FM2 3.3 - Toppling and sliding

You already know

For coplanar equilibrium, resolve forces and take moments. Weight acts vertically through the actual mass centre G. From A-level Mathematics, friction satisfies $|f|\le\mu R$, with equality at impending sliding. The normal reaction acts within the contact region; its position is found from moments, not fixed at the base centre.

Weight-only toppling on an incline

Let G be a perpendicular height h above the base plane. Let d be the distance, along the base, from its perpendicular projection to the downhill supporting edge. For $h>0$, $d>0$, and sufficient friction to prevent earlier sliding,

For a centred body with base width $2b$, $d=b$. For a circular base tilted in an axial section, d is the appropriate radius when G is on its axis. Use the actual centre height, not the full body height.

Not in the formula booklet: learn it.

Worked example 1: a uniform block on a sufficiently rough plane

Question. A uniform cuboid has a rectangular section of base width $4\,\mathrm{m}$ and height $6\,\mathrm{m}$. Its base rests on a plane that is slowly tilted. Friction is sufficient to prevent sliding. Find the inclination at which the cuboid is on the point of toppling.

Plan. Use centre height $h=3$ and downhill distance $d=2$, then place the weight line through the downhill edge.

Why this step? Toppling depends on the height of G, halfway up a uniform cuboid. The full height $6$ is not the weight's moment arm.

Answer. The critical inclination is $33.7^\circ$, to three significant figures.

Check. At that angle, $h\tan\alpha=3(2/3)=2$, exactly the distance to the downhill edge. The friction assumption must allow $f/R=\tan\alpha=2/3$.

Worked example 2: raise the centre without changing the base

Question. The same cuboid-shaped body now has its centre $4\,\mathrm{m}$ above the base centre. Sliding is still prevented. Find its critical toppling angle.

Plan. Keep the downhill base distance $2$, replacing only the centre height.

Why this step? Raising G makes the weight line move farther downhill for the same small tilt, so the supporting edge is reached earlier.

Answer. The body reaches impending toppling at $26.6^\circ$.

Check. This is smaller than $33.7^\circ$. The required friction ratio at the boundary is now $1/2$; a larger total mass would cancel from the moment equation again.

Watch out

Wrong: using a base diameter instead of its radius, a full body height instead of G's height, or a centre distance from the wrong end. Right: draw the section containing G and the downhill supporting edge, and label d and h directly on it.

Which event occurs first?

For a body with centre height h, downhill base distance d, no other applied force, and a slowly increasing inclination $0\le\alpha<90^\circ$,

Compare the angles, or equivalently compare $\mu$ and $d/h$:

The later candidate assumes equilibrium could continue that far. Once an earlier boundary is reached, it is not a prediction of the body's subsequent motion.

Not in the formula booklet: learn it.

Worked example 3: a block slides before its pressure reaches the edge

Question. The uniform width-$4\,\mathrm{m}$, height-$6\,\mathrm{m}$ cuboid rests on a plane slowly tilted from horizontal. The coefficient of friction is $0.4$. Determine which event occurs first and its angle.

Plan. Use $d/h=2/3$ for toppling and $\mu=0.4$ for sliding, then compare.

Why this step? Reaching the toppling geometry is irrelevant if the plane cannot supply enough friction to hold the body until then.

Answer. Sliding occurs first, at $21.8^\circ$.

Check. At this angle, the normal resultant is $h\tan\alpha=3(0.4)=1.2\,\mathrm{m}$ downhill of the base centre. Since $1.2<2$, the supporting edge has not been reached.

Worked example 4: the rougher plane changes the first event

Question. The same cuboid is on a plane with coefficient of friction $0.8$, slowly tilted from horizontal. Find the first event and its angle.

Plan. Retain the toppling angle, recalculate the sliding candidate, and test the contact condition at the earlier one.

Why this step? Friction has changed, so the sliding boundary moves. The mass distribution and supporting width have not changed, so the toppling boundary stays fixed.

Answer. Toppling occurs first, at $33.7^\circ$.

Check. At impending toppling, $f/R=2/3<0.8$. Friction is below its limiting value; writing $f=0.8R$ at this event would be incorrect.

Watch out

Wrong: imposing limiting friction and impending toppling in every problem. Right: calculate the two candidates separately. Both equalities hold together only when the model gives simultaneous limiting events.