MG5 - Centre of mass and equilibrium | OCR A-Level Further Mathematics B (MEI)

The centre of mass tells you where a body's weight acts when calculating moments. You will use it to find a suspended body's orientation, test whether supports can hold a horizontal body, and decide whether an inclined body first slides or topples.

A freely suspended body settles below its support

Can turning the same prism around on a slope change its first failure from toppling to sliding? The key is where its centre of mass lies relative to its supports. Begin with the simpler situation of a body hanging freely from one point.

You already know

Rigid-body equilibrium requires zero resultant force and zero resultant moment. A body's weight may be taken to act through its centre of mass $G$ . The centre of a composite body comes from mass-weighted part centres.

Suppose a body hangs from a light vertical wire attached at $P$ , with no other applied force apart from gravity. Tension acts through $P$ , so weight must also have zero moment about $P$ : its vertical line must pass through $P$ .

Stable suspension from one point

For a rigid body freely suspended at $P$ , under its weight and the supporting tension alone, stable hanging equilibrium places $G$ vertically below $P$ . The body-fixed direction $\overrightarrow{PG}$ therefore becomes the downward vertical.

Not in the formula booklet: learn it.

To find an orientation, recall the scalar-product angle formula, which is not in the formula booklet and must be learned. For a non-zero body-fixed vector $\mathbf a$ , its angle $\theta$ with downward vertical satisfies

\cos\theta=\frac{\mathbf a\mathbin{\cdot}\overrightarrow{PG}} {|\mathbf a|\,|\overrightarrow{PG}|}

provided $G\ne P$ .

The point directly above $P$ also gives zero weight moment, but that inverted position is unstable: a small displacement lets the weight turn the body away. The usual freely hanging position has $G$ below $P$ . If $G=P$ , weight gives no preferred orientation, and the vector-angle formula above cannot be used.

[DIAGRAM: asset_name: Suspension turns PG into the vertical; asset_slug: h645_b068_suspension; caption: Body coordinates identify PG before the body turns and stable suspension makes that direction vertical; description: A triangular lamina with vertices P zero zero, A six zero and B zero three has G two one marked. A second drawing shows the same triangle rotated so PG points vertically down. The angle between PA and downward vertical is labelled theta, with tangent theta equals one half.]
Body coordinates identify PG before the body turns and stable suspension makes that direction vertical

Worked example 1: a triangular sign

Question. A uniform triangular lamina has vertices $P(0,0)$ , $A(6,0)$ , $B(0,3)$ , in centimetres. It hangs freely from $P$ in stable equilibrium. Find the angle that the direction $PA$ makes with the downward vertical.

Plan. Find $G$ , then calculate the angle between $PA$ and $PG$ in the original body coordinates.

\begin{aligned} G&=\left(\frac{0+6+0}{3},\frac{0+0+3}{3}\right)=(2,1)\\ \overrightarrow{PG}&=(2,1),\qquad \overrightarrow{PA}=(6,0)\\ \tan\theta&=\frac12,\qquad \theta=26.565\ldots^\circ \end{aligned}

Why this step? The lamina rotates, but angles between its own lines do not change. There is no need to find its new coordinates.

Answer. The direction $PA$ makes $26.6^\circ$ with the downward vertical.

Check. The scalar product gives $\cos\theta=12/(6\sqrt5)=2/\sqrt5$ , giving the same angle.

Keep the same horizontal projection of $PG$ , but add a third coordinate. This changes the spatial angle to the chosen edge.

Worked example 2: a suspended spatial frame

Question. In body-fixed coordinates, a rigid frame has suspension point $P(0,0,0)$ , point $A(6,0,0)$ , and centre of mass $G(2,1,2)$ . It hangs freely from $P$ in stable equilibrium. Find the angle between $PA$ and the downward vertical.

Plan. Use the spatial angle between $(6,0,0)$ and $(2,1,2)$ .

\begin{aligned} |PG|&=\sqrt{2^2+1^2+2^2}=3\\ \cos\theta&=\frac{(6,0,0)\mathbin{\cdot}(2,1,2)}{6(3)} =\frac{12}{18}=\frac23\\ \theta&=48.189\ldots^\circ \end{aligned}

Why this step? The distance perpendicular to $PA$ involves both the $y$ - and $z$ -offsets. The planar calculation $\tan\theta=1/2$ would discard part of the geometry.

Answer. The angle is $48.2^\circ$ .

Check. The perpendicular distance from $G$ to $PA$ is $\sqrt5\,\mathrm{cm}$ , so $\tan\theta=\sqrt5/2$ , consistent with $\cos\theta=2/3$ .

The direction of the question matters. An angle with the horizontal is complementary to an acute angle with the vertical in a vertical plane; it is not the same angle.

Watch out

Draw the weight line through $G$ and the support point. For stable free suspension, follow $P$ towards $G$ to identify downward vertical; reversing that vector gives the supplementary angle.

Retain the frame's horizontal projection while changing its third coordinate again.

A horizontal body's weight must fall within its supports

A body resting on supports underneath it needs non-negative upward reactions. If a calculation requires a support to pull down, that assumed contact cannot hold. This gives a geometric test: with gravity and upward vertical support reactions only, the vertical projection of $G$ must lie in the region spanned by the supporting contact points.

For two point supports that region is their line segment. For three non-collinear point supports it is their triangle, including its boundary for limiting equilibrium. With a broad base it is the supporting footprint. On a boundary, a reaction may be zero and toppling may be impending.

Vertical reactions at horizontal support points

Use horizontal plan coordinates $(x_i,y_i)$ for supports and $(\bar x,\bar y)$ for the projection of $G$ . For weight $W$ and upward reactions $R_i$ ,

\sum_i R_i=W,\qquad \sum_i R_i x_i=W\bar x,\qquad \sum_i R_i y_i=W\bar y

The last two equations are moment balances about two perpendicular horizontal axes. Every contact retained in the model must have $R_i\ge0$ .

Not in the formula booklet: learn it.

Dividing these equations by $W$ writes the projection of $G$ as an average of the support points, with non-negative weights $R_i/W$ adding to $1$ . This explains the support-region test rather than merely memorising it.

[DIAGRAM: asset_name: Three supports seen from above; asset_slug: h645_b068_supports; caption: The drawing is a horizontal plan; the reactions act vertically out of it; description: In a horizontal plan, B is zero six, C is six six and D is d zero, with G at three two. The support triangle is shaded. A separate plan shows the limiting D positions one point five zero and four point five zero, with the corresponding boundary lines through G.]
The drawing is a horizontal plan

Worked example 3: three vertical reactions

Question. A horizontal rigid lamina of weight $120\,\mathrm N$ has centre of mass whose projection is $G(3,2)$ . It rests on supports at $B(0,6)$ , $C(6,6)$ , $D(3,0)$ , with coordinates in centimetres. Find the reactions.

Plan. Balance vertical forces and take moments about the horizontal $x$ - and $y$ -axes.

\begin{aligned} R_B+R_C+R_D&=120 && \text{vertical force balance}\\ 6R_B+6R_C&=120(2) && \text{moments about the }x\text{-axis}\\ R_B+R_C&=40,\qquad R_D=80\\ 6R_C+3(80)&=120(3) && \text{moments about the }y\text{-axis}\\ R_C&=20,\qquad R_B=20 \end{aligned}

Why this step? The plan coordinates are perpendicular lever arms for vertical forces. Treating the plan as a vertical drawing would assign the wrong directions to the reactions.

Answer. $R_B=20\,\mathrm N,\ R_C=20\,\mathrm N,\ R_D=80\,\mathrm N$ .

Check. The reactions sum to $120\,\mathrm N$ , are all positive, and give both required moments: $6(40)=240$ and $6(20)+3(80)=360\,\mathrm{N\,cm}$ .

Move only $D$ , to $(4,0)$ . Its share of the load remains fixed by the $y$ -balance, but the other two shares must change.

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MG5 - Centre of mass and equilibrium

A freely suspended body settles below its support

A horizontal body's weight must fall within its supports

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