Problem Statement
Solve the Newton’s Laws / mechanics problem: Solve the Newton’s Laws / mechanics problem: A ball is thrown horizontally inside a train moving with acceleration $w_0$ relative to the ground. In the train’s frame, the ball appears to have an extra horizontal deceleration. Find the equation of motion in the train’s frame and the trajectory of the
Given Information
- Mass(es), forces, angles, and coefficients of friction as given
- $g = 9.8\,\text{m/s}^2$ (acceleration due to gravity)
Physical Concepts & Formulas
Newton’s three laws form the complete foundation of classical mechanics. The second law $\vec{F}_{\text{net}} = m\vec{a}$ is the workhorse: draw a free-body diagram for each object, resolve forces into components along chosen axes, and write $\sum F_x = ma_x$, $\sum F_y = ma_y$. For systems connected by strings over pulleys, the tension is common to both sides of a massless string. Friction force $f = \mu N$ opposes relative sliding and is proportional to the normal force, not the contact area.
- $\vec{F}_{\text{net}} = m\vec{a}$ — Newton’s second law
- $f_k = \mu_k N$ — kinetic friction
- $f_{s,\max} = \mu_s N$ — maximum static friction
- $N = mg\cos\theta$ — normal force on incline of angle $\theta$
- $a = g\sin\theta – \mu_k g\cos\theta$ — acceleration on rough incline
Step-by-Step Solution
Step 1 — Free-body diagram: Draw all forces on each object separately.
Step 2 — Choose coordinate axes: Align one axis along the direction of motion.
Step 3 — Apply Newton’s 2nd Law component by component:
$$\sum F_x = ma_x\quad,\quad \sum F_y = ma_y = 0\quad\text{(if no vertical acceleration)}$$
Step 4 — Constraint equations: For Atwood machine: $a_1 = -a_2 = a$, same tension $T$.
Step 5 — Solve the system of equations for $a$ and $T$.
Worked Calculation
Substituting all values with units:
Atwood machine: $m_1 = 3\,\text{kg}$, $m_2 = 5\,\text{kg}$:
$$a = \frac{(m_2-m_1)g}{m_1+m_2} = \frac{(5-3)\times9.8}{8} = \frac{19.6}{8} = 2.45\,\text{m/s}^2$$
$$T = \frac{2m_1 m_2 g}{m_1+m_2} = \frac{2\times3\times5\times9.8}{8} = \frac{294}{8} = 36.75\,\text{N}$$
Answer
$$\boxed{a = \dfrac{(m_2-m_1)g}{m_1+m_2}}$$
Physical Interpretation
In the Atwood machine, if $m_1 = m_2$ then $a = 0$ (equilibrium). The acceleration grows as the mass difference increases and approaches $g$ when one mass is negligible. The tension is always between $m_1 g$ and $m_2 g$ — the string must support the lighter side while slowing the heavier one. Real pulleys have mass (moment of inertia) which reduces $a$ slightly.
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