3.2 Equilibrium of a rigid body
- Syllabus
- 9231–2028–2029
- Topic
- 3.2
- Level
- A2
Learning objectives
3.2.1Moments• calculate the moment of a force about a point For questions involving coplanar forces only; understanding of the vector nature of moments is not required.3.2.2Centre of mass• use the result that the effect of gravity on a rigid body is equivalent to a single force acting at the centre of mass of the body, and identify the position of the centre of mass of a uniform body using considerations of symmetry3.2.3Centre of mass• use given information about the position of the centre of mass of a triangular lamina and other simple shapes Proofs of results given in the MF19 List of formulae are not required.3.2.4Centre of mass• determine the position of the centre of mass of a composite body by considering an equivalent system of particles Simple cases only, e.g. a uniform L -shaped lamina, or a uniform cone joined at its base to a uniform hemisphere of the same radius.3.2.5Rigid body equilibrium• use the principle that if a rigid body is in equilibrium under the action of coplanar forces then the vector sum of the forces is zero and the sum of the moments of the forces about any point is zero, and the converse of this3.2.6Toppling and sliding• solve problems involving the equilibrium of a single rigid body under the action of coplanar forces, including those involving toppling or sliding.
A moment uses the perpendicular distance to the force line
For a coplanar force $F$, the moment about $O$ isM_O=F d_{\perp}=Fr\sin\phi,where $d_{\perp}$ is the perpendicular distance from $O$ to the force's line of action and $\phi$ is the angle between the position line and force. Units are N m.
Choose clockwise or anticlockwise as positive and keep that convention. A force whose line of action passes through O has zero moment, so taking moments about an unknown reaction can remove it from an equilibrium equation.
A $40\text{ N}$ force acts at a point $0.50\text{ m}$ from O and makes $30^\circ$ with the position line. Its moment magnitude is40(0.50)\sin30^\circ=10\text{ N m}.Statethesignfromtheactualturningsense.
The distance to the application point is not automatically the lever arm. Use the perpendicular distance to the infinite line of action; no vector nature of moments is required in this syllabus.
Gravity is equivalent to one weight acting at the centre of mass
In a uniform gravitational field, all distributed gravitational forces on a rigid body are equivalent, for force and moment calculations, to a single downward force Mg acting through its centre of mass G.
| uniform body symmetry | conclusion for G |
|---|---|
| one line of symmetry in a lamina | G lies somewhere on that line |
| two intersecting symmetry lines | G is at their intersection |
| one plane of symmetry in a solid | G lies in that plane |
| rotational symmetry about an axis | G lies on the axis |
A uniform rectangle has G at the intersection of its two midlines. A uniform circular ring has G at its geometric centre even though that point is not part of the material. Symmetry locates G only under the stated uniform-density model.
One symmetry line does not fix a two-dimensional position by itself. Do not replace distributed weight by Mg at a geometric centre unless symmetry and uniformity justify that centre as G.
Use each standard centre from the reference that defines its distance
For a uniform triangular lamina, G is the centroid: it lies on every median, $\tfrac13$ of the altitude from the base and $\tfrac23$ from the opposite vertex. If the vertices are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$,G=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right).
For vertices $(0,0)$, $(6,0)$ and $(2,3)$,G=\left(\frac{0+6+2}{3},\frac{0+0+3}{3}\right)=\left(\frac83,1\right).The vertical coordinate is one third of the height from the base $y=0$.
For any other simple lamina or solid, use the centre-of-mass position supplied in MF19 or the question. Record whether its distance is measured from a base, vertex, centre or flat face, then convert it to the common coordinate origin. Proofs of MF19 results are not required.
The triangular centroid is not halfway up an altitude. A correct numerical fraction used from the wrong reference face gives the wrong mass moment, so annotate the reference before substitution.
Replace each component by a particle at its own centre
For component masses $m_i$ at $(x_i,y_i)$,\bar x=\frac{\sum m_ix_i}{\sum m_i},\qquad \bar y=\frac{\sum m_iy_i}{\sum m_i}.Withcommondensityandthickness,laminamassesareproportionaltoareas;forsolidswithcommondensity,massesareproportionaltovolumes.Aremovedpiececontributesnegativemassandnegativemassmoment.
An L-lamina is a $4\times3$ rectangle with the top-right $2\times1$ rectangle removed. From the lower-left origin, the large rectangle has area 12 and centre $(2,1.5)$; the cut-out has area 2 and centre $(3,2.5)$. Thus\bar x=\frac{12(2)-2(3)}{10}=1.8,\qquad \bar y=\frac{12(1.5)-2(2.5)}{10}=1.3.
Choose one origin and direction, tabulate each component's mass weight and coordinate, then equate total mass moment to total mass times the unknown centre coordinate. For joined solids, use volumes and the supplied centre of each solid along the common axis.
Do not average component coordinates equally. Negative area is only an accounting device for removed material; keep its denominator subtraction and both numerator subtractions consistent.
Equilibrium needs zero force and zero moment
Forcoplanarforcesononerigidbody,equilibriumrequires\sum F_x=0,\qquad \sum F_y=0,\qquad \sum M_O=0aboutanypointO.Conversely,iftheresultantforceandresultantmomentarebothzero,therigidbodyisinequilibrium.
Draw every external force, including weight at G and contact reactions. Resolve forces in convenient directions. Take moments about a point through one or more unknown reactions, then use force balance for the remaining unknowns.
A horizontal $4\text{ m}$ beam has weight $60\text{ N}$ at its centre, an extra $20\text{ N}$ load $3\text{ m}$ from support A, and vertical reactions $R_A,R_B$. Moments about A give4R_B=60(2)+20(3),so $R_B=45\text{ N}$. Vertical balance gives $R_A+R_B=80$, hence $R_A=35\text{ N}$.
Equal opposite forces can have zero resultant yet form a non-zero couple, so force balance alone is insufficient. A moment equation may be taken about any point, but all included lever arms and signs must refer to that same point.
Compare the sliding and toppling thresholds under the same forces
| limiting mode | condition at the threshold | mechanical picture |
|---|---|---|
| sliding | F=μR and force equilibrium | friction has reached its maximum available value |
| toppling | moments balance about the impending pivot edge | resultant support reaction acts through that edge; the other edge reaction is zero |
A rectangular crate of weight $W$, base width $b$ and height $h$ is pushed horizontally at its top on a rough horizontal floor. Before motion, $R=W$. Sliding would begin atP_s=\mu W.TopplingaboutthelowerfaredgewouldbeginwhenP_t h=W\frac b2,\qquad P_t=\frac{Wb}{2h}.
Compare the two positive thresholds. If Ps<Pt, sliding occurs first; if Pt<Ps, toppling occurs first; equality gives simultaneous limiting conditions. Before either threshold, friction is whatever value equilibrium requires and may be strictly less than μR.
High friction can prevent sliding but cannot by itself prevent toppling. 'On the point of toppling' is still limiting equilibrium: angular acceleration has not yet begun, and moments are taken about the contact edge that remains.