M1: Forces and Equilibrium

The Art of Staying Still: Mastering Forces ⚖️

Introduction

1. Introduction

Welcome to the world of Mechanics! Ever wondered how a bridge stays up or why an object on a slope doesn't always slide down? It's all about forces. In this chapter, we're going to dive into the core principles that govern how objects interact. We'll start by learning how to break down forces into manageable pieces by resolving them . Then, we'll see what happens when all these forces perfectly balance out, a state we call Equilibrium . Of course, the real world isn't frictionless, so we'll tackle the gritty details of Friction and the crucial concept of Limiting Equilibrium . Finally, we'll wrap up with one of the most fundamental laws of physics: Newton's Third Law, which explains the 'action-reaction' pairs you see everywhere. Let's get this done!

F

F_x

Worked example

Let's Find Out Where This Thing is *Actually* Going... 🧭

F_1

1
$First things first, we need to break each force down into its horizontal (x) and vertical (y) components. This is our core strategy. Always draw a quick sketch to visualize the forces!$
$\text{For } F_1: F_{1x} = 12 \cos(20^\circ), \quad F_{1y} = 12 \sin(20^\circ) \\ \text{For } F_2: F_{2x} = 15 \cos(110^\circ), \quad F_{2y} = 15 \sin(110^\circ)$
2
$F_{2x}$
$F_{1x} \approx 11.276 \text{ N}, \quad F_{1y} \approx 4.104 \text{ N} \\ F_{2x} \approx -5.130 \text{ N}, \quad F_{2y} \approx 14.095 \text{ N}$
3
$F_1$
$R_x = F_{1x} + F_{2x} = 11.276 + (-5.130) = 6.146 \text{ N} \\ R_y = F_{1y} + F_{2y} = 4.104 + 14.095 = 18.199 \text{ N}$
4
$R_x$
$|R| = \sqrt{R_x^2 + R_y^2} = \sqrt{(6.146)^2 + (18.199)^2} \\ |R| \approx \sqrt{37.77 + 331.20} \approx 19.21 \text{ N}$
5
$\theta$
$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{R_y}{R_x} = \frac{18.199}{6.146} \\ \theta = \arctan\left(\frac{18.199}{6.146}\right) \approx 71.3^\circ$
6
$Always finish with a clear, concluding statement, quoting your answers to an appropriate degree of accuracy (usually 3 s.f. in exams).$
$The resultant force has a magnitude of 19.2 N and acts at an angle of 71.3° to the positive x-axis.$

Answer

The resultant force has a magnitude of 19.2 N and acts at an angle of 71.3° to the positive x-axis.

\sum \vec{F} = \vec{0}

\sum \vec{F} = \vec{0}

Worked example

Solving the Classic 'Hanging Object' Puzzle 🧐

30^\circ

1
$W$
$W = 8 \times 9.8 = 78.4 \text{ N}$
2
$T_1$
$\sum F_x = 0 \implies T_2 \cos(60^\circ) - T_1 \cos(30^\circ) = 0$
3
$Next, we do the same for the vertical direction. Both strings have upward vertical components, which together must balance the downward force of the weight.$
$\sum F_y = 0 \implies T_1 \sin(30^\circ) + T_2 \sin(60^\circ) - 78.4 = 0$
4
$T_2$
$T_2 (0.5) = T_1 \left(\frac{\sqrt{3}}{2}\right) \implies T_2 = T_1 \sqrt{3}$
5
$T_2$
$T_1(0.5) + (T_1\sqrt{3})\left(\frac{\sqrt{3}}{2}\right) = 78.4 \implies 0.5T_1 + 1.5T_1 = 78.4 \implies 2T_1 = 78.4 \implies T_1 = 39.2 \text{ N}$
6
$T_1$
$T_2 = 39.2 \times \sqrt{3} \approx 67.89... \implies T_2 = 67.9 \text{ N} \quad (3 \text{ s.f.})$

Answer

T_2 = 39.2 \times \sqrt{3} \approx 67.89... \implies T_2 = 67.9 \text{ N} \quad (3 \text{ s.f.})

F

F = \mu R

Worked example

The One Where the Box Won't Budge 📦

30^\circ

1
$R$
$R - mg\cos(30^\circ) = 0 \implies R = 98\cos(30^\circ) = 84.87 \text{ N}$
2
$F_{max}$
$F_{max} = \mu R = 0.2 \times 84.87 = 16.97 \text{ N}$
3
$P$
$P_{min} + F_{max} - mg\sin(30^\circ) = 0 \\ P_{min} = 98\sin(30^\circ) - 16.97 \\ P_{min} = 49 - 16.97 = 32.03 \text{ N}$
4
$P$
$P_{max} - F_{max} - mg\sin(30^\circ) = 0 \\ P_{max} = 98\sin(30^\circ) + 16.97 \\ P_{max} = 49 + 16.97 = 65.97 \text{ N}$
5
$P$
$32.0 \le P \le 66.0 \text{ (to 3 s.f.)}$

Answer

32.0 \le P \le 66.0 \text{ (to 3 s.f.)}

m

F=ma

Worked example

Let's see this in action... literally! 🚀

m_P = 2

1
$a$
$F = (m_P + m_Q)a \\ 32 = (2 + 6)a \\ 32 = 8a \\ a = \frac{32}{8} = 4 \text{ ms}^{-2}$
2
$R$
$[IMAGE_PLACEHOLDER_2: Free Body Diagram of Block Q. A single horizontal arrow points right, labelled 'R'. The mass '6 kg' is shown inside the block.]$
3
$R$
$F_{net, Q} = m_Q a \\ R = 6 \times 4 \\ R = 24 \text{ N}$
4
$F$
$F_{net, P} = m_P a \\ 32 - R = 2 \times 4 \\ 32 - R = 8 \\ R = 32 - 8 = 24 \text{ N}$
5
$The results match! This confirms our acceleration is correct and demonstrates the action-reaction pair beautifully. The force P exerts on Q is 24 N, and the force Q exerts back on P is also 24 N. You've successfully analysed the system. Nicely done!$
$\text{Acceleration, } a = 4 \text{ ms}^{-2} \\ \text{Contact Force, } R = 24 \text{ N}$

Answer

\text{Acceleration, } a = 4 \text{ ms}^{-2} \\ \text{Contact Force, } R = 24 \text{ N}

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