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AS 4

Forces, density and pressure

AS Level · syllabus topic 4 · theory on Papers 1 and 2; the practical skills behind it on Paper 3

This is the topic that decides whether a body stays still. Equilibrium questions are generous with marks and unforgiving about method: they reward a clear free-body diagram and punish a guessed one. Weight, friction and drag were dealt with under topic 3 and are assumed here.

🎯What you need to be able to do

  • Define and use density and pressure, and derive and use \( p = \rho gh \) for the pressure difference in a fluid.
  • Explain the origin of upthrust acting on a body in a fluid, and use Archimedes’ principle, \( F = \rho gV \).
  • Define and use the moment of a force, and the torque of a couple.
  • State and apply the principle of moments.
  • State and apply the two conditions for equilibrium: zero resultant force and zero resultant moment about any point.
  • Use a vector triangle to represent coplanar forces in equilibrium, and resolve forces into perpendicular components.
  • Understand and use centre of gravity.

📚The physics

Density and pressure. \( \rho = m/V \); \( p = F/A \), where the force must be perpendicular to the area. The pressure at depth \(h\) in a fluid of density \(\rho\) exceeds the pressure at the surface by \( \rho gh \). Notice what is missing from that expression: the shape of the container and the total weight of fluid. A narrow tube and a wide tank filled to the same depth have the same pressure at the bottom.

Upthrust is not a new kind of force. It is the resultant of the pressure forces on a submerged body, and it exists because the pressure on the bottom face is greater than on the top. For a rectangular block of cross-section \(A\) and height \(h\) the difference is \( \rho gh \times A = \rho gV \), which is the weight of fluid displaced — Archimedes’ principle, derived rather than remembered.

A block submerged in water with small pressure arrows pushing down on its top face, larger arrows pushing up on its bottom face and equal arrows on the sides. The resultant is rho g h A, the weight of fluid displaced.
The pressure on the bottom face exceeds that on the top by \( \rho gh \); times the area \( A \) this is \( \rho gV \), the weight of fluid displaced.

Moments. The moment of a force about a point is the force multiplied by the perpendicular distance from the point to the line of action of the force. A couple is a pair of equal, antiparallel forces whose lines of action do not coincide; its torque is one force times the perpendicular separation, and it is the same about every point.

Left: a force applied at an angle to the end of a bar, with its line of action extended and the perpendicular distance d from the pivot drawn at right angles to it. Right: a couple, two equal and opposite forces a distance d apart.
Moment \( = Fd \) with \( d \) perpendicular to the line of action; a couple has zero resultant force and torque \( Fd \) about every point.

Equilibrium requires two conditions together. First, the vector sum of the forces is zero — equivalently, the resolved components in any two perpendicular directions each sum to zero, or the force vectors form a closed polygon. Second, the sum of the moments about any point is zero. Because the second condition holds about any point, you may choose the point that eliminates the force you know least about; taking moments about the line of action of an unknown force removes it from the equation entirely.

Centre of gravity is the single point at which the whole weight of a body may be taken to act. For a uniform body in a uniform field it coincides with the centre of mass and lies at the geometrical centre.

✏️Worked example

A uniform plank of weight 220 N and length 4.0 m rests horizontally on two supports, one at the left-hand end and one 3.0 m from that end. A child of weight 340 N stands 3.5 m from the left-hand end. Find the force from each support.

Take moments about the left-hand support, which removes the unknown force \( R_1 \) from the equation. The plank is uniform, so its weight acts at 2.0 m.

Clockwise moments \( = 220 \times 2.0 + 340 \times 3.5 = 440 + 1190 = 1630 \) N m. Anticlockwise moment \( = R_2 \times 3.0 \).

\( R_2 = 1630/3.0 = 543 \) N \( \approx 540 \) N.

Now use zero resultant force vertically: \( R_1 + R_2 = 220 + 340 = 560 \) N, so \( R_1 = 560 - 543 = 17 \) N.

Free-body diagram of the 4.0 metre plank: support forces of 17 newtons at the left end and 543 newtons at 3.0 metres pointing up; the plank's 220 newton weight at 2.0 metres and the child's 340 newtons at 3.5 metres pointing down.
The worked example as a free-body diagram. The plank’s weight acts at its own midpoint, 2.0 m, whatever the supports do.
The mark people actually lose here is forgetting the second condition. Plenty of candidates take moments, find \( R_2 \), and stop — or they find \( R_1 \) by taking moments a second time and make an arithmetic slip. Worse, some place the plank’s weight at 1.5 m because the second support is at 3.0 m. The weight of a uniform plank acts at its own midpoint, 2.0 m, regardless of where the supports are. Mark the centre of gravity on your diagram before you write a single moment.

🔭See it happen

A metre rule, a pivot and a set of masses will demonstrate the principle of moments in five minutes, but the more instructive version is to hang the rule off-centre and ask students to predict the reading on a newton-meter before they take it. For upthrust, weigh an object in air and then submerged: the loss in weight is the upthrust, and it equals the weight of water that overflows.

🔗Go deeper — other people’s work

The links below are not mine. They are here because they are good, and they may move or disappear without warning.

  • PhET, Balancing Act and Buoyancy — both give immediate numerical feedback on predictions.
  • The Physics Classroom, “Static Equilibrium” — strong on free-body diagrams.
  • Isaac Physics, “Statics” problem set.