Comprehensive Study Notes: Newton's Laws and Dynamics for BAU Admission Exam

Core Principles of Dynamics and Newton’s Laws of Motion

• Dynamics Defined: Dynamics is the branch of physics that explains how forces change the motion of objects. It focus on the relationship between the motion of an object and the forces acting upon it.

• Newton’s First Law (Law of Inertia): An object remains at rest or moves with constant velocity in a straight line unless a non-zero net external force acts on it.

  • Inertia: The tendency of matter to resist a change in velocity.

  • Mass: The quantitative measure of an object's inertia.

  • Condition for Equilibrium: $\Sigma F = 0 \implies a = 0 \implies \text{velocity is constant}$.

• Newton’s Second Law: The acceleration of an object relates to the net external force and the mass of the object. Acceleration points in the same direction as the net force.

  • Vector Equation: $\Sigma F = m \mathbf{a}$.

  • Component Form: The law must be applied to individual axes: $\Sigma F_x = m a_x$ and $\Sigma F_y = m a_y$.

  • Proportions: For a fixed mass, increasing the net force increases acceleration. For a fixed net force, increasing mass decreases acceleration.

• Newton’s Third Law: If object A exerts a force on object B, object B simultaneously exerts a force of equal magnitude and opposite direction on object A.

  • Mathematical Form: $F_{(\text{A on B})} = -F_{(\text{B on A})}$.

  • Characteristics: These forces represent the same type of interaction, exist at the same instant, and act on different objects. Because they act on different bodies, they never cancel each other out on a single free-body diagram.

System Identification and Problem-Solving Workflow

• System of Interest: Before analyzing forces, identify the specific object or group of objects being studied.

  • Single Object System: Choose one object when solving for its specific acceleration, normal force, tension, or friction.

  • Multiple Object System: Treat several objects as one system when internal tensions or contact forces are not needed for the solution.

  • Connected Bodies: Use a system approach for total acceleration, followed by separate free-body diagrams for internal forces.

• Force Definition: A push or pull capable of changing an object’s velocity. It is a vector quantity with magnitude and direction.

• Net Force ($F_{\text{net}}$): The vector sum of all forces acting on a system, denoted as $\Sigma F$.

• Reliable Workflow for Newton-Law Problems:

  1. Identify the system of interest.

  2. Draw a complete Free-Body Diagram (FBD).

  3. Select suitable, often rotated, coordinate axes.

  4. Apply Newton's Second Law consistently along those axes.

Free-Body Diagrams (FBD)

• Purpose: An FBD isolates one system and illustrates every external force acting upon it. This is often the most critical step in solving dynamics problems.

• Execution Steps:

  1. Represent the isolated object as a point or simple box.

  2. Draw all external forces acting on the object (never include forces the object exerts on others).

  3. Label forces by type: weight ($W$), normal ($N$), tension ($T$), friction ($f$), applied force ($F_a$), or spring force ($F_s$).

  4. Establish axes. On inclines, rotate axes so one is parallel to the slope.

  5. Exclude velocity ($v$), acceleration ($a$), or the net force ($F_{\text{net}}$) from the force arrows.

• Force-Specific Checks:

  • Weight ($W$): Always directed vertically downward toward the center of the Earth, regardless of the surface orientation.

  • Normal Force ($N$): Always perpendicular to the contact surface.

  • Tension ($T$): Always points away from the object along the length of the string or rope.

  • Friction ($f$): Always parallel to the contact surface, opposing relative motion or the tendency to slip.

Essential Force Equations and High-Yield Formula Sheet

• Weight: $W = mg$.

• Normal Force on Horizontal Surface with Upward/Downward Forces:

  • Upward external force ($F_{\text{up}}$): $N = mg - F_{\text{up}}$.

  • Downward external force ($F_{\text{down}}$): $N = mg + F_{\text{down}}$.

• Friction:

  • Static Friction ($f_s$): $0 \leq f_s \leq \mu_s N$. It adjusts to the value required for equilibrium up to a maximum threshold ($f_{s,\text{max}}$).

  • Kinetic Friction ($f_k$): $f_k = \mu_k N$. This opposes motion once sliding has begun.

• Spring Force (Hooke’s Law): $F_s = -kx$, where $x$ is the displacement from equilibrium.

• Inclined Planes:

  • Parallel weight component: $mg \sin(\theta)$.

  • Perpendicular weight component: $mg \cos(\theta)$.

  • Frictionless acceleration: $a = g \sin(\theta)$.

  • Acceleration with sliding friction (downwards): $a = g(\sin(\theta) - \mu_k \cos(\theta))$.

• Connected Objects (Ideal Atwood Machine):

  • Acceleration ($m_2 > m_1$): $a = \frac{(m_2 - m_1)g}{m_1 + m_2}$.

  • Tension: $T = m_1(g + a) = m_2(g - a)$.

• Elevator Dynamics (Apparent Weight):

  • Upward acceleration ($a$): $N = m(g + a)$.

  • Downward acceleration ($a$): $N = m(g - a)$.

  • Free fall: $N = 0$ (apparent weightlessness).

Units and Vector Conventions

• Standard Units:

  • Force: Newtons ($N$), where $1\,N = 1\,kg \cdot m/s^2$.

  • Mass: Kilograms ($kg$).

• Gravitational Acceleration ($g$): Use $9.81\,m/s^2$ unless the question or answer options imply using $g = 10\,m/s^2$.

• Directional Notation: Choose positive axes before writing equations. Forces opposite the positive axis are assigned negative components. A negative force component indicates direction, while the magnitude itself is never negative.

• Constant Speed: This only implies zero acceleration if the direction of motion is also constant (motion in a straight line).

Detailed Analysis of Friction

• Mechanism: Friction acts parallel to the contact surface and opposes relative sliding or the tendency to slide. Its direction is determined by the expected relative motion, not necessarily the direction of the applied force.

• Static Friction ($f_s$): Prevents surfaces from beginning to slide. It is not always equal to $\mu_s N$; it is only equal to this value when the object is on the verge of slipping.

• Kinetic Friction ($f_k$): Once surfaces are in relative motion, kinetic friction replaces static friction. Generally, $\mu_k < \mu_s$.

Inclined Plane Dynamics

• Axis Selection: The x-axis should be parallel to the slope; the y-axis should be perpendicular to the slope.

• Normal Force on Incline: For a simple incline without other external forces, $N = mg \cos(\theta)$.

• Mass Independence: Acceleration on a frictionless incline ($a = g \sin(\theta)$) does not depend on the mass of the object.

Connected Objects and Pulleys

• Ideal Systems: Characterized by a massless, inextensible string and a frictionless pulley.

• String Tension: Tension has the same magnitude throughout a single continuous string in an ideal system.

• Constraint of Motion: Objects connected by a taut, inextensible string share the same acceleration magnitude.

Elevators and Apparent Weight

• Apparent Weight Measurement: A weighing scale measures the normal force ($N$) it exerts on a person, not the gravitational force directly.

• Situational Normal Forces:

  • Rest or Constant Velocity: $N = mg$.

  • Acceleration Upwards: $N > mg$ (Person feels heavier).

  • Acceleration Downwards: $N < mg$ (Person feels lighter).

Integrated Dynamics and Kinematics

• Multi-Stage Strategy:

  1. Dynamics Stage: Use $\Sigma F = ma$ to determine the acceleration of the object.

  2. Kinematics Stage: Use constant-acceleration equations (e.g., $v = v_0 + at$) to find time, displacement, or velocity, provided the net force remains constant.

Common Mistakes to Avoid

• Using an individual force instead of the net force ($\Sigma F$) in the second law equation.

• Including internal forces (forces the object exerts on the environment) in an FBD.

• Incorrectly assuming $N = mg$ in every scenario (e.g., on an incline or with additional vertical forces).

• Setting static friction equal to $\mu_s N$ when the applied force is lower than the maximum threshold.

• Ignoring the direction of motion when determining the direction of the friction force.

• Confusing weight and normal force as a Newton-third-law action-reaction pair. They act on the same body and are not necessarily equal.

• Treating $\Sigma F = 0$ only because speed is constant, even if the object is turning (changing direction).

• Applying kinematics equations before confirming that the acceleration is constant.

Worked Examples in Dynamics

• Example 1: Balanced Forces and Inertia:

  • Scenario: A $12\,kg$ box is pushed right with $50\,N$ and has $50\,N$ friction to the left. Initial velocity is $3.0\,m/s$ right.

  • Solution: $\Sigma F_x = 50 - 50 = 0 \implies a = 0$. The velocity remains unchanged.

  • Result: Box continues at $3.0\,m/s$ to the right.

• Example 2: Net Force and Acceleration:

  • Scenario: A $5.0\,kg$ object is acted on by $28\,N$ right and $8\,N$ left.

  • Solution: $\Sigma F_x = 28 - 8 = 20\,N$. $\mathbf{a} = 20 / 5.0 = 4.0\,m/s^2$.

  • Result: $4.0\,m/s^2$ to the right.

• Example 3: Identification of Third-Law Pairs:

  • Scenario: A book rests on a table. Identify the partner to the normal force exerted by the table on the book.

  • Solution: Swap the interacting objects. The table acts on the book; therefore, the book acts on the table.

  • Result: The normal force exerted by the book on the table.

• Example 4: Vertical Forces and Normal Force:

  • Scenario: A $10\,kg$ box sits on a floor with a $30\,N$ upward force applied. Use $g = 10\,m/s^2$.

  • Solution: Vertically, $\Sigma F_y = N + 30 - 100 = 0$.

  • Result: $N = 70\,N$.

• Example 5: Friction and Acceleration:

  • Scenario: A $6.0\,kg$ block pulled by $28\,N$ on a rough surface ($\mu_k = 0.25$). Use $g = 9.81\,m/s^2$.

  • Solution: $N = 6.0 \times 9.81 = 58.86\,N$. $f_k = 0.25 \times 58.86 = 14.72\,N$. $\Sigma F_x = 28 - 14.72 = 13.28\,N$. $a = 13.28 / 6.0 = 2.21\,m/s^2$.

  • Result: $2.21\,m/s^2$ in the direction of the pull.

• Example 6: Static vs. Kinetic Friction:

  • Scenario: A $4.0\,kg$ block on a surface ($\mu_s = 0.50$, $\mu_k = 0.30$). Use $g = 10\,m/s^2$. Find friction for (a) $12\,N$ and (b) $25\,N$ applied forces.

  • Solution: $N = 40\,N$. Maximum static friction $f_{s,\text{max}} = 0.50 \times 40 = 20\,N$.

  • (a) Applied $12\,N < 20\,N$. Block stays at rest. $f_s = 12\,N$.

  • (b) Applied $25\,N > 20\,N$. Block slides. $f_k = 0.30 \times 40 = 12\,N$.

• Example 7: Frictionless Incline:

  • Scenario: Block slides down a frictionless $20^{\circ}$ incline. Use $g = 9.81\,m/s^2$.

  • Solution: $a = g \sin(20^{\circ}) = 9.81 \times 0.342 = 3.35\,m/s^2$.

  • Result: $3.35\,m/s^2$ down the slope.

• Example 8: Incline with Friction:

  • Scenario: $3.0\,kg$ block on $30^{\circ}$ incline with $\mu_k = 0.10$. Use $g = 9.81\,m/s^2$.

  • Solution: $a = g(\sin(30^{\circ}) - 0.10 \cos(30^{\circ})) = 9.81(0.5 - 0.0866)$.

  • Result: $4.06\,m/s^2$ down the slope.

• Example 9: Atwood Machine Calculation:

  • Scenario: $m_1 = 2.0\,kg$, $m_2 = 7.0\,kg$ over an ideal pulley. Use $g = 9.81\,m/s^2$.

  • Solution: $a = (7 - 2) \times 9.81 / (7 + 2) = 5.45\,m/s^2$. $T = 2.0(9.81 + 5.45) = 2.0(15.26) = 30.52\,N$.

  • Result: $a = 5.45\,m/s^2$, $T = 30.5\,N$.

• Example 10: Contact Forces:

  • Scenario: $2.0\,kg$ and $3.0\,kg$ blocks touching. A $30\,N$ force pushes the $3.0\,kg$ block.

  • Solution: Total mass $= 5.0\,kg$. $a = 30 / 5 = 6.0\,m/s^2$. For the $2.0\,kg$ block, internal contact force $\Sigma F = ma = 2.0 \times 6.0 = 12\,N$.

  • Result: $a = 6.0\,m/s^2$, Contact force $= 12\,N$.

• Example 11: Elevator Apparent Weight:

  • Scenario: $60\,kg$ person in an elevator. Use $g = 9.81\,m/s^2$. Find scale reading for (a) upward $2.0\,m/s^2$ and (b) downward $2.0\,m/s^2$.

  • Solution (a): $N = 60(9.81 + 2.0) = 708.6\,N$.

  • Solution (b): $N = 60(9.81 - 2.0) = 468.6\,N$.

• Example 12: Integrated Kinematics:

  • Scenario: $1000\,kg$ car, driving force $4000\,N$, resistive force $1000\,N$. Starts from rest. Speed after $8.0\,s$?

  • Solution: $\Sigma F = 4000 - 1000 = 3000\,N$. $a = 3000 / 1000 = 3.0\,m/s^2$. $v = 0 + (3.0 \times 8.0) = 24\,m/s$.

Questions & Discussion

1. Question: An object moves at constant velocity in a straight line. Which statement must be true? (A) No forces act (B) Net force is zero (C) Acceleration is increasing (D) Its speed is zero.

   • Response: B. Constant velocity means zero acceleration, so the net external force is zero. Forces may still be present but are balanced.

2. Question: A net force of $20\,N$ acts on a $5.0\,kg$ object. Find its acceleration.

   • Response: B. $a = F_{\text{net}}/m = 20/5 = 4.0\,m/s^2$.

3. Question: A $10\,kg$ object has $50\,N$ right and $20\,N$ left. Find acceleration.

   • Response: A. $F_{\text{net}} = 50 - 20 = 30\,N$ right. $a = 30/10 = 3.0\,m/s^2$ right.

4. Question: The weight of a $6.0\,kg$ object is (using $g=10$)?

   • Response: C. $W = mg = 6 \times 10 = 60\,N$.

5. Question: A $50\,kg$ person accelerates upward in an elevator at $2.0\,m/s^2$. What does the scale read?

   • Response: C. $N = m(g + a) = 50(10 + 2) = 600\,N$.

6. Question: Which is a Newton-third-law pair? (A) Weight/normal force (B) Force of Earth on book/force of book on Earth (C) Friction/applied force on one box.

   • Response: B. These forces involve the same pair of objects, are equal/opposite, and act on different bodies.

7. Question: A $4.0\,kg$ block is pulled by $20\,N$ horizontally with $\mu_k = 0.25$. What is the acceleration?

   • Response: B. $N = 40\,N$, $f_k = 0.25(40) = 10\,N$. $F_{\text{net}} = 20 - 10 = 10\,N$. $a = 10/4 = 2.5\,m/s^2$.

8. Question: A $15\,N$ horizontal force is applied to a stationary block ($f_{s,\text{max}} = 20\,N$). What is the static friction?

   • Response: C. Static friction adjusts to match the applied force until the maximum is reached, thus it is $15\,N$.

9. Question: A block slides on a frictionless $30^{\circ}$ incline. What is its acceleration?

   • Response: B. $a = g \sin(30^{\circ}) = 10(0.5) = 5.0\,m/s^2$.

10. Question: A block slides down a $37^{\circ}$ incline with $\mu_k = 0.20$. $\sin(37^{\circ}) = 0.60$, $\cos(37^{\circ}) = 0.80$. Find acceleration.

    • Response: B. $a = g(\sin(\theta) - \mu_k \cos(\theta)) = 10(0.60 - 0.20 \times 0.80) = 10(0.60 - 0.16) = 4.4\,m/s^2$.

11. Question: A $10\,kg$ box on a floor is pulled upward by $60\,N$ but remains in contact. Find normal force.

    • Response: A. $N + 60 - 100 = 0 \implies N = 40\,N$.

12. Question: A $10\,kg$ box is pushed downward by $30\,N$ on a floor. Find normal force.

    • Response: C. $N - 100 - 30 = 0 \implies N = 133\,N$.

13. Question: An ideal Atwood machine has masses $2\,kg$ and $3\,kg$. Find acceleration.

    • Response: B. $a = (3 - 2)10 / (3 + 2) = 10/5 = 2.0\,m/s^2$.

14. Question: For the same $2\,kg$ and $3\,kg$ Atwood machine, find tension.

    • Response: B. For the $2\,kg$ mass, $T = m(g + a) = 2(10 + 2) = 24\,N$.

15. Question: A $3.0\,kg$ mass hangs at rest from a rope. Find tension.

    • Response: C. $T = mg = 30\,N$.

16. Question: A $60\,kg$ person in an elevator accelerates downward at $2.0\,m/s^2$. Find scale reading.

    • Response: C. $N = m(g - a) = 60(10 - 2) = 480\,N$.

17. Question: During free fall, the apparent weight of a person is?

    • Response: A. In free fall, normal force is zero, so apparent weight is zero.

18. Question: Which statement about the normal force is correct? (A) Always equal to mg (B) Always vertical (C) Perpendicular to contact surface (D) Partner of weight.

    • Response: C. This is the definition of the normal force.

19. Question: Two blocks of different masses slide on the same frictionless incline. Their accelerations are?

    • Response: C. Accelerations are equal since $a = g \sin(\theta)$, which is independent of mass.

20. Question: A spring with $k = 200\,N/m$ is stretched by $0.050\,m$. Find spring-force magnitude.

    • Response: B. $F = kx = 200 \times 0.050 = 10\,N$.