Forces: motion of a body on a rough surface, connected particles, equilibrium of rigid bodies

Mechanics (M2) – Forces

1️⃣ Motion of a Body on a Rough Surface

Imagine sliding a book across a table. The table exerts a normal force $N$ perpendicular to the surface, while friction $f$ opposes the motion. The friction can be static (when the book is at rest) or kinetic (when it is sliding).

• Static friction: $f_s \le \mu_s N$ – it can adjust up to a maximum value.
• Kinetic friction: $f_k = \mu_k N$ – constant once motion starts.

When a horizontal force $F$ is applied, the net horizontal force is $F - f$. Using Newton’s second law:

$$F_{\text{net}} = ma = F - f$$

Example: A 2 kg box on a floor with $\mu_k = 0.4$. If you push it with $F = 10$ N, the acceleration is:

$$a = \frac{F - \mu_k mg}{m} = \frac{10 - 0.4 \times 2 \times 9.81}{2} \approx 1.96\;\text{m/s}^2$$

Parameter	Symbol	Formula
Normal force	$N$	$N = mg$ (if horizontal surface)
Static friction	$f_s$	$f_s \le \mu_s N$
Kinetic friction	$f_k$	$f_k = \mu_k N$

2️⃣ Connected Particles – Systems of Particles

When two or more particles are connected (e.g., by a string or a rigid rod), the forces on each particle affect the whole system. We often use the centre of mass to analyse motion.

For a system of $n$ particles with masses $m_i$ and velocities $\mathbf{v}_i$, the centre of mass velocity is:

$$\mathbf{V}_{\text{cm}} = \frac{\sum_{i=1}^{n} m_i \mathbf{v}_i}{\sum_{i=1}^{n} m_i}$$

Example: Two blocks, $m_1 = 3$ kg and $m_2 = 5$ kg, are connected by a light string and pulled with a force $F = 20$ N. The acceleration of the system is:

$$a = \frac{F}{m_1 + m_2} = \frac{20}{8} = 2.5\;\text{m/s}^2$$

Each block experiences tension $T$ in the string:

$$T = m_1 a = 3 \times 2.5 = 7.5\;\text{N}$$

Analogy: Think of two kids on a seesaw. Pulling one side moves both, but the heavier side moves less.

3️⃣ Equilibrium of Rigid Bodies

A rigid body is in equilibrium when the sum of all forces and the sum of all moments about any point are zero.

1. Force equilibrium: $\displaystyle \sum \mathbf{F} = \mathbf{0}$
2. Moment equilibrium: $\displaystyle \sum \mathbf{M}_O = \mathbf{0}$ (about any point $O$)

Example: A horizontal beam of length $L = 4$ m is supported at its ends by two pins. A weight $W = 200$ N is hung at the centre. To keep the beam horizontal, the reaction forces at the pins must satisfy:

$$R_1 + R_2 = W$$ $$R_1 \times L = W \times \frac{L}{2}$$

Solving gives $R_1 = R_2 = 100$ N.

Condition	Equation	Interpretation
Force equilibrium	$\sum \mathbf{F} = \mathbf{0}$	Net horizontal and vertical forces cancel.
Moment equilibrium	$\sum \mathbf{M}_O = \mathbf{0}$	Net torque about any point is zero.

🛠️ Tip: Always choose a convenient point $O$ when calculating moments – often a support point where the reaction force is unknown.

4️⃣ Quick Review & Practice Questions

• Calculate the kinetic friction on a 10 kg crate with $\mu_k = 0.3$ on a horizontal floor.
• Two masses $2$ kg and $4$ kg are connected by a string over a frictionless pulley. If a $10$ N force pulls the $2$ kg mass, find the acceleration of the system.
• A beam of length $5$ m has a weight of $300$ N at one end. If a support is placed $2$ m from that end, what is the reaction force at the support?

Remember: Think of forces as pushes and pulls. Visualise each force acting at a point, and check that they all balance out for equilibrium or produce the expected motion.