HC Verma Chapter 4 Problem 9 — Tension in lift cable during acceleration

Problem Statement

Solve the Newton’s Laws / mechanics problem: Solve the Newton’s Laws / mechanics problem: A 60 kg person stands in an elevator that accelerates upward at 2 m/s². Find the apparent weight (normal force). ($g = 10$ m/s²) $N – mg = ma \Rightarrow N = m(g+a)$ Step 1: $N = m(g+a) = 60(10+2) = 60 \times 12 = 720$ N. $$\boxed{N = 720\text{ N (apparen

Given Information

• See problem statement for all given quantities.

Physical Concepts & Formulas

Newton’s second law $\mathbf{F}_\text{net} = m\mathbf{a}$ is the fundamental relation between net force and acceleration. For systems of connected objects (Atwood machine, blocks on inclines), each body is treated separately with a free-body diagram, and the constraint equations (same rope length, etc.) link the accelerations.

• $\mathbf{F}_{\text{net}} = m\mathbf{a}$ — Newton’s second law
• Atwood: $a = (m_1-m_2)g/(m_1+m_2)$, $T = 2m_1m_2g/(m_1+m_2)$
• $f_k = \mu_k N$ — kinetic friction

Step-by-Step Solution

Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Worked Calculation

$$\boxed{N = 720\text{ N (apparen

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:

$$

$$

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: $m_1 = 3\,\text{kg}$, $m_2 = 5\,\text{kg}$:

$$

$$

$$

Answer

$$\boxed{a = \dfrac{(m_2-m_1)g}{m_1+m_2}}$$

Physical Interpretation

The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.