Newton's Laws Applications: Friction, Circular Motion, and Drag Forces Study Guide

Applications of Newton’s Laws: Friction, Circular Motion, and Drag

This chapter expands on Newton's laws to cover practical and complex physical situations, including friction, uniform and nonuniform circular motion, and velocity-dependent drag forces.
Friction and circular motion are fundamental to understanding the mechanics of everyday objects and celestial bodies, while drag forces provide an advanced look at motion through fluids.

Understanding Friction at the Microscopic Level

Friction acts between two solid surfaces in contact.
Origin of Friction: Even surfaces that appear smooth are microscopically rough. These microscopic bumps and ridges impede motion when one surface tries to slide over another.
Atomic "Welds": One leading theory suggests that atoms on the peaks of surface bumps come so close to atoms on the opposing surface that they form brief, tiny welds or chemical bonds.
Jerky Motion: The observed "jerky" nature of sliding is attributed to the continuous making and breaking of these microscopic atomic bonds.
Rolling Friction: This occurs when a round object rolls across a surface. While still present, rolling friction is generally significantly less than sliding friction.

Kinetic and Static Friction Principles

Kinetic Friction: Sliding friction that acts when an object is in motion relative to a surface. Its direction is always opposite to the object's velocity.
Proportionality: The magnitude of kinetic friction ( $F_{\text{fr}}$ ) is approximately proportional to the magnitude of the normal force ( $F_N$ ).
Normal Force ( $F_N$ ): The force exerted perpendicular to the common surface of contact between two objects.
Area Independence: Experiments show that the friction force depends very little on the total surface area of contact (e.g., a book sliding on its face vs. its spine experiences roughly the same friction).
Kinetic Friction Equation: $F_{\text{fr}} = \mu_k F_N$ .
Coefficient of Kinetic Friction ( $\mu_k$ ): A dimensionless constant whose value depends on the material and roughness of the surfaces. It is generally independent of sliding speed and contact area.
Static Friction: A force parallel to surfaces that prevents them from sliding when a horizontal force is applied to an object at rest.
Variable Nature: Static friction increases to exactly match the applied force ( $F_A$ ) as long as the object remains at rest, such that the net force remains zero.
Maximum Static Friction: Once a threshold is reached, the object begins to move, and kinetic friction takes over. This threshold is $F_{\text{fr}}^{\text{max}} = \mu_s F_N$ .
Static Friction Inequality: $F_{\text{fr}} \le \mu_s F_N$ .
Comparison: In most cases, it is harder to start an object moving than to keep it moving because $\mu_s > \mu_k$ .
Historical Credit: Leonardo da Vinci ( $1452-1519$ ) was the first to establish the relationship $F_{\text{fr}} \propto F_N$ .

Table 5-1: Approximate Coefficients of Friction

Wood on wood: $\mu_k = 0.2$ , $\mu_s = 0.4$
Ice on ice: $\mu_k = 0.03$ , $\mu_s = 0.1$
Metal on metal (lubricated): $\mu_k = 0.07$ , $\mu_s = 0.1$
Steel on steel (unlubricated): $\mu_k = 0.6$ , $\mu_s = 0.7$
Rubber on dry concrete: $\mu_k = 0.8$ , $\mu_s = 1.0$
Rubber on wet concrete: $\mu_k = 0.5$ , $\mu_s = 0.7$
Rubber on other solid surfaces: $\mu_k \approx 1$ , $\mu_s = 1-4$
Teflon on Teflon (in air): $\mu_k = 0.04$ , $\mu_s = 0.04$
Teflon on steel (in air): $\mu_k = 0.04$ , $\mu_s = 0.04$
Lubricated ball bearings: $\mu_k < 0.01$ , $\mu_s < 0.01$
Synovial joints (human limbs): $\mu_k < 0.01$ , $\mu_s < 0.01$

Example 5-1: Static and Kinetic Friction on a 10.0-kg Box

Setup: A $10.0\,kg$ box rests on a horizontal floor ( $\mu_s = 0.40$ , $\mu_k = 0.30$ ).
Normal Force: $\Sigma F_y = F_N - mg = 0 \rightarrow F_N = (10.0\,kg)(9.80\,m/s^2) = 98.0\,N$ .
Maximum Static Friction: $\mu_s F_N = (0.40)(98.0\,N) = 39.2\,N$ (rounded to $39\,N$ ).
Case (a): If $F_A = 0$ , then $F_{\text{fr}} = 0$ .
Case (b): If $F_A = 10\,N$ , box remains at rest; $\Sigma F_x = 10\,N - F_{\text{fr}} = 0$ , so $F_{\text{fr}} = 10\,N$ .
Case (c): If $F_A = 20\,N$ , box remains at rest; $F_{\text{fr}} = 20\,N$ .
Case (d): If $F_A = 38\,N$ , box remains at rest; $F_{\text{fr}} = 38\,N$ .
Case (e): If $F_A = 40\,N$ , it exceeds the maximum static friction of $39.2\,N$ . The box moves.
Kinetic friction in Case (e): $F_{\text{fr}} = \mu_k F_N = (0.30)(98.0\,N) = 29.4\,N$ (rounded to $29\,N$ ).
Resulting Acceleration: $\Sigma F = 40\,N - 29.4\,N = 10.6\,N$ . Acceleration $a = \frac{10.6\,N}{10.0\,kg} = 1.06\,m/s^2$ , rounded to $1.1\,m/s^2$ .

Practical Applications and Reductions of Friction

Friction as a Hindrance: Causes heating, wear, and mechanical binding.
Reduction Techniques:
- Lubricants like oil.
- Maintaining a thin layer of gas/air between surfaces (used in air tracks and air tables).
- Magnetic Levitation: Suspending objects using magnetic fields to eliminate surface contact.
Friction as a Benefit:
- Essential for walking (specifically static friction between shoes and ground).
- Necessary for car movement and stability; driving on ice illustrates the danger of low friction.

Example 5-2: Holding a Box Against a Wall

Concept: Pressing a box horizontally against a wall to prevent it from sliding down.
Mechanism: The applied horizontal force creates a horizontal normal force from the wall ( $F_N = F_{\text{applied}}$ ). This normal force determines the maximum static friction ( $F_{\text{fr}} = \mu_s F_N$ ) which acts upward to balance gravity ( $mg$ ).
Failure Condition: If the applied force is too low, $mg > \mu_s F_N$ , and the box slips.
Exercise A: If $\mu_s = 0.40$ and $mg = 20\,N$ , what minimum horizontal force $F$ is required?
- Calculation: $F_{\text{fr}} = mg = 20\,N$ . Since $F_{\text{fr}} = \mu_s F_N$ , then $F_N = \frac{20\,N}{0.40} = 50\,N$ .

Example 5-3: Pulling a Box with Friction at an Angle

Setup: A $10.0\,kg$ box pulled by $F_p = 40.0\,N$ at $30.0^{\circ}$ above horizontal ( $\mu_k = 0.30$ ).
Components of Applied Force:
- $F_{px} = (40.0\,N) \cos(30.0^{\circ}) = 34.6\,N$
- $F_{py} = (40.0\,N) \sin(30.0^{\circ}) = 20.0\,N$
Vertical Equation: $F_N - mg + F_{py} = 0 \rightarrow F_N = 98.0\,N - 20.0\,N = 78.0\,N$ .
Kinetic Friction: $F_{\text{fr}} = \mu_k F_N = (0.30)(78.0\,N) = 23.4\,N$ .
Horizontal Acceleration: $a_x = \frac{F_{px} - F_{\text{fr}}}{m} = \frac{34.6\,N - 23.4\,N}{10.0\,kg} = 1.12\,m/s^2$ , rounded to $1.1\,m/s^2$ for significant figures.
Exercise B: If $\mu_k F_N > F_{px}$ , one would conclude the box does not move or decelerates if already in motion.

Example 5-4: Pushing vs. Pulling a Sled

Scenario: Moving a sled on flat snow at an angle.
Pushing: Your force has a downward component, which increases the normal force ( $F_N > mg$ ). This increases the friction force.
Pulling: Your force has an upward component, which decreases the normal force ( $F_N < mg$ ). This decreases the friction force.
Conclusion: Pulling is more efficient because it requires less force to overcome friction.

Example 5-5: Two Boxes, a Pulley, and Kinetic Friction

Setup: Box A ( $5.0\,kg$ ) on a table connected to Box B ( $2.0\,kg$ ) hanging vertically. $\mu_k = 0.20$ .
Box A Forces:
- $F_N = m_Ag = (5.0\,kg)(9.8\,m/s^2) = 49\,N$
- $F_{\text{fr}} = \mu_k F_N = (0.20)(49\,N) = 9.8\,N$
- Newton's Second Law: $F_T - F_{\text{fr}} = m_Aa$
Box B Forces:
- $m_Bg = (2.0\,kg)(9.8\,m/s^2) = 19.6\,N$
- Newton's Second Law: $m_Bg - F_T = m_Ba$
Solving for Acceleration ( $a$ ):
- $(m_Bg - m_Ba) - F_{\text{fr}} = m_Aa \rightarrow a = \frac{m_Bg - F_{\text{fr}}}{m_A + m_B}$
- $a = \frac{19.6\,N - 9.8\,N}{5.0\,kg + 2.0\,kg} = 1.4\,m/s^2$
Tension ( $F_T$ ): $F_T = 9.8\,N + (5.0\,kg)(1.4\,m/s^2) = 17.0\,N$ . Note that $F_T \neq m_Bg$ when the system accelerates.

Example 5-6: Forces and Acceleration of a Skier on a Slope

Setup: Skier on a $30^{\circ}$ slope with $\mu_k = 0.10$ .
Coordinate System: $x$ -axis is downslope; $y$ -axis is perpendicular to the slope.
Weight Components:
- $F_{gx} = mg \sin(\theta)$
- $F_{gy} = mg \cos(\theta)$
Normal Force: $\Sigma F_y = F_N - mg \cos(\theta) = 0 \rightarrow F_N = mg \cos(\theta)$ .
Downslope Acceleration: $\Sigma F_x = mg \sin(\theta) - \mu_k (mg \cos(\theta)) = ma_x$ .
Mass Cancellation: $a_x = g (\sin(\theta) - \mu_k \cos(\theta)) = (9.8\,m/s^2) [\sin(30^{\circ}) - (0.10) \cos(30^{\circ})]$ .
Numerical Result: $a = 0.41g = 4.0\,m/s^2$ .
Skiing Physics: High speed friction is very low because heat melts snow, allowing skis to float on water. Air resistance (drag) becomes the dominant slowing force, reduced by "tucking."
Caution: Always remember that $F_N$ is perpendicular to the slope while gravity acts vertically downward.

Example 5-7: Systems with Ramps, Pulleys, and Two Boxes

Setup: Box A ( $10.0\,kg$ ) on a $37^{\circ}$ incline connected to Box B ( $m_B$ ) which hangs freely ( $\mu_s = 0.40$ , $\mu_k = 0.30$ ).
Static Case Range:
- Minimum $m_B$ (Prevent slide down): $m_B = m_A (\sin(\theta) - \mu_s \cos(\theta)) = 10.0\,kg (\sin(37^{\circ}) - 0.40 \cos(37^{\circ})) = 2.8\,kg$ .
- Maximum $m_B$ (Prevent pull up): $m_B = m_A (\sin(\theta) + \mu_s \cos(\theta)) = 10.0\,kg (\sin(37^{\circ}) + 0.40 \cos(37^{\circ})) = 9.2\,kg$ .
- Range for stability: $2.8\,kg < m_B < 9.2\,kg$ .
Kinetic Case: If $m_B = 10.0\,kg$ , Box A slides up.
- Acceleration formula: $a = \frac{m_Bg - m_Ag \sin(\theta) - \mu_k m_Ag \cos(\theta)}{m_A + m_B}$ .
- Result: $a = \frac{(10.0)(9.8)(1 - \sin(37^{\circ}) - 0.30 \cos(37^{\circ}))}{20.0} = 0.78\,m/s^2$ .

Kinematics of Uniform Circular Motion

Definition: Motion in a circle at a constant speed ( $v$ ). While speed is constant, velocity is not because its direction changes continuously.
Acceleration ( $\mathbf{a}$ ): Since velocity change is an acceleration, revolving objects are always accelerating.
Direction: The acceleration vector points toward the center of the circle, called centripetal or radial acceleration ( $a_R$ ).
Geometric Similarity Derivation: Using similar triangles from velocity vectors ( $\Delta v/v$ ) and position vectors ( $\Delta l/r$ ), we find:
- $a_R = \lim_{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t} = \frac{v^2}{r}$ .
Vector Relationship: Velocity is always tangential to the path, meaning $\mathbf{a}_R$ and $\mathbf{v}$ are perpendicular ( $90^{\circ}$ ) at every point in uniform circular motion.
Caution: Speed is constant, but acceleration is NOT zero.

Fundamental Relationships in Circular Motion: Period and Frequency

Frequency ( $f$ ): Number of revolutions per second. Unit: $s^{-1}$ or $Hz$ .
Period ( $T$ ): Time required for one complete revolution.
Mathematical relationship: $T = \frac{1}{f}$ .
Orbital Speed: Over one period, an object travels the circumference ( $2\pi r$ ).
- $v = \frac{2\pi r}{T}$ .

Examples 5-8 and 5-9: Revolving Ball and Moon's Acceleration

Example 5-8: Ball on a string ( $r = 0.600\,m$ ) making $2.00$ revolutions per second.
- $T = 0.500\,s$
- $v = \frac{2\pi (0.600\,m)}{0.500\,s} = 7.54\,m/s$
- $a_R = \frac{(7.54\,m/s)^2}{0.600\,m} = 94.7\,m/s^2$
Exercise D: Doubling the radius while keeping the period same changes acceleration by a factor of $2$ (since $a_R = \frac{4\pi^2 r}{T^2}$ , $a_R \propto r$ ).
Example 5-9: Moon orbiting Earth ( $r = 3.84 \times 10^8\,m$ , $T = 27.3\,days = 2.36 \times 10^6\,s$ ).
- $a_R = \frac{4\pi^2 (3.84 \times 10^8\,m)}{(2.36 \times 10^6\,s)^2} = 0.00272\,m/s^2$ .
- In terms of gravity: $a_R = 2.78 \times 10^{-4}g$ .

Centrifugation and the Ultracentrifuge

Purpose: Sediment materials and separate particles (e.g., macromolecules) using high speeds.
Mechanics: The fluid exerts a centripetal force, but if results are insufficient, particles move toward the bottom of the tube.
Example 5-10: Ultracentrifuge at $50,000\,rpm$ and $r = 6.00\,cm$ .
- $f = \frac{50,000}{60} = 833\,rev/s$
- $T = 1.20 \times 10^{-3}\,s$
- $v = \frac{2\pi (0.0600\,m)}{1.20 \times 10^{-3}\,s} = 3.14 \times 10^2\,m/s$
- $a_R = \frac{(314)^2}{0.0600} = 1.64 \times 10^6\,m/s^2 \approx 167,000g$ .
Other applications: Astronaut training at several $g$ 's and Uranium enrichment ( $U-235$ vs $U-238$ ).

Dynamics of Uniform Circular Motion

Centripetal Force: A net force is required to provide centripetal acceleration ( $a_R$ ) per Newton's Second Law:
- $\Sigma F_R = ma_R = m \frac{v^2}{r}$ .
Nature of the Force: "Centripetal" is not a special type of force; it is a direction label for established forces like tension, gravity, friction, or normal force that act toward the center.
Newton's First Law: Without this net force, the object would move in a straight line, not a circle.

The Misconception of Centrifugal Force

Centrifugal ("center-fleeing") force is NOT a real force acting on the revolving object.
Explanation: When swinging a ball, your hand feels a pull because the string pulls on your hand (Newton’s Third Law). To the revolving ball, the only net force is the inward tension from the string.
Evidence: If the string breaks, the ball does not fly radially outward; it flies off tangentially in a straight line (inertia).

Example 5-11: Horizontal Force on a Revolving Ball

Estimate the force for a $0.150\,kg$ ball, $r = 0.600\,m$ , $2.00\,rev/s$ .
Vertical Complication: A ball cannot revolve in a perfectly horizontal circle because some vertical component of tension is needed to balance gravity ( $mg$ ).
Scaling: $mg = 1.5\,N$ . The calculated centripetal force is $F_T \approx 14\,N$ . Because $mg$ is small ( $\approx 1/10$ ) relative to $F_T$ , we can estimate the tension as roughly horizontal.

Example 5-12: Vertical Circular Motion

Setup: $0.150\,kg$ ball on a $1.10\,m$ cord in a vertical circle.
Case (a) - At the Top (Point 1): Gravity and tension both act downward ( $+$ direction).
- $\Sigma F_R = F_{T1} + mg = m\frac{v_1^2}{r}$ .
- Minimum Speed: At the minimum speed, the cord goes limp (tension $F_{T1} = 0$ ).
- $mg = m\frac{v^2}{r} \rightarrow v = \sqrt{gr} = \sqrt{(9.80)(1.10)} = 3.28\,m/s$ .
Case (b) - At the Bottom (Point 2): Tension acts upward ( $+$ ), gravity downward ( $-$ ).
- Assume $v_2 = 2 v_{min} = 6.566\,m/s$ .
- $\Sigma F_R = F_{T2} - mg = m\frac{v_2^2}{r}$ .
- $F_{T2} = m (\frac{v_2^2}{r} + g) = 0.150 (\frac{6.566^2}{1.10} + 9.80) = 7.35\,N$ .

Problem-Solving Strategy for Uniform Circular Motion

Draw a free-body diagram showing all physical sources of force.
Identify which forces or components act radially (toward/away from the center).
Sum the radial forces ( $\Sigma F_R$ ) and set them equal to $ma_R = m\frac{v^2}{r}$ .
Choose a coordinate system with one axis along the acceleration direction.

Dynamics of Highway Curves: Banked and Unbanked

Curved Motion: To round a curve, a car must have a force acting toward the center. On a flat road, this is provided by static friction between the tires and pavement.
Static vs Kinetic Friction: As long as the tires roll without slipping, static friction applies. If speed is too high, the required force exceeds $\mu_s F_N$ , and the car skids.
Skidding: Once skidding, kinetic friction (which is smaller and acts opposite to velocity) takes over, causing the car to leave its circular path.
Anti-lock Brakes (ABS): Designed to prevent wheels from locking, maintaining static friction for better directional control.
Banked Curves: The road is tilted at an angle $\theta$ . The normal force ( $F_N$ ) has a horizontal component $F_N \sin(\theta)$ that provides centripetal acceleration.
Design Speed: The speed at which no friction is required to stay on the path.
- $F_N \sin(\theta) = m \frac{v^2}{r}$
- $F_N \cos(\theta) = mg$
- $\tan(\theta) = \frac{v^2}{rg}$ .

Example 5-14: Analyzing Skidding on a Curve

Conditions: $1000\,kg$ car, $r = 50\,m$ , $v = 15\,m/s$ .
Required Force: $F_R = m\frac{v^2}{r} = 1000 \frac{15^2}{50} = 4500\,N$ .
Case (a) - Dry Pavement ( $\mu_s = 0.60$ ):
- Max static friction: $\mu_s F_N = (0.60)(9800\,N) = 5880\,N$ . Since $5880 > 4500$ , the car stays on track.
Case (b) - Icy Pavement ( $\mu_s = 0.25$ ):
- Max static friction: $\mu_s F_N = (0.25)(9800\,N) = 2450\,N$ . Since $2450 < 4500$ , the car will skid.

Nonuniform Circular Motion and Tangential Acceleration

Non-Centric Net Force: If the net force acts at an angle, it has two components:
- Radial Component ( $F_R$ ): Responsible for centripetal (radial) acceleration ( $a_R = v^2/r$ ).
- Tangential Component ( $F_{\text{tan}}$ ): Responsible for tangential acceleration ( $a_{\text{tan}} = dv/dt$ ) which changes the object's speed.
Total Acceleration: The vector sum of both components.
- $\mathbf{a} = \mathbf{a}_R + \mathbf{a}_{\text{tan}}$
- $a = \sqrt{a_R^2 + a_{\text{tan}}^2}$ .
Direction: $a_{\text{tan}}$ is parallel to velocity if speed is increasing, and antiparallel if speed is decreasing.

Example 5-15: Acceleration Components of a Race Car

Setup: Car starts from rest, reaches $35\,m/s$ in $11\,s$ on track with $r = 250\,m$ .
At instant $v = 15\,m/s$ :
- Tangential Acceleration: $a_{\text{tan}} = \frac{\Delta v}{\Delta t} = \frac{35 - 0}{11} = 3.2\,m/s^2$ .
- Radial Acceleration: $a_R = \frac{15^2}{250} = 0.90\,m/s^2$ .

Analysis of Curvilinear Motion

General Curved Paths: Any curved path can be treated locally as an arc of a circle with a specific radius of curvature ( $r$ ).
The same principles of tangential ( $a_{\text{tan}} = dv/dt$ ) and radial ( $a_R = v^2/r$ ) acceleration apply to any point $P$ along the curve.

Velocity-Dependent Forces: Drag and Terminal Velocity

Overview: Unlike surface friction, drag in fluids (gases/liquids) depends on velocity.
Low-Speed Drag: For small objects or viscous liquids, drag force ( $F_D$ ) is proportional to velocity:
- $F_D = -bv$ (where $b$ is a constant based on viscosity and shape).
High-Speed Drag: For objects like planes or skydivers, drag is proportional to velocity squared:
- $F_D \propto v^2$ .
Dynamics of Falling: For an object falling from rest with $F_D = -bv$ :
- $\Sigma F = mg - bv = m\frac{dv}{dt}$ .
Terminal Velocity ( $v_T$ ): As speed increases, drag increases until it equals gravity. At this point, net force and acceleration are zero.
- $mg - bv_T = 0 \rightarrow v_T = \frac{mg}{b}$ .

Mathematical Derivation for Falling with Linear Drag

Differential Equation: $\frac{dv}{dt} = g - \frac{b}{m}v$ .
Variable Separation: $\int_0^v \frac{dv}{g - \frac{bv}{m}} = \int_0^t dt$ .
Integration Result: $\ln(1 - \frac{bv}{mg}) = -\frac{bt}{m}$ .
Velocity Function: $v(t) = \frac{mg}{b} (1 - e^{-\frac{bt}{m}})$ .
Time Constant: When $t = \frac{m}{b}$ , the velocity reaches approximately $63\%$ of the terminal velocity (since $e^{-1} \approx 0.37$ ).
Initial State: At $t=0$ , $v=0$ and $a=g$ .
Asymptote: As $t \rightarrow \infty$ , $v \rightarrow v_T = \frac{mg}{b}$ .