IITJEE: Newton’s Laws & Dynamic Free Body Diagrams Visualizer

Ashish Kumar Bahekar

Published on

October 8, 2026

IITJEE

Apex class • IITJEE Force & Friction Lab

Newton’s Laws & Dynamic Free Body Diagrams

Toggle configurations to see how Tension, Friction, and Net Forces automatically rebalance!

⚙️ 1. Atwood Machine (Pulleys) 📐 2. Inclined Plane & Friction Trap

⚙️ Select Mass Configuration on Pulley (m₁ vs m₂):

5kg & 5kg (Equilibrium) 6kg & 4kg (a = 2 m/s²) 8kg & 2kg (a = 6 m/s²)

Driving Force (Net)

0 N 20 N (60N – 40N) 60 N (80N – 20N)

F_net = (m₁ – m₂)g

Total Inertia (Mass)

10 kg Total

m_total = (m₁ + m₂)

Acceleration (a = F/m)

0.0 m/s² 2.0 m/s² 6.0 m/s²

Both blocks share same ‘a’

String Tension (T)

50.0 N 48.0 N 32.0 N

T = 2m₁m₂g / (m₁ + m₂)

💡 JEE Shortcut: Think of an Atwood Machine as a straight line! Unfold the string: the forward pull is $m_1 g$, backward pull is $m_2 g$. Thus, $a = \frac{\text{Net Pull}}{\text{Total Mass}}$.

📐 Mass = 5kg | Angle = 37° (mg·sin37° = 30N) | Select Surface:

1. Smooth (μ = 0) 2. Kinetic Rough (μ_k = 0.5) 3. High Friction (μ_s = 0.8) ⚠️ Trap

Driving Force Down Incline

30.0 N

F_d = mg sin(37°)

Friction Limit vs Actual

0 N (Smooth) Actual = 20.0 N (Kinetic) Max = 32 N | Actual = 30 N

μ = 0 f_k = μ_k N (0.5 × 40) Since 30N < 32N limit, f_s = 30N!

Net Force (F_net)

30.0 N Down 10.0 N Down (30 – 20) 0 N (Forces Balanced)

F_net = Driving – Friction

Acceleration (a = F/m)

6.0 m/s² 2.0 m/s² 0.0 m/s² (Block at Rest)

Mass = 5kg

⚠️ The Deadliest JEE Friction Trap: In Case 3, the formula gives $f_{s,\max} = \mu_s N = 0.8 \times 40 = 32\text{N}$. But the driving force is only $30\text{N}$. Friction does NOT pull the block backward! Static friction is self-adjusting; it will only apply exactly $30\text{N}$ to keep the net force at zero.