Chapter 5: Applying Newton’s Laws – Statics, Dynamics, and Friction

Applying Newton’s Laws

Overview

This chapter focuses on the application of Newton’s laws to solve problems in statics (equilibrium) and dynamics (motion). It also introduces particular types of forces, especially friction, and explores conditions for equilibrium, contact forces, elastic forces, and circular motion.

Goals for Chapter 5

• Use Newton’s First Law for bodies in equilibrium.

• Apply Newton’s Second Law for accelerating bodies.

• Study types of friction and fluid resistance.

• Solve problems involving circular motion.

• Understand conditions that establish equilibrium.

• Analyze contact forces and elastic (spring) forces.

Applications of Newton’s Laws

• Equilibrium (including static friction)

• Normal forces and tension

• Dynamic (kinetic) friction

• Connected objects (systems)

• Circular motion

Translational Equilibrium

Definition and Conditions

An object is in translational equilibrium when the vector sum of all forces acting on it is zero, resulting in zero acceleration.

• Necessary conditions:

$\sum \vec{F} = 0$

$\sum \vec{F}_x = 0 \quad \text{and} \quad \sum \vec{F}_y = 0$

Example: One-Dimensional Equilibrium

• A gymnast hangs from a rope. Her weight is 500 N, and the rope’s weight is 100 N.

• Forces on gymnast: $T_1 = 500\,\text{N}$

• Forces on rope and ceiling: $T_2 = T_1 + w = 600\,\text{N}$

Example: Mass Suspended on a String

• Apply Newton’s Second Law: $m\vec{a} = \vec{F} = \vec{T} + \vec{W}$

• In equilibrium: $T = mg$

Statics in Two Dimensions

Example: Car Engine Hanging from Chains

• Three chains support a car engine. Find tension in each chain.

• Key equations:

  • $T_3 = \dfrac{w}{\sin 60^\circ} = 1.155w$

  • $T_2 = \dfrac{w \cos 60^\circ}{\sin 60^\circ} = 0.577w$

  • $T_1 = w$

Equilibrium of Hanging Chain

• Tension at endpoints depends on chain’s flatness.

• Actual curve is a catenary, but endpoint tension is unaffected by shape.

Body in Static Equilibrium

• Free-body diagrams show all forces acting on a point.

• For equilibrium: $\sum \vec{F}_i = 0 \Rightarrow \vec{T}_1 + \vec{T}_2 + \vec{W} = 0$

• Component equations allow solving for string tensions.

Equilibrium on Inclined Planes and Ramps

Example: Car on a Ramp

• Ramp angle $\alpha = 26^\circ$, car weight $w = 1.76 \times 10^4\,\text{N}$

• Tension in cable: $T = w \sin \alpha = 7.72 \times 10^3\,\text{N}$

• Normal force: $n = w \cos \alpha = 1.58 \times 10^4\,\text{N}$

Elastic Forces and Hooke’s Law

Ideal Springs

• Force exerted by a spring is proportional to its displacement from equilibrium.

• Hooke’s Law:

$F = -kx$

• $k$ is the spring constant.

• Direction of force is toward equilibrium (restoring force).

Example: Spring Balance

• Given $k = 1200\,\text{N/m}$, a 1.50 kg fish stretches the spring:

• $mg = kx \Rightarrow x = \dfrac{mg}{k} = 0.0123\,\text{m} = 1.23\,\text{cm}$

Free-Body Diagrams

• Show all forces acting on a body.

• Important: $ma$ (inertia) does not appear as a force in the diagram.

• Normal force must be perpendicular to the surface.

Connected Objects and Systems

Strings and Pulleys

• Objects connected by ideal strings/pulleys have the same acceleration.

• Tension is found by analyzing each object separately.

Example: Two Blocks Connected by a Cord

• Block $m_1$ on a frictionless table, block $m_2$ hanging.

• Equations:

  • $m_1 a_{1x} = T$

  • $m_2 a_{2y} = T - m_2 g$

  • Constraint: $a_{1x} = -a_{2y}$

• Solution:

  • $a_{1x} = \dfrac{m_2}{m_1 + m_2} g$

  • $T = \dfrac{m_1 m_2}{m_1 + m_2} g$

The Atwood Machine

• Two masses $m_1$ and $m_2$ connected by a massless string over a massless pulley.

• Equations:

  • $m_1 a_{1y} = T - m_1 g$

  • $m_2 a_{2y} = T - m_2 g$

  • Constraint: $a_{1y} = -a_{2y}$

• Solution:

  • $T = \dfrac{2 m_1 m_2}{m_1 + m_2} g$

  • $a_{1y} = \dfrac{m_2 - m_1}{m_1 + m_2} g$

Contact Forces and Friction

Normal and Frictional Forces

• Contact forces arise from microscopic compressions.

• Normal force acts perpendicular to the contact surface.

• Frictional force acts parallel to the surface, opposing motion.

Frictional Forces

• Microscopically complex, but macroscopically simple.

• Empirically, friction force is proportional to normal force:

$|\vec{f}| \propto |\vec{N}|$

• No dependence on surface area; friction depends only on mass (via normal force).

Types of Friction

• Static friction ($f_s$): Prevents motion up to a maximum value.

• Kinetic friction ($f_k$): Opposes motion once sliding begins.

$\text{Object at rest:}\quad f_s = \text{force of static friction}$ $\text{Object in motion:}\quad f_k = \text{force of kinetic friction}$

Kinetic Friction

• Magnitude: $f_k = \mu_k N$

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

• Direction opposes motion.

Static Friction

• Magnitude: $f_s \leq \mu_s N$

• $\mu_s$ is the coefficient of static friction.

• Direction opposes the initiation of motion.

Summary Table: Friction Coefficients

Surface Type	Static Friction ($\mu_s$)	Kinetic Friction ($\mu_k$)
Glass on glass	0.5	0.4
Steel on steel	0.8	0.6
Rubber on concrete	1.0	0.8
Ice on ice	0.1	0.03
Additional info: Values are approximate and depend on surface conditions.

Problem-Solving Strategies

• Draw diagrams and coordinate axes.

• Choose a specific body and draw its free-body diagram.

• Calculate vector sum of forces (resolve into components).

• Set equations for equilibrium ($\sum \vec{F} = 0$) or dynamics ($\sum \vec{F} = m\vec{a}$).

• Repeat for all relevant objects; use constraints for connected systems.

Additional info:

• Some equations and values have been expanded for clarity and completeness.

• Table entries for friction coefficients are inferred from standard physics references.