🚀 Newton's Second Law
⛰️ Inclined Plane (Frictionless)

The Laws of Motion

Dynamics is the study of how forces influence the motion of objects. Isaac Newton established three fundamental laws that describe the relationship between a body and the forces acting upon it.

ΣF = ma

1. Newton's First Law (Inertia)

An object will remain at rest or move at a constant velocity unless acted upon by a resultant force. This implies that if all forces are balanced, the acceleration is zero.

2. Newton's Second Law

The resultant force acting on an object is directly proportional to the rate of change of its momentum. For an object with constant mass, this simplifies to the famous F = ma.

3. Friction and Normal Contact Force

Friction is a force that opposes motion. On a surface, the maximum static friction is proportional to the normal contact force (R).

Fᵣ = μR
Exam Tip: When resolving forces on an inclined plane, the component of gravity acting down the slope is mg sin(θ), and the component normal to the slope is mg cos(θ).

Deep Dive: Worked Examples

✅ Example 1: Horizontal Resultant Force

A crate of mass 50 kg is pulled by a horizontal force of 200 N. A constant resistive force of 50 N acts against it. Calculate the acceleration.

Step 1: Resultant Force F_res = 200 - 50 = 150 N
Step 2: Apply a = F/m a = 150 / 50 = 3.0 m/s²

✅ Example 2: Vertical Lift (Tension)

A cargo lift of mass 800 kg accelerates upwards at 1.2 m/s². Calculate the tension in the cable.

Step 1: Free Body Diagram Analysis Resultant Force = Tension (T) - Weight (mg)
Step 2: Formula T - mg = ma → T = m(g + a)
Step 3: Solve T = 800(9.81 + 1.2) = 800(11.01) = 8808 N

✅ Example 3: Inclined Plane with Friction

A block of mass 10 kg sits on a 25° slope. If it is on the verge of sliding, what is the coefficient of static friction μ?

Step 1: Balance Forces Parallel to slope: mg sin θ = F_fric
Normal to slope: mg cos θ = R
Step 2: Apply F_fric = μR mg sin θ = μ (mg cos θ) → μ = tan θ
Step 3: Solve μ = tan(25°) ≈ 0.466

✅ Example 4: Connected Masses (Atwood Machine)

Two masses 3 kg and 5 kg are connected via a pulley. Calculate the acceleration of the system.

Step 1: Net Force F_net = (5 × 9.81) - (3 × 9.81) = 19.62 N
Step 2: Total Mass m_total = 3 + 5 = 8 kg
Step 3: Solve a = 19.62 / 8 = 2.45 m/s²

✅ Example 5: Braking Car (Kinematics + Dynamics)

A 1200 kg car traveling at 30 m/s brakes with a force of 6000 N. How far does it travel before stopping?

Step 1: Acceleration (Deceleration) a = -6000 / 1200 = -5.0 m/s²
Step 2: Kinematics (v² = u² + 2as) 0² = 30² + 2(-5)s
Step 3: Solve for s 0 = 900 - 10s → s = 90 m