CourseCourseQuestionQuestionDocumentDocument

4. Forces, density and pressure

Syllabus
9702–2028–2029
Section
4
Level
AS

4.1 Turning effects of forces

Syllabus
9702–2028–2029
Topic
4.1
Level
AS

Treat an object’s weight as acting at its centre of gravity

The centre of gravity is the point through which the resultant weight of an object may be taken to act.

For a uniform symmetric object it is often at the geometric centre; for a non-uniform object it depends on the mass distribution. Use the point to calculate moments.

A uniform metre rule has its weight acting at the 50 cm mark, so its own moment about that point is zero.

The centre of gravity need not lie inside an irregular body, and it is not automatically the centre of the drawing.

Moment equals force times perpendicular distance from the pivot

moment=Fdperpendicularmoment=F d_{perpendicular}

The moment of a force about a pivot is force multiplied by the shortest perpendicular distance from the pivot to the force's line of action. State whether the turning sense is clockwise or anticlockwise.

Choose the pivot, extend the force's line of action if needed, find its perpendicular distance and multiply by force. A line of action through the pivot gives zero moment.

A 20 N force with perpendicular distance 0.30 m from a hinge produces a moment 20 × 0.30 = 6.0 N m.

Distance to the point where the force is applied is not necessarily the perpendicular lever arm. Torque of a couple is defined separately using force-line separation.

A couple is two equal opposite forces whose separation creates a pure turning effect

A couple consists of two equal, parallel forces acting in opposite directions on different lines. Its resultant force is zero, but it produces a moment.

The turning effect is independent of the chosen pivot. Use the perpendicular separation of the two lines of action and identify the sense of rotation.

Two 15 N forces separated by 0.20 m form a couple with moment 3.0 N m, even though their resultant force is zero.

A pair of equal forces only forms a couple when their lines of action are separated; collinear opposite forces cancel without rotation.

The moment of a couple is force multiplied by the perpendicular separation of its forces

For a couple, moment τ=F s, where F is the magnitude of either force and s is the perpendicular distance between their lines of action.

Use one force magnitude, not the sum of both forces. The couple moment is the same about any pivot because the net force is zero.

Increasing the handle separation from 0.10 m to 0.25 m at 12 N increases the turning moment from 1.2 to 3.0 N m.

Do not use the distance from a pivot to one force when the problem asks for a couple; use the force-line separation.

4.2 Equilibrium of forces

Syllabus
9702–2028–2029
Topic
4.2
Level
AS

In rotational equilibrium, clockwise and anticlockwise moments about a pivot balance

The principle of moments states that if a body is in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments.

Choose a convenient pivot, use perpendicular distances and take a consistent sign convention. A force through the pivot has zero moment.

A 20 N load 0.40 m from a pivot is balanced by a 10 N load at 0.80 m on the opposite side.

Moment balance alone does not guarantee translational equilibrium; the resultant force must also be zero.

Equilibrium requires zero resultant force and zero resultant torque

ΣF=0andΣτ=0ΣF = 0 and Στ = 0

When both conditions hold, there is no linear acceleration and no angular acceleration. The object may be stationary or may move with constant velocity without changing its rotational state.

Resolve forces in each independent direction, then take moments about a convenient point. Satisfying either condition alone is insufficient.

Balanced clockwise and anticlockwise moments do not stop sideways acceleration if a horizontal resultant remains. Conversely, zero resultant force can coexist with a non-zero couple and angular acceleration.

Equilibrium means no acceleration, not necessarily no motion. Always test translation and rotation separately.

A vector triangle can show three coplanar forces in equilibrium

For three coplanar forces in equilibrium, placing their vectors head-to-tail forms a closed triangle because their vector sum is zero.

Keep each vector’s direction and scale, then use the triangle or sine rule to relate unknown magnitudes. A closed shape is a force-balance test.

For a symmetric load supported by two equal strings, the two tension vectors and the weight close to a triangle.

A force triangle represents vector addition, not the physical shape of the object or the path of motion.

4.3 Density and pressure

Syllabus
9702–2028–2029
Topic
4.3
Level
AS

Density is mass per unit volume and links material amount to size

Density is ρ=m/V. It is a property of a material sample at specified conditions, while mass and volume describe the particular object.

Use consistent units and distinguish an object’s external volume from the volume of material if voids or hollow regions are present.

A 0.54 kg metal block occupying 2.0×10⁻⁴ m³ has density 2700 kg m⁻³, consistent with aluminium.

A larger sample of the same uniform material is not automatically denser; changing mass and volume proportionally leaves ρ unchanged.

Pressure is force per unit area acting normally on a surface

Pressure is p=F/A, where F is the normal force distributed over area A. Its SI unit is the pascal, N m⁻².

Use the force perpendicular to the surface and the actual contact area. The same force produces greater pressure over a smaller area.

A 600 N person standing on 0.030 m² exerts an average pressure of 2.0×10⁴ Pa on the floor.

Pressure is not the total force, and forces parallel to a surface do not contribute to normal pressure.

Derive hydrostatic pressure from a fluid column

Consider a stationary fluid column of uniform density ρ, cross-sectional area A and vertical height difference Δh. The pressure difference supports the column's weight.

  1. Column volume V = AΔh. 2. From ρ = m/V, its mass m = ρAΔh. 3. Its weight F = mg = ρAΔh g. 4. From Δp = F/A, Δp = (ρAΔh g)/A = ρgΔh.

Δp=ρgΔhΔp = ρgΔh

This is a pressure difference between two levels in a fluid at rest with uniform density and uniform g. Atmospheric pressure cancels when it acts equally at both levels; container shape does not enter the derivation.

The pressure difference between two levels in a uniform fluid is ∆p=ρg∆h

Between two points separated vertically by ∆h in a fluid of uniform density, the pressure difference is ∆p=ρg∆h.

Only the vertical separation matters. Include a sign or state which point is deeper; pressure increases downward.

A 0.50 m level difference in oil of density 800 kg m⁻³ gives ∆p≈3.92 kPa at g=9.8 m s⁻².

Do not use the total container height when comparing two points, and do not reverse the pressure order with the depth sign.

Upthrust comes from the pressure difference between the lower and upper surfaces of an immersed object

Fluid pressure is greater at the lower surface than at the upper surface, giving a resultant upward force called upthrust.

The net force is caused by the pressure gradient, not by a mysterious property of the object. Compare upthrust with weight to predict rise, sink or equilibrium.

A submerged block has larger pressure on its bottom face than its top face, so the vertical pressure forces do not cancel.

Upthrust is not always equal to weight; equality occurs only when the object has no vertical acceleration.

The upthrust on a displaced volume of fluid is F=ρgV

For a fully or partly immersed object, upthrust equals the weight of displaced fluid: F=ρ_fluid g V_displaced.

Use the fluid density and displaced volume, not the object’s total volume unless it is fully submerged. Compare the result with object weight.

Displacing 2.0×10⁻³ m³ of water gives F≈19.6 N at g=9.8 m s⁻².

Upthrust depends on displaced volume and fluid density, not directly on the object’s material or mass.