System of Particles and Rotational Motion · Lesson 10 of 15
Centre of Gravity and Equilibrium Problems
“Locate weight resultants and solve support-reaction and balance problems with complete force diagrams.”
• Distinguish centre of gravity from centre of mass. • Explain balancing and suspension methods. • Solve beam, ladder and angled-string equilibrium. • Include all support forces and correct moment arms.
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.
Quiz
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.
Where does a uniform rod’s weight act?
The weight resultant acts at CG, which coincides with the midpoint in uniform gravity.
The vertical component of tension at angle θ to vertical is what?
The angle is measured from the vertical component direction.
A freely suspended irregular plate settles with CG where?
Then weight has no moment about the suspension point.
Key Takeaways
• CG concerns the resultant gravitational effect; CM concerns mass distribution. • Draw every external force before writing equilibrium equations. • Use perpendicular lever arms, particularly for ladders and angled strings. • Check support directions and available friction after solving.