Dynamics: Newton's Laws

Comprehensive guide to Newton's three laws of motion for AP Physics 1, including free-body diagrams, friction, and problem-solving strategies.

# Dynamics: Newton's Laws — AP Physics 1

Dynamics extends kinematics by explaining why objects accelerate — the answer is forces. Newton's three laws of motion are the backbone of classical mechanics and are heavily tested on AP Physics 1. You must be able to draw free-body diagrams, apply Newton's second law, and analyze systems involving friction, tension, and normal forces.

Key Concepts

Newton's First Law (Law of Inertia)

An object at rest stays at rest, and an object in motion stays in motion with constant velocity, unless acted upon by a net external force.

Newton's Second Law

Fnet=ma\vec{F}_{\text{net}} = m\vec{a}

The net force on an object equals its mass times its acceleration. This is the most important equation in mechanics.

Newton's Third Law

For every action, there is an equal and opposite reaction. If object A exerts a force on object B, then B exerts a force of equal magnitude and opposite direction on A.

Common Forces

  • Weight (Gravity): Fg=mgF_g = mg (directed downward)
  • Normal Force (FNF_N): perpendicular to the contact surface
  • Friction: opposes relative motion or tendency of motion
    • Static: fsμsFNf_s \leq \mu_s F_N
    • Kinetic: fk=μkFNf_k = \mu_k F_N
  • Tension (TT): pulling force through a string, rope, or cable
  • Applied Force (FappF_{\text{app}}): any push or pull by an external agent

Free-Body Diagrams (FBD)

  1. Identify the object of interest.
  2. Draw all forces acting on that object.
  3. Choose a coordinate system (often tilted for inclines).
  4. Resolve forces into components.
  5. Apply ΣFx=max\Sigma F_x = ma_x and ΣFy=may\Sigma F_y = ma_y.

Inclined Planes

For an object on a frictionless incline at angle θ\theta:

  • Along the incline: mgsinθ=mamg\sin\theta = ma
  • Perpendicular: FN=mgcosθF_N = mg\cos\theta

Atwood Machine

Two masses m1m_1 and m2m_2 connected by a string over a pulley: a=(m1m2)gm1+m2,T=2m1m2gm1+m2a = \frac{(m_1 - m_2)g}{m_1 + m_2}, \quad T = \frac{2m_1 m_2 g}{m_1 + m_2}

Worked Example

Problem: A 5 kg5\ \text{kg} block is pushed across a horizontal floor with a 40 N40\ \text{N} horizontal force. The coefficient of kinetic friction is μk=0.3\mu_k = 0.3. Find the acceleration.

Solution:

Step 1: Normal force. On a horizontal surface, FN=mg=5(9.8)=49 NF_N = mg = 5(9.8) = 49\ \text{N}.

Step 2: Friction force. fk=μkFN=0.3(49)=14.7 Nf_k = \mu_k F_N = 0.3(49) = 14.7\ \text{N}.

Step 3: Apply Newton's second law (horizontal): Fappfk=maF_{\text{app}} - f_k = ma 4014.7=5a40 - 14.7 = 5a a=25.35=5.06 m/s2a = \frac{25.3}{5} = 5.06\ \text{m/s}^2

Practice Questions

  1. 1. A 10 kg10\ \text{kg} box sits on a surface with μs=0.5\mu_s = 0.5. What is the maximum horizontal force you can apply before it starts moving?

    fs,max=μsmg=0.5(10)(9.8)=49 Nf_{s,\max} = \mu_s mg = 0.5(10)(9.8) = 49\ \text{N}.

    2. Two blocks (3 kg3\ \text{kg} and 5 kg5\ \text{kg}) are connected by a string on a frictionless surface. A 24 N24\ \text{N} force pulls the 5 kg5\ \text{kg} block. What is the tension in the string?

    a=24/(3+5)=3 m/s2a = 24/(3+5) = 3\ \text{m/s}^2. Tension: T=3×3=9 NT = 3 \times 3 = 9\ \text{N}.

    3. A 2 kg2\ \text{kg} block slides down a frictionless 30°30° incline. What is its acceleration?

    a=gsin30°=9.8(0.5)=4.9 m/s2a = g\sin 30° = 9.8(0.5) = 4.9\ \text{m/s}^2.

    4. In an Atwood machine with m1=4 kgm_1 = 4\ \text{kg} and m2=6 kgm_2 = 6\ \text{kg}, find the acceleration and tension.

    a=(64)(9.8)/(6+4)=1.96 m/s2a = (6-4)(9.8)/(6+4) = 1.96\ \text{m/s}^2. T=2(4)(6)(9.8)/(10)=47.04 NT = 2(4)(6)(9.8)/(10) = 47.04\ \text{N}.

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Summary

  • Newton's first law defines inertia; net force causes acceleration.
  • Fnet=maF_{\text{net}} = ma is applied component-by-component after drawing an FBD.
  • Friction depends on the normal force and the coefficient of friction.
  • Newton's third law pairs act on different objects — never cancel each other on the same FBD.

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