Maths › Further Mechanics 2 › Equilibrium, suspension, toppling and sliding
Equilibrium, suspension, toppling and sliding
Knowing where the centre of mass is only matters because of what it predicts: which way a hanging body tilts, and whether a body on a slope slides away or falls over first.
Builds on Centres of mass of plane figures and frameworks and Friction and inclined planes.
IN THIS TOPIC
- Find the angle at which a suspended lamina hangs.
- Determine the angle at which a body on a rough slope topples.
- Decide whether toppling or sliding happens first.
COMMON MISCONCEPTION
A body on a slope topples as soon as its centre of mass is higher than the lower edge of its base.
Hanging from a point
A body suspended freely from a point hangs so that its centre of mass is vertically below that point. Nothing else is needed. Draw the line from the pivot to G, and the angle any edge makes with the vertical is the angle between that edge and that line.
In practice you find the horizontal and vertical displacements from the pivot to G in the body's own frame, then take an inverse tangent. Which way round the division goes decides which angle you get, so say clearly whether you are giving the angle to the vertical or the angle to an edge, and sketch the tilted body to confirm the sense.
WORKED EXAMPLE
Hanging a lamina
The lamina from earlier, an 8 by 6 rectangle with a 3 by 2 corner removed, has G at (4.36, 3.29). It is hung from the corner (8, 6). Find the angle the long edge makes with the vertical.
Displacements from the pivot to G, in the body's frame: 8 − 4.36 = 3.64 along the long edge, and 6 − 3.29 = 2.71 along the short one.
The line from the pivot to G becomes vertical. The long edge lies at arctan(2.71/3.64) = 36.7° to it.
The short edge is therefore at 53.3° to the vertical, and quoting that instead is the standard error here.
A check on the sketch: the pivot at (8, 6) and the far corner at (0, 0) sit almost exactly on the line through G, so that corner hangs nearly straight below the pivot.
Topple or slide
A body on a rough slope stays put until one of two things happens. It slides when the slope reaches the friction limit, tan θ = μ. It topples when the vertical through the centre of mass passes outside the base, which for a block of base 2a and height 2h happens at tan θ = a/h.
Whichever angle is smaller happens first. The height of the centre of mass above the ground settles nothing on its own. What matters is whether the line of the weight still lands inside the base, since only then can the normal reaction act at a point that balances the moments.
GUIDED PRACTICE
Which happens first
A uniform block 0.4 m wide and 1 m tall stands on a rough plane with coefficient of friction 0.3, which is slowly tilted. Find whether it topples or slides, and at what angle.
Show the working
Toppling: the base half-width is 0.2 and the height of G is 0.5, so tan θ = 0.2/0.5 = 0.4 and θ = 21.8°.
Sliding: tan θ = μ = 0.3, so θ = 16.7°.
The smaller angle is reached first, so the block slides at 16.7° and never topples.
A wider or shorter block, or a rougher surface, would reverse the verdict.
ASSESSMENT FOCUS
- For a suspended body, draw the vertical through the pivot and mark G on it before calculating anything.
- Say which angle you are giving: to the vertical, to the horizontal, or to a named edge.
- For toppling, use half the base width over the height of G, not the full width.
- Compare both angles and state explicitly which is smaller and therefore which happens.
- A body already resting on a slope needs its own diagram, since the base and the vertical are no longer at right angles.
CHECK YOURSELF
A uniform cube of side 0.6 m rests on a rough plane with μ = 0.8. Does it topple or slide as the plane is tilted?
Show a hint
Compare tan θ for each.
Show the answer
Toppling needs tan θ = 0.3/0.3 = 1, so 45°. Sliding needs tan θ = 0.8, so 38.7°. The smaller comes first, so it slides.
A freely suspended body hangs with its centre of mass vertically below the point of suspension.
On a slope, sliding needs tan θ = μ and toppling needs tan θ = a/h, with a the base half-width and h the height of G.
Whichever of those two angles is smaller is the one that happens.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the equilibrium, suspension, toppling and sliding questions page.
CHECK YOUR PROGRESS
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- Find the angle at which a suspended lamina hangs.
- Determine the angle at which a body on a rough slope topples.
- Decide whether toppling or sliding happens first.
Open the full revision checklist to see every objective in the course in one place.