Equilibrium of a Rigid Body – Student-Friendly Notes
1 Mechanical equilibrium in plain words
A rigid body is in perfect balance when it neither speeds up in a straight line nor starts spinning faster or slower. In symbols:
- \(\displaystyle \sum \vec F_i = \vec 0\) — no net push ⇒ no linear acceleration :contentReference[oaicite:0]{index=0}
- \(\displaystyle \sum \vec \tau_i = \vec 0\) — no net twist ⇒ no angular acceleration
These two rules give six separate “component” equations (three for forces, three for torques). If every force lies in one plane, only three conditions matter: two for forces inside the plane and one torque perpendicular to it :contentReference[oaicite:2]{index=2}.
2 Helpful facts about those rules
- When the first rule holds, shifting your torque origin does not disturb the second rule; the balance of twists is origin-independent :contentReference[oaicite:3]{index=3}.
- A body can be partly balanced:
• Translational equilibrium but not rotational ⇒ pure rotation without sliding.
• Rotational equilibrium but not translational ⇒ slides without spinning :contentReference[oaicite:4]{index=4}.
3 Couple (pure twist)
A couple is a pair of equal forces pointing in opposite directions along different lines. It makes the body spin but does not shove it sideways :contentReference[oaicite:5]{index=5}.
The moment (twist) of a couple is \(\vec r_{BA}\times\vec F\) and is the same about every point — handy when choosing an origin :contentReference[oaicite:6]{index=6}.
Daily‐life examples: turning a jar lid or the Earth’s field twisting a compass needle :contentReference[oaicite:7]{index=7}.
4 Principle of moments (levers)
An ideal lever is a light rod pivoted at a fulcrum. For balance:
\(\displaystyle d_1 F_1 = d_2 F_2\) ( load-arm × load = effort-arm × effort ) :contentReference[oaicite:8]{index=8}
The mechanical advantage is \(\displaystyle \text{M.A.}= \dfrac{F_1}{F_2}= \dfrac{d_2}{d_1}\). A longer effort arm lets a small force lift a hefty weight :contentReference[oaicite:9]{index=9}.
5 Centre of gravity (CG)
The CG is the unique point where the total gravitational twist of all tiny masses is zero:
\(\displaystyle \sum_i \bigl(\vec r_i\times m_i\vec g\bigr)=\vec 0\) ⇒ \(\displaystyle \sum_i m_i\vec r_i = \vec 0\) :contentReference[oaicite:10]{index=10}
- In uniform gravity, CG and centre of mass overlap.
- Find CG by hanging the object from two or more points; the hanging lines cross at CG :contentReference[oaicite:11]{index=11}.
6 Worked-out illustrations
6.1 Balancing a loaded bar
A 4 kg, 70 cm bar rests on two knife-edges 10 cm from each end. A 6 kg load hangs 30 cm from one end. Applying the balance rules:
- \(R_1+R_2 = (4+6)g\)
- \(-0.25R_1 +0.05(6g)+0.25R_2 = 0\)
Gives \(R_1 \approx 55\;\text{N},\; R_2 \approx 43\;\text{N}\) :contentReference[oaicite:12]{index=12}.
6.2 Leaning ladder on a smooth wall
A 3 m, 20 kg ladder rests with its foot 1 m from a frictionless wall. Solving the three equilibrium equations leads to:
- Normal from floor \(N = 196\;\text{N}\)
- Wall force \(F_1 = 34.6\;\text{N}\)
- Total floor reaction \(F_2 \approx 199\;\text{N}\) at \(80^{\circ}\) above the floor :contentReference[oaicite:13]{index=13}
7 Quick comparison with a single particle
A lone particle can only satisfy \(\sum\vec F_i=\vec0\); it has no extension, so torques about it vanish automatically :contentReference[oaicite:14]{index=14}.
High-Yield NEET Takeaways (3-5-1 method)
- The twin balance laws: \(\displaystyle \sum\vec F=0\) and \(\displaystyle \sum\vec \tau=0\).
- Lever rule \(d_1F_1=d_2F_2\) and mechanical advantage.
- Concept of a couple — pure rotation without translation.
- Centre of gravity vs. centre of mass and how to locate CG.
- Typical equilibrium numericals (bars, ladders) solved with force-and-moment equations.
