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Equilibrium, suspension, toppling and sliding questions
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.
6 original questions · 22 marks · the equilibrium, suspension, toppling and sliding notes · Further Mechanics 2
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State where the centre of mass lies when a body is suspended freely, and why.
Worked answer
Vertically below the point of suspension. If it were not, the weight would have a moment about the pivot and the body would turn until it was. B1 for the position, B1 for the moment argument. That single fact fixes the angle at which any suspended lamina hangs.A lamina has its centre of mass 2 units horizontally and 3 units vertically from the corner it is hung from. Find the angle the vertical edge makes with the vertical.
Worked answer
The line from the pivot to G makes an angle arctan(2/3) with the vertical, so the edge tilts by 33.7°. B1 for taking the line to G as vertical, M1 for tan θ = 2/3, A1 for the angle. Dividing the other way round would give 56.3°, the angle to the horizontal, so state which angle you have found.A uniform block 0.5 m wide and 1.2 m tall stands on a rough plane with coefficient of friction 0.5. The plane is slowly tilted. Find whether it topples or slides, and at what angle.
Worked answer
Toppling: the base half-width is 0.25 and the height of G is 0.6, so tan θ = 0.25/0.6 = 0.4167 and θ = 22.6°. Sliding: tan θ = μ = 0.5, so θ = 26.6°. The smaller angle comes first, so the block topples at 22.6°. M1 for tan θ = 0.25/0.6, A1 for 22.6°, B1 for the sliding angle, A1 for choosing the smaller of the two.A uniform block 0.5 m wide and 1.2 m tall stands on a rough plane which is slowly tilted. Find the coefficient of friction for which toppling and sliding would happen at the same angle, and state what happens for coefficients above and below that value.
Worked answer
Toppling begins when tan θ = a/h, where a = 0.25 is the half-width of the base and h = 0.6 is the height of the centre of mass. Sliding begins when tan θ = μ.
The two coincide when μ = a/h = 0.25/0.6 = 0.4167. Setting the two tangents equal, rather than the two angles, is the quickest route and is worth the method mark.
Below that value the friction limit is reached at the smaller angle, so the block slides. Above it the line of the weight leaves the base first, so the block topples. M1 for equating the two tangents, A1 for 0.4167, B1 B1 for the two cases.
A smoother surface therefore protects a tall block from falling over, which runs against most people's intuition.Explain why the toppling condition is about the line of the weight rather than the height of the centre of mass.
Worked answer
The normal reaction acts somewhere within the area of contact. For equilibrium its moment about the centre of mass must balance, which is only possible if the vertical through the centre of mass meets the base. Once it falls outside, no position of the reaction can balance the moments and the block turns. Height matters only through where that vertical line lands. B1 for the reaction lying within the base, B1 for moments balancing only while the vertical through G meets it, B1 for the turn once it does not.A uniform block of weight 200 N stands on rough horizontal ground. Its base is a square of side 0.6 m and its height is 1.0 m, and the coefficient of friction between the block and the ground is 0.4. A horizontal force P is applied at the middle of the top edge and slowly increased. Determine whether the block slides or topples first, and find the value of P at which it does so.
Worked answer
Test the two possibilities separately and compare the forces they need. Working out only one of them is what loses this question.
Sliding. Vertically R = 200. The block slides when P reaches the limiting friction μR = 0.4(200) = 80 N.
Toppling. The block turns about the bottom edge furthest from P. At that instant the normal reaction has moved to that edge, so it has no moment about it, and only P and the weight remain. Taking moments about the edge: P acts at the top, a distance 1.0 m up, and the weight acts through the centre, 0.3 m horizontally from the edge.
So P(1.0) = 200(0.3), giving P = 60 N.
Sixty newtons is reached before eighty, so the block topples first, at P = 60 N.
B1 for R = 200, M1 A1 for the sliding value of 80 N, M1 for moments about the edge, A1 for P = 60 N, A1 for the comparison.
Saying where the normal reaction acts is worth a mark on its own, and taking moments about the centre of mass instead of the edge is the usual wrong start.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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