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MHT-CET Physics · Laws of Motion

Equilibrium and the Centre of Mass

A body is in equilibrium when both the net force and the net torque on it are zero; the centre of mass of a set of particles is their mass-weighted average position, which for two particles lies on the line joining them, nearer the heavier.

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Why this matters

5 PYQs, one HARD: a metre scale balanced on a wedge, the reaction at one of two knife-edges, three weights held by strings over pulleys, and where the centre of mass of two particles lies and how it stays fixed. One card.

Concept 1 of 1: Moments, Reactions and the Centre of Mass

For balance, take moments about a point where an unknown force acts, so it drops out. A metre scale on a wedge at its centre balances when w × (its distance) = W × (the other distance). A rod on two knife-edges r apart, with its weight x from A, has reaction W(r − x)/r at A. At a knot held by strings the vector sum of the tensions is zero: two equal tensions T at θ to the vertical hold up 2T cos θ. The centre of mass of two particles is at (m₁x₁ + m₂x₂)/(m₁ + m₂), on the line between them; to keep it fixed, moving m₁ by d toward it must be balanced by moving m₂ by (m₁/m₂)d toward it too.

Definition

  • Moments: ∑τ=0\sum \tau = 0 (w at 20 cm, 25 g at 74 cm about 50 cm ⇒ w = 20 g).
  • Two supports r apart, CM at x from A: NA=W(r−x)rN_A = \dfrac{W(r - x)}{r}.
  • Knot: ∑T⃗=0\sum \vec{T} = 0 (two Mg at θ holding √2 Mg ⇒ θ = 45°).
  • xcm=m1x1+m2x2m1+m2x_{cm} = \dfrac{m_1x_1 + m_2x_2}{m_1 + m_2}; fixed CM: m1Δx1+m2Δx2=0m_1\Delta x_1 + m_2\Delta x_2 = 0.

Equilibrium and CM

∑F⃗=0,∑τ=0,xcm=m1x1+m2x2m1+m2\sum\vec{F} = 0, \quad \sum\tau = 0, \qquad x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}

Worked example

A uniform 2 m plank of weight 100 N rests on supports at its ends. A 300 N load sits 0.5 m from the left end. Reactions?
Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 2 · Q30Easy

Example 1 · Laws of Motion · Equilibrium, Centre of Mass, and Friction

A metal rod of weight 'W' is supported by two parallel knife-edges A and B. The rod is in equilibrium in horizontal position. The distance between two knife-edges is 'r'. The centre of mass of the rod is at a distance 'x' from A. The normal reaction on A is

Taking moments about the wrong point

Take moments about the support whose reaction you do NOT want. For the reaction at A, take moments about B: N_A·r = W(r − x).

Placing the centre of mass at the midpoint

Only for equal masses. For unequal masses it sits closer to the heavier one, still on the line joining them.
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Summary — formulas & gotchas at a glance

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Formulas (1)

  • Moments, Reactions and the Centre of Mass

    Equilibrium and CM

    ∑F⃗=0,∑τ=0,xcm=m1x1+m2x2m1+m2\sum\vec{F} = 0, \quad \sum\tau = 0, \qquad x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}

Watch out for (2)

Test yourself on Laws of Motion

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