Understanding Static Equilibrium with Real Examples

✅ Lesson Overview

This lesson delves into the principles of static equilibrium, focusing on analyzing a ladder leaning against a frictionless wall. By applying force and torque conditions, you'll learn to determine the forces exerted by the wall and the ground on the ladder.

✅ What You’ll Learn

• Application of Newton’s laws to static systems.
• Construction and analysis of free body diagrams (FBDs).
• Calculation of torque and understanding of moment arms.
• Strategic selection of pivot points to simplify torque equations.
• Determination of normal and frictional forces in ladder scenarios.
• Integration of equilibrium conditions in real-world contexts.

✅ Key Concepts Covered

• Static equilibrium
• Torque
• Free body diagram
• Moment arm
• Normal force
• Static friction
• Center of mass
• Pivot point selection

✅ Why This Lesson Matters

Mastering static equilibrium is essential for solving problems where objects remain at rest under multiple forces. This knowledge is pivotal for success in AP Physics and IIT JEE exams and has practical applications in engineering and safety assessments, such as evaluating the stability of ladders and structures.

✅ Prerequisite or Follow-Up Lessons

• Prerequisite: Newton’s Laws of Motion
• Follow-Up: Rotational Dynamics and Angular Momentum

📖 Full Lesson Transcript – Balancing a Leaning Ladder

The full transcript of this lesson is provided below to enhance accessibility and support your study and revision efforts.

Introduction

In this lesson, we explore how to apply the conditions of static equilibrium—zero net force and zero net torque—to solve physics problems involving a ladder leaning against a frictionless wall. We'll break down the problem step by step to understand the forces at play and how to calculate them efficiently.

Problem Setup

Consider a ladder resting against a frictionless wall (imagine it's been freshly waxed). The ground provides friction, preventing the ladder from slipping. The ladder's center of mass is located one-third up from the bottom. A firefighter climbs the ladder, and her center of mass is at a distance of L/2 from the bottom. Our goal is to find the magnitudes of the forces exerted on the ladder by the wall and the ground.

Step 1: Choose the System

We consider the ladder and the firefighter as a single system.

Step 2: Draw the Free Body Diagram (FBD)

Identify and label all the forces acting on the system:

• Weight of the firefighter (Mg) acting downward at a distance L/2 from the bottom.
• Weight of the ladder (mg) acting downward at a distance L/3 from the bottom.
• Normal force from the wall (F_w) acting horizontally at the top of the ladder.
• Normal force from the ground (F_pN) acting vertically upward at the base.
• Frictional force from the ground (F_pf) acting horizontally at the base.

Step 3: Apply Equilibrium Conditions

For static equilibrium:

• Sum of forces in the x-direction: ΣFₓ = 0
• Sum of forces in the y-direction: ΣFᵧ = 0
• Sum of torques about any point: Στ = 0

Step 4: Choose the Pivot Point

Selecting the base of the ladder as the pivot point simplifies calculations because the normal and frictional forces from the ground act through this point, contributing zero torque.

Step 5: Calculate Torques

Using τ = r⊥F, where r⊥ is the perpendicular distance from the pivot to the line of action of the force:

• Torque due to F_w: –F_w * h (negative because it tends to rotate the ladder clockwise)
• Torque due to Mg: Mg * (a/2)
• Torque due to mg: mg * (a/3)

Setting the sum of torques to zero:

–F_w * h + Mg * (a/2) + mg * (a/3) = 0

Solving for F_w:

F_w = [g * a * (M/2 + m/3)] / h

Step 6: Calculate Horizontal and Vertical Forces

From ΣFₓ = 0:

F_w – F_pf = 0 ⇒ F_pf = F_w

From ΣFᵧ = 0:

F_pN – Mg – mg = 0 ⇒ F_pN = (M + m) * g

Step 7: Plug in Values

Given:

• a = √(L² – h²) = 7.58 m
• M = 72 kg
• m = 45 kg
• g = 9.8 m/s²