4. Forces, density and pressure

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.

The moment (torque) about a pivot is τ=F d⊥, where d⊥ is the shortest perpendicular distance from the pivot to the force’s line of action.

Choose the pivot, resolve the turning sense, and use the perpendicular distance rather than the distance to the point where the force is applied.

A 20 N force acting 0.30 m perpendicularly from a hinge gives a 6.0 N m moment; a force through the hinge gives zero moment.

A large force can have a small moment if its line of action is close to the pivot; distance along the rod is not always d⊥.

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.

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.

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.

Static equilibrium requires ΣF=0 and Στ=0. The body has no linear acceleration and no angular acceleration in the chosen frame.

Check force balance in each independent direction, then check moments about a convenient point. Satisfying one condition alone is insufficient.

A beam can have clockwise and anticlockwise moments balanced yet still accelerate sideways if a horizontal resultant force remains.

Zero resultant force does not prevent rotation, and zero resultant torque does not prevent translation.

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.

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 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.

For a static fluid, the pressure increase caused by a depth h is p=ρgh, from weight per unit area of the fluid column.

Derive it using ρ=m/V and p=F/A; state whether the pressure is gauge pressure or includes atmospheric pressure.

Water 3.0 m below the surface adds about 2.94×10⁴ Pa gauge pressure when ρ=1000 kg m⁻³ and g=9.8 m s⁻².

Depth, not container shape, sets the pressure in a connected static fluid; the equation is not valid when density varies without adjustment.

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.

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.

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.