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Topic 1 of 11

Newton's Laws & Constraints

Classical mechanics begins with Newton's laws, but real systems involve constraints — surfaces, ropes, and rigid connections that restrict how objects move. Setting up constraint equations and applying Newton's second law to each body is the foundation of the analytical approach that leads to Lagrangian mechanics.

Key Concepts

Newton's Second Law

The net force on a body equals its mass times its acceleration:

\vec{F}_{\rm net}=m\vec{a}

. In problems with multiple bodies, draw a free-body diagram for each and write

F=ma

along every independent direction.

Constraint Forces

Normal forces, tension, and reaction forces arise to enforce geometric constraints (e.g., a block stays on a surface). They do no work when the constraint is ideal (frictionless) and are determined by the equations of motion, not prescribed independently.

Atwood Machine

Two masses

m_1 > m_2

connected over a massless, frictionless pulley. The constraint is that both strings have the same length, so

|a_1|=|a_2|=a

. Applying

F=ma

to each mass gives

a=(m_1-m_2)g/(m_1+m_2)

Friction

Kinetic friction

f_k=\mu_k N

opposes motion; static friction

f_s\le\mu_s N

prevents motion. The normal force

N

is determined from the perpendicular force balance, not from the weight alone on inclines or accelerating surfaces.

Key Equations

Newton's Second Law

\vec{F}_{\rm net} = m\vec{a}

Vector form; apply along each independent direction.

Atwood acceleration

a = \frac{(m_1 - m_2)g}{m_1 + m_2}

For two masses over a massless pulley; tension T = 2m₁m₂g/(m₁+m₂).

Incline acceleration

a = g(\sin\theta - \mu_k\cos\theta)

Block sliding down incline at angle θ with kinetic friction μₖ.

Worked Example

Atwood Machine with Friction

Problem

Mass $m_1=6$ kg sits on a frictionless table connected by a light string over a massless pulley to hanging mass $m_2=4$ kg. Find the acceleration and the tension in the string.

Solution

Apply $F=ma$ to $m_2$ (downward positive): $m_2 g - T = m_2 a$

Apply $F=ma$ to $m_1$ (horizontal, toward pulley): $T = m_1 a$

Adding the two equations: $m_2 g = (m_1+m_2)a$

a = \frac{m_2 g}{m_1+m_2} = \frac{4\times9.8}{10} = 3.92\text{ m/s}^2

T = m_1 a = 6\times3.92 = 23.5\text{ N}

Answer a = 3.92 m/s², T = 23.5 N.

Practice

Exercises

7 problems

1 of 7

A block slides from rest down a frictionless incline at $\theta=30°$ . What is its speed after $t=4.0$ s?

Your answer

m/s

2 of 7

An Atwood machine has $m_1=7.0$ kg and $m_2=3.0$ kg. What is the magnitude of the acceleration (in m/s²)?

Your answer

m/s²

3 of 7

In a table-pulley system, $m_1=4.0$ kg hangs and $m_2=6.0$ kg rests on a frictionless table. Find the tension (in N) in the connecting string.

Unlock Exercise 3