System of Particles and Rotational Motion · Lesson 6 of 9
Rigid-Body Equilibrium and Centre of Gravity
“Combine force and moment balance with couples, levers, centre of gravity and the PDF’s supported-body problems.”
• Use force and torque balance as independent equilibrium conditions. • Derive the origin independence of a couple and the principle of moments. • Relate centre of gravity to centre of mass in uniform gravity. • Solve the bar, ladder, angled-string, car-load and metre-stick problems.
Equilibrium, couples and levers
Two balances must be checked
For a body at rest to remain at rest in the rigid-body problems here, both the resultant force and resultant torque must vanish. Force balance controls CM acceleration; moment balance controls rotational change. A particle model only needs the concurrent force balance, but an extended body can turn even with zero resultant force. Mechanical equilibrium need not mean rest if the body already has a steady allowed motion; static equilibrium is the at-rest case used in our support calculations.
Two equal parallel forces in the same direction, applied symmetrically at opposite rod ends, can have cancelling moments about the CM but a nonzero resultant force: translation changes. Reverse one force and the resultant force becomes zero, while the moments reinforce: rotation changes. This shows why the conditions are independent. Do not treat torque balance alone as proof of complete equilibrium.
A couple
A couple consists of two equal, opposite forces with distinct parallel lines of action. The net force vanishes but the torque magnitude is Fd, where d is the perpendicular separation. Turning a bottle lid with opposite finger forces is a familiar example. A compass needle in a uniform magnetic field also experiences a turning effect, represented at this level by a force-pair model, until aligned.
This relation follows by replacing every rᵢ with rᵢ − a and distributing the cross products in the torque sum. It explains why, once force balance holds, any convenient pivot can be chosen for the moment equation. Choosing a support point often removes an unknown support force from that equation. It does not remove the need to check the force equations afterward.
The principle of moments
An ideal lever is a light rod pivoted at a fulcrum. A load F₁ at perpendicular arm d₁ and an opposing effort F₂ at arm d₂ balance when their moments agree. A support reaction at the fulcrum supplies force balance but has zero moment about the fulcrum. If the rod’s own weight is not negligible, include it at its CM.
Centre of gravity and equilibrium applications
Where the gravitational turning effect balances
The centre of gravity, or CG, is the point through which the effective resultant weight can be taken to act in the usual approximately uniform gravitational field. About that point the total gravitational torque vanishes. In uniform gravity, torque about the CM is Σ(rᵢ′ × mᵢg) = (Σmᵢrᵢ′) × g = 0, because the mass-weighted relative positions sum to zero. Thus CG and CM coincide. CM is defined by mass distribution, whereas CG concerns gravitational loading. In a field varying appreciably across a large body they need not coincide, and a simple single CG description may require care.
Balance a thin cardboard horizontally on a pencil tip. At the correct support point, the upward reaction balances its weight and gravitational torque vanishes. If supported away from its CG, it tends to tilt. Alternatively, suspend the cardboard freely from a hole, mark the plumb vertical through the suspension point, and repeat from another hole. In settled equilibrium the CG lies vertically below the suspension point; otherwise weight produces a turning moment. Intersecting the two marked verticals locates the CG. For a small cardboard in uniform gravity this also locates the CM.
A systematic equilibrium method
First isolate the body and draw only external forces on it. Place its own weight at its CG, add any loads, and include each support reaction. Resolve angled forces into components. Choose a convenient moment origin, use perpendicular lever arms and assign one rotational sense as positive. Solve force and moment equations together, then check the result: a support unable to pull cannot provide a physically required negative upward reaction.
Problem
A uniform 70 cm bar has mass 4 kg. Supports are at 10 cm and 60 cm. A 6 kg load hangs at 30 cm. Find the reactions, using g = 9.8 m/s².
- 1.The bar’s weight 39.2 N acts at 35 cm, and the load 58.8 N acts at 30 cm. Vertical balance gives R₁ + R₂ = 98 N.
- 2.Take moments about the left support: R₂(0.50) = 39.2(0.25) + 58.8(0.20) = 21.56 N m.
- 3.Thus R₂ = 43.12 N and R₁ = 54.88 N. The reaction nearer the applied load is larger.
- 4.Check moments about the bar centre: −0.25R₁ + 0.25R₂ + 0.05(58.8) = 0.
Problem
A uniform 3 m ladder of mass 20 kg has its foot 1 m from a frictionless wall. Find the wall and floor reactions.
- 1.The contact height is √(3² − 1²) = √8 m. The wall reaction H is horizontal; floor normal N is upward; floor friction f acts toward the wall. Weight W = 196 N acts halfway along the ladder.
- 2.Vertical balance gives N = 196 N; horizontal balance gives f = H.
- 3.Taking moments about the foot: H√8 = W(0.5), so H = 34.65 N. The horizontal lever arm of the weight is 0.5 m, not half the ladder’s slanted length.
- 4.The total floor reaction is √(196² + 34.65²) ≈ 199.0 N and is inclined about 80.0° above the horizontal.
- 5.The calculation presumes adequate floor friction. It requires μₛ at least f/N ≈ 0.177; without sufficient friction no static solution is possible.
Problem
A 2 m nonuniform bar of weight W hangs from strings at 36.9° and 53.1° to the vertical at its left and right ends. Locate its CG.
- 1.Using sin36.9° ≈ 0.6 and cos36.9° ≈ 0.8, horizontal balance gives 0.6T_L = 0.8T_R.
- 2.Vertical balance gives 0.8T_L + 0.6T_R = W. Therefore T_L = 0.8W and T_R = 0.6W.
- 3.The right vertical component is 0.36W. Moments about the left end give (0.36W)(2) = Wd.
- 4.Thus d = 0.72 m from the left end. The string angle is measured from vertical, so its vertical component uses cosine.
Problem
A 1800 kg car has axle spacing 1.8 m and CG 1.05 m behind the front axle. Find each wheel’s reaction on level ground, assuming lateral symmetry.
- 1.Weight is 17640 N. Taking moments about the front axle gives R_rear(1.8) = 17640(1.05), so rear-axle reaction is 10290 N.
- 2.Front-axle reaction is 17640 − 10290 = 7350 N. Each rear wheel carries 5145 N and each front wheel 3675 N.
- 3.Axle totals must be divided by two only because the left–right loading is assumed symmetric.
Problem
A uniform metre stick balances at 45 cm when two 5 g coins are stacked at 12 cm. Find its mass.
- 1.The stick’s own CM remains at 50 cm. The added coin mass is 10 g at 12 cm.
- 2.Moment balance about 45 cm gives 10(45 − 12) = M(50 − 45), with matching gram and centimetre units on both sides.
- 3.M = 330/5 = 66 g. Gravity cancels from the equation.
Check your understanding
Quiz
Zero resultant force alone guarantees what?
Rotation may still change if a couple acts.
A couple has which net force?
Equal opposite forces cancel as vectors.
A larger effort arm gives what ideal mechanical advantage?
MA equals effort arm divided by load arm.
CG coincides with CM under which assumption?
A common g factors out of the gravitational torque sum.
For a smooth vertical wall, its reaction on a ladder is what?
A smooth contact supplies only the normal force.
Key Takeaways
• A supported rigid body needs both resultant force and resultant torque to vanish for static equilibrium. • A couple has zero force and a nonzero origin-independent moment. • A lever’s perpendicular arm ratio gives its mechanical advantage. • CG and CM coincide in uniform gravity; draw all external forces before taking moments.