistance varies continuously with the square of the speed ($v^2$), evaluating the trajectory requires this calculus-based formulation.