College Physics Study Notes: Application of Newton's Laws

Two-Dimensional Vectors and Force Components

• When applying Newton's second law to situations where forces do not lie strictly along a single coordinate axis, vector quantities must be resolved into perpendicular component vectors along chosen axes.

• Vector Component Resolution:

  • Any arbitrary vector, such as a displacement or force vector $\vec{F}$, can be represented as the vector sum of two mutually perpendicular vectors: an $\text{x}$-component vector $\vec{F}_x$ and a $\text{y}$-component vector $\vec{F}_y$.
  • Graphically, vector resolution is the inverse of vector addition. For example, traveling along Route 1 (a direct path represented by vector $\vec{C}$) reaches the exact same final destination as traveling along Route 2 (two sequential perpendicular paths represented by displacement vectors $\vec{A}$ and $\vec{B}$), such that:

$\vec{C} = \vec{A} + \vec{B}$

• Scalar Components vs. Vector Components:

  • Vector components ($\vec{F}_x$ and $\vec{F}_y$) possess both magnitude and direction.
  • Scalar components ($F_x$ and $F_y$) are signed real numbers (scalars) indicating magnitude and directional orientation along the coordinate axes:
  • A scalar component is positive if its corresponding vector component points in the positive axis direction.
  • A scalar component is negative if its corresponding vector component points in the negative axis direction.
  • Example: A force vector $\vec{F}$ of magnitude $5\,\text{N}$ directed down and to the left at $37^\circ$ below the negative $\text{x}$-axis in the third quadrant has scalar components $F_x = -4\,\text{N}$ and $F_y = +3\,\text{N}$ if the vertical component points upward and horizontal component points leftward.

• Calculating Scalar Components from Magnitude and Angle:

  • To calculate scalar components from a vector magnitude $F$ and angle $\theta$ ($90^\circ$ or less) measured with respect to the $\pm\text{x}$-axis:

$F_x = \pm F \cos(\theta)$

$F_y = \pm F \sin(\theta)$

• Physics Tool Box 4.1: Systematic Protocol for Determining Scalar Components:
  1. Draw an arrow representing the original vector $\vec{F}$.
  2. Choose and draw perpendicular coordinate axes (e.g., an $\text{x}$-axis and a $\text{y}$-axis).
  3. Represent vector $\vec{F}$ as the vector sum of two perpendicular component vectors $\vec{F}_x$ and $\vec{F}_y$.
  4. Apply right-triangle trigonometry to compute scalar component magnitudes $F \cos(\theta)$ and $F \sin(\theta)$, assigning a positive or negative algebraic sign based on axis orientation.
  5. Note: If the angle $\theta$ is given with respect to the $\text{y}$-axis instead of the $\text{x}$-axis, trigonometric definitions adjust accordingly ($F_x = \pm F \sin(\theta)$ and $F_y = \pm F \cos(\theta)$).

Newton's Second Law in Component Form

• Scalar Component Equations of Motion:
  • Because orthogonal dimensions operate independently, Newton's second law ($\vec{a} = \frac{\sum \vec{F}}{m}$) decomposes into two uncoupled scalar algebraic equations along the $\text{x}$ and $\text{y}$ axes:

$a_{Sx} = \frac{F_{1\,\text{on}\,S\,x} + F_{2\,\text{on}\,S\,x} + F_{3\,\text{on}\,S\,x} + \dots}{m_S}$

$a_{Sy} = \frac{F_{1\,\text{on}\,S\,y} + F_{2\,\text{on}\,S\,y} + F_{3\,\text{on}\,S\,y} + \dots}{m_S}$

• Static Equilibrium in Two Dimensions:
  • For a system $K$ that remains at rest or moves with constant velocity ($\vec{a} = 0$), the net force scalar components along both axes must independently sum to zero:

$a_{Kx} = \frac{T_{1\,\text{on}\,K\,x} + T_{2\,\text{on}\,K\,x} + T_{3\,\text{on}\,K\,x}}{m_K} = 0$

$a_{Ky} = \frac{T_{1\,\text{on}\,K\,y} + T_{2\,\text{on}\,K\,y} + T_{3\,\text{on}\,K\,y}}{m_K} = 0$

• Analyzing Given Mathematical Component Equations (Review Question 4.2):
  • Consider a system of mass $m = 50\,\text{kg}$ governed by the component equations:

$x\text{-component}: 1.0\,\text{m/s}^2 = \frac{(60\,\text{N}) \cos(30^\circ)}{50\,\text{kg}}$

$y\text{-component}: 0\,\text{m/s}^2 = \frac{(50\,\text{kg}) \times (9.8\,\text{N/kg}) + (60\,\text{N}) \sin(30^\circ) + (-520\,\text{N})}{50\,\text{kg}}$

• Physical Interpretation:
  • The system is a $50\,\text{kg}$ object (such as a sled or crate) accelerating horizontally along a flat surface at $1.0\,\text{m/s}^2$.
  • A pulling force of $60\,\text{N}$ is applied at an angle of $30^\circ$ above the horizontal.
  • The vertical forces acting on the system are downward Earth gravity $F_{g} = (50\,\text{kg})(9.8\,\text{N/kg}) = 490\,\text{N}$, upward vertical pull component $(60\,\text{N})\sin(30^\circ) = 30\,\text{N}$, and downward normal/exerted vertical interaction force totaling $-520\,\text{N}$, balancing vertical acceleration to zero.

Static and Kinetic Friction

• When two interacting surfaces are in contact, the total force $\vec{F}_{S\,\text{on}\,B}$ exerted by surface $S$ on object $B$ decomposes into two perpendicular component vectors:

  • Normal Force ($\vec{N}$): The component vector perpendicular to the contact surface.
  • Friction Force ($\vec{f}$): The component vector parallel to the contact surface that opposes relative sliding motion or the tendency of motion.

• Observational Experiment 4.1: Characteristics of Friction:

  • Condition 1: A block rests on a desk with no horizontal pull. The surface exerts only an upward perpendicular normal force matching gravity ($\vec{F}_{E\,\text{on}\,B}$).
  • Condition 2: A spring scale pulls lightly horizontally on the block; the block does not move. The surface exerts a static friction force $\vec{f}_s$ parallel to the surface equal in magnitude and opposite in direction to the pulling force $\vec{F}_{Sp\,\text{on}\,B}$.
  • Condition 3: The spring scale pulls harder; the block remains stationary. The magnitude of static friction increases matching the pulling force; the normal force remains unchanged, causing the resultant total contact force angle to shift.
  • Condition 4: The spring scale pulls even harder until the block just begins to move. Static friction reaches its absolute threshold, termed the maximum static friction force ($f_{s\,\text{max}}$).

• Static Friction Force Properties:

  • Acts on stationary objects to prevent relative motion.
  • Self-adjusting in magnitude from zero up to $f_{s\,\text{max}}$.
  • Once the applied parallel pulling force exceeds $f_{s\,\text{max}}$, the object breaks static contact and begins sliding.
  • Once sliding occurs, the resistive contact force transitions to kinetic friction force ($\vec{f}_k$).

• Biomechanical Application of Static Friction (Human Locomotion):

  • Friction is essential for walking:
  1. As the front foot lands, the foot pushes forward on the ground; the ground exerts a backward static friction force $\vec{f}_{s\,\text{S on F}}$ on the front shoe to decelerate forward foot speed.
  2. When the body is directly above the foot, static friction is momentarily zero ($\vec{f}_s = 0$).
  3. As the rear foot pushes backward against the ground, the ground pushes forward on the rear foot via static friction $\vec{f}_{s\,\text{S on F}}$, propelling the human forward.

• Observational Experiment 4.2: Factors Affecting Maximum Static Friction:

  • Experiment 1 (Surface Roughness): Pulling a smooth plastic block across glass, wood, and rubber mat. Result: $f_{s\,\text{max}}$ is greatest on rubber, lower on wood, lowest on glass ($f_{s\,\text{Rubber}} > f_{s\,\text{Wood}} > f_{s\,\text{Glass}}$).
  • Experiment 2 (Contact Area): Pulling a brick-shaped block on faces of three distinct surface areas across the same rubber mat. Result: Scale reading at breakaway is identical for all three areas ($f_{s\,\text{Area 1}} = f_{s\,\text{Area 2}} = f_{s\,\text{Area 3}}$). Maximum static friction is independent of macroscopic contact area.
  • Experiment 3 (Block Mass): Pulling plastic blocks of $1.0\,\text{kg}$, $2.0\,\text{kg}$, and $3.0\,\text{kg}$ across wood. Result: Scale readings at breakaway scale linearly with mass ($f_{s\,(3.0\,\text{kg})} = 3 f_{s\,(1.0\,\text{kg})}$).

• Testing Experiment 4.3 & Table 4.4: Normal Force vs. Mass Hypothesis:

  • Hypothesis Testing: To determine if $f_{s\,\text{max}}$ depends directly on mass or on normal force, a $1.0\,\text{kg}$ block was pulled while external downward forces were applied using a wheeled vertical spring apparatus (keeping mass constant at $1.0\,\text{kg}$).
  • Data (Table 4.4):

• Conclusion: Maximum static friction force is directly proportional to the normal force ($N$), not the mass itself.

• Mathematical Model for Static Friction:
  • The constant ratio of maximum static friction force to normal force defines the coefficient of static friction ($\mu_s$):

$\mu_s = \frac{f_{s\,\text{max}}}{N}$

• The general inequality for static friction is:

$0 \le f_{s\,\text{S on O}} \le \mu_s N_{\text{S on O}}$

• Mathematical Model for Kinetic Friction:
  • When an object slides, kinetic friction force is directly proportional to normal force and independent of sliding speed over typical ranges:

$f_{k\,\text{S on O}} = \mu_k N_{\text{S on O}}$

• where $\mu_k$ is the coefficient of kinetic friction.

  • Coefficients of Friction Table (Table 4.5):

• Microscopic Origin of Friction:

  • On microscopic scales, all surfaces display irregularities, peaks, and valleys (asperities). Friction arises from mechanical interlocking and microscopic chemical bonding between these surface asperities.

• Sample Problem (Refrigerator Friction Analysis):

  • Problem: Determine the friction force exerted by the floor on a $100\,\text{kg}$ stationary refrigerator ($\mu_s = 0.35$, $\mu_k = 0.30$) when pushed horizontally.
  • Analysis:
  • Normal force $N = m g = (100\,\text{kg})(9.8\,\text{m/s}^2) = 980\,\text{N}$.
  • Maximum static friction $f_{s\,\text{max}} = \mu_s N = (0.35)(980\,\text{N}) = 343\,\text{N}$.
  • If no horizontal push is applied ($F_{\text{push}} = 0$), static friction is $0\,\text{N}$.
  • If pushed with $F_{\text{push}} < 343\,\text{N}$, static friction exactly equals $F_{\text{push}}$.
  • If pushed with $F_{\text{push}} > 343\,\text{N}$, the refrigerator slides and kinetic friction becomes $f_k = \mu_k N = (0.30)(980\,\text{N}) = 294\,\text{N}$.

Problem-Solving Framework for Two-Dimensional Forces

1. Sketch and Translate:

   • Sketch the physical arrangement.
   • Define the physical system of interest.
   • Choose coordinate axes such that one axis aligns with the direction of acceleration (or anticipated motion) and the perpendicular axis aligns with zero acceleration.
   • Annotate known quantities and identify the target unknown variable.

2. Simplify and Diagram:

   • Simplify system components (e.g., model objects as rigid point masses, evaluate if friction can be neglected).
   • Construct a motion diagram showing velocity and acceleration vectors.
   • Construct a force diagram (free-body diagram).
   • Ensure force vectors graphically sum to produce a net force vector matching acceleration direction.

3. Represent Mathematically:

   • Write Newton's second law in component form ($\sum F_x = m a_x$ and $\sum F_y = m a_y$).
   • Express individual force component vectors using trigonometric relations ($F_x = \pm F \cos(\theta)$, $F_y = \pm F \sin(\theta)$) and friction laws ($f_s \le \mu_s N$, $f_k = \mu_k N$).
   • Apply relevant kinematic equations for motion with constant acceleration.

4. Solve and Evaluate:

   • Solve algebraic equations for the unknown quantities.
   • Substitute quantitative values with units.
   • Check limiting cases, units, algebraic sign consistency, and physical magnitude feasibility.

Projectile Motion

• Definition: A projectile is an object launched horizontally or at an angle relative to the horizontal that moves under the sole influence of gravity (neglecting air resistance).

• Testing Experiment 4.6: Independence of Motion Components:

  • Setup: Ball 1 is released from rest dropped vertically. Simultaneously, Ball 2 is launched horizontally by a compressed spring from the same initial height.
  • Hypothesis/Prediction: Because horizontal force does not affect vertical gravity, vertical motion is completely independent of horizontal motion. Both balls start with zero initial vertical speed ($v_{0y} = 0$) and should land simultaneously.
  • Outcome: Both balls strike the floor at the exact same instant, proving that vertical free-fall acceleration operates independently of horizontal velocity.

• Kinematic Equations for Projectile Motion (assuming upward $\text{y}$-axis, $a_x = 0$, $a_y = -g = -9.8\,\text{m/s}^2$):
  • Initial velocity components for launch speed $v_0$ at angle $\theta$ above horizontal:

$v_{0x} = v_0 \cos(\theta)$

$v_{0y} = v_0 \sin(\theta)$

• Horizontal Motion ($\text{x}$-direction, $a_x = 0$):

$v_x = v_{0x} = v_0 \cos(\theta) = \text{constant}$

$x = x_0 + v_{0x} t = x_0 + (v_0 \cos(\theta)) t$

• Vertical Motion ($\text{y}$-direction, $a_y = -g$):

$v_y = v_{0y} + a_y t = v_0 \sin(\theta) - g t$

$y = y_0 + v_{0y} t + \frac{1}{2} a_y t^2 = y_0 + (v_0 \sin(\theta)) t - \frac{1}{2} g t^2$

• Why Initial Velocity Must Be Resolved into Components:
  • Resolving initial velocity into components is necessary because gravity acts exclusively along the vertical axis ($a_y = -g$), modifying $v_y$ over time, whereas zero horizontal net force ($a_x = 0$) leaves $v_x$ strictly constant throughout the entire trajectory.

Automotive Dynamics: Starting, Rolling, and Stopping

• Rolling Tire Mechanics:
  • For a rolling tire moving without skidding, the instantaneous point of contact between the tire rubber and the road surface is at rest relative to the road ($v = 0$ at contact point).
  • Therefore, the friction force operating between a non-skidding rolling tire and the road is static friction, not kinetic friction.

• Vehicle Acceleration (Starting / Speeding Up):

  • When the engine turns the drive axle, it rotates the tire faster.
  • The tire tread pushes backward against the road surface ($\vec{f}_{s\,\text{Tire on Road}}$).
  • By Newton's third law, the road exerts an equal and opposite forward static friction force on the tire ($\vec{f}_{s\,\text{Road on Tire}}$).
  • This forward external static friction force accelerates the vehicle forward.

• Vehicle Deceleration (Braking / Stopping):

  • When brakes are applied, the wheel rotation slows down relative to forward vehicle motion.
  • The tire tread pulls/pushes forward against the road surface.
  • The road responds by exerting a backward static friction force on the tire ($\vec{f}_{s\,\text{Road on Tire}}$), decelerating the vehicle.

• Internal vs. External Forces Misconception:

  • Misconception: The car engine exerts a force directly on the car that drives it forward.
  • Correction: Engine torque and internal gears exert internal forces within the system. Internal forces cannot produce center-of-mass acceleration. The engine rotates the wheels, but it is the external static friction force exerted by the road surface on the tires that causes the car to accelerate or stop.