Questions on equilibrium, suspension, toppling and sliding

Question 1

State where the centre of mass lies when a body is suspended freely, and why.

Accepted 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.

Question 2

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.

Accepted 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.

Question 3

A uniform solid cylinder of radius 0.2 m and height 0.9 m stands on one end on a rough plane. The plane is tilted slowly, and the friction is large enough to prevent sliding. Find the angle of the plane at which the cylinder topples.

Accepted Answer

The cylinder topples when the vertical through its centre of mass passes outside the lower edge of the base. The centre of mass is at half the height, 0.45 m up the axis, and the base extends 0.2 m from the axis to the edge, so at the toppling angle tan θ = 0.2/0.45 = 0.444 and θ = 24.0°. M1 for tan θ = radius over half-height, A1 for 0.444, A1 for 24.0°. The half-height is what enters, not the full height, and the radius is what enters, not the diameter: the angle is fixed by the position of G relative to the edge it tips over.

Question 4

State the two conditions for a rigid body acted on by coplanar forces to be in equilibrium.

Accepted Answer

The forces must have no resultant, so the components in two perpendicular directions each sum to zero; and the moments of the forces about any one point must sum to zero. B1 for the zero resultant force, B1 for the zero resultant moment. Once the first condition holds, the moment sum is the same about every point, which is why the point can be chosen to make an unknown force drop out. Both conditions are needed: two equal and opposite forces on different lines have zero resultant and still turn the body.

Question 5

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.

Accepted 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.

Question 6

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.

Accepted 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.
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.

Question 7

Explain why the toppling condition is about the line of the weight rather than the height of the centre of mass.

Accepted 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.

Question 8

Two forces of 8 N act on a rigid body in opposite directions along parallel horizontal lines 0.35 m apart. Find the moment of the couple they form, and show that the moment about a point 0.5 m above the lower line is the same.

Accepted Answer

The two forces are equal and opposite, so their resultant is zero and the pair is a couple. B1 for naming the resultant zero. Its moment is Fd = 8 × 0.35 = 2.8 N m. M1 A1. About the higher point, the lower force acts 0.5 m below it and gives 8 × 0.5 = 4 N m one way; the upper force acts 0.5 − 0.35 = 0.15 m below it and gives 8 × 0.15 = 1.2 N m the other. The total is 4 − 1.2 = 2.8 N m again. A1 for the check. Neither separate moment matched the first calculation; only their difference does.

Question 9

A uniform rectangular block with a base 0.4 m wide and a height of 0.6 m rests in equilibrium on a rough plane inclined at 25° to the horizontal, with its base flat on the plane. Find the least possible coefficient of friction, and confirm that the block is not on the point of toppling.

Accepted Answer

Resolving along and perpendicular to the plane for a block of weight W: F = W sin 25° and R = W cos 25°. Since F ≤ μR, μ ≥ tan 25° = 0.466, so the least coefficient is 0.466. For toppling, the vertical through G leaves the base when tan θ = (0.2)/(0.3) = 0.667, that is at 33.7°; the plane is at 25°, below that, so the block does not topple. M1 for resolving in both directions, A1 for μ ≥ tan 25° = 0.466, M1 for comparing tan 25° with the half-width over half-height ratio, A1 for the conclusion. With μ exactly 0.466 the block is on the point of sliding; a squatter block with a wider base would need the same μ, since the sliding condition does not involve the shape at all.

Question 10

A uniform cube of side 0.5 m and weight 100 N rests on rough horizontal ground, with coefficient of friction 0.6. A horizontal force is applied to a vertical face at a height h above the ground and increased slowly until the cube moves. Find the greatest value of h for which the cube slides rather than topples.

Accepted Answer

Sliding: R = 100 N, so the force needed is μR = 0.6 × 100 = 60 N, whatever the height. Toppling: at the instant of tipping the reaction acts at the far bottom edge, and moments about that edge give P × h = 100 × 0.25, so the force needed is P = 25/h. The cube slides first when 60 < 25/h, that is h < 25/60 = 0.417 m. B1 for the sliding force 60 N, M1 for moments about the bottom edge, A1 for P = 25/h, A1 for h < 0.417 m. The higher the push, the smaller the force that tips the cube, so a low push slides it and a high push tips it, with the changeover five sixths of the way up the face.

Question 11

A uniform L-shaped lamina is made of a 6 by 2 rectangle with corners (0, 0) and (6, 2) and a 2 by 6 rectangle with corners (0, 2) and (2, 8); its centre of mass is at (2, 3). The lamina is suspended freely from the corner (0, 8). Find the angle that the edge from (0, 8) to (0, 0) makes with the vertical.

Accepted Answer

The lamina hangs with G vertically below the pivot. In the lamina's own frame, G is 2 units across from the pivot and 8 − 3 = 5 units down from it. The line from the pivot to G becomes vertical, and the edge from (0, 8) to (0, 0) makes angle arctan(2/5) with it: 21.8°. B1 for G below the point of suspension, M1 for the offsets 2 across and 5 down, A1 for tan θ = 2/5, A1 for 21.8°. The edge swings away from the side where the mass is, so the bottom of that edge moves to the left of the pivot; the horizontal edge along y = 8 tilts by the same 21.8° from the horizontal.

Question 12

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.

Accepted Answer

Test the two possibilities separately and compare the forces they need.
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.

Question 13

Explain why any system of coplanar forces acting on a rigid body can be reduced to a single force through a point of your choosing together with a couple, and describe what the body does when that force is zero but the couple is not.

Accepted Answer

A force may be moved onto a parallel line through any chosen point, provided a couple is added to pay for the move, the couple's moment being the force times the distance the line shifted. Doing that to every force in turn piles them all through one point, where they add to a single resultant, and the couples add to a single couple. B1 for the move-plus-couple idea. Equilibrium then asks for both to vanish: no resultant force and no resultant couple, which is why the moment condition can be checked about any point at all. B1 for both conditions. When the force is zero and the couple is not, the body turns without being pushed anywhere: a turning effect with no resultant force is exactly what a couple is. B1 for the turn without translation.

Question 14

Three forces act on a rigid lamina in the plane of x and y axes: 5 N in the positive x direction at the point (0, 2), 3 N in the positive y direction at the point (4, 0), and 2 N in the negative x direction at the point (0, −2), distances in metres. Reduce the system to a single force through the origin together with a couple, and find where the line of action of a single resultant force with no couple would cross the x-axis.

Accepted Answer

Resultant force: components (5 − 2, 3) = 3i + 3j N, magnitude 3√2 = 4.24 N at 45° to the x-axis.
Moments about O, anticlockwise positive: the 5 N force acts 2 m above O and pushes to the right, giving 5 × 2 = 10 N m clockwise; the 3 N force acts 4 m to the right and pushes upward, giving 12 N m anticlockwise; the 2 N force acts 2 m below O and pushes to the left, giving 2 × 2 = 4 N m clockwise. Total: 12 − 10 − 4 = −2, i.e. 2 N m clockwise.
So the system is equivalent to 3i + 3j N through O together with a couple of 2 N m clockwise.
A single resultant 3i + 3j through the point (d, 0) has moment 3d about O, so 3d = −2 and d = −2/3: the line of action crosses the x-axis at (−0.667, 0).
M1 for adding the components, A1 for 3i + 3j, M1 for the moments about O with consistent signs, A1 for 2 N m clockwise, M1 for equating the moment of the single resultant to the couple, A1 for d = −2/3. Moving a force onto a parallel line through O leaves a couple equal to its moment about O, which is why the three moments added together are the couple that goes with the resultant.

Question 15

A uniform rectangular lamina ABCD has AB = 8 cm and BC = 6 cm. A particle of mass half that of the lamina is attached at the corner B. The loaded lamina is suspended freely from A. Find the position of the centre of mass and the angle that AB makes with the vertical.

Accepted Answer

Take A as the origin with AB along the x-axis, so B is (8, 0) and the lamina's centre is (4, 3). With the lamina of mass 2m and the particle m: x(G) = (2m × 4 + m × 8)/3m = 16/3 and y(G) = (2m × 3 + m × 0)/3m = 2, so G is at (16/3, 2), about (5.33, 2).
Hung from A, the line AG is vertical. AB makes angle arctan(2 ÷ 16/3) = arctan(3/8) with AG, so AB is at 20.6° to the vertical.
M1 for the mass-weighted average with the lamina as 2m, A1 for (16/3, 2), B1 for AG vertical, M1 for tan θ = 2/(16/3), A1 for 20.6°. The uniform rectangle alone would hang with AB at arctan(3/4) = 36.9° to the vertical; the particle at B pulls G towards AB and brings that edge closer to vertical.