Dynamics, forces and pressure

4. Forces, Density and Pressure

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4. Forces, Density and Pressure — CAMBRIDGE AS & A LEVEL PHYSICS 9702 — COMPLETE NOTES notes: Use centre of gravity, moments, couples and equilibrium conditions, and calculate density, pressure, hydrostatic pressure and upthrust.Centre of gravity — the point at which the weight o…

Centre of gravity versus centre of mass
Torque of a couple
Common forces
To draw the triangle
Unit conversions
Calculation

Section 1 of 15Centre of gravity

1Centre of gravity

Centre of gravity the point at which the weight of an object may be considered to act.
For a uniform symmetrical object it is at the point of symmetry, e.g. the centre of a sphere.
It can lie outside the body and can move when the shape or orientation changes;
- leaning forward lowers a person's centre of gravity.

2Centre of gravity versus centre of mass

Centre of gravity versus centre of mass they are the same in a uniform field,
but the centre of mass does not depend on the field whereas the centre of gravity does.
- (the Moon's centre of gravity is pulled slightly towards Earth)

Section 2 of 15Stability

More stable	Less stable
Wide base	Narrow base
Low centre of gravity	High centre of gravity
Centre of gravity above the base	Centre of gravity outside the base, so it topples

Stability

An object is stable while its centre of gravity lies vertically above its base.

The most stable objects have wide bases and low centres of gravity.

Section 3 of 15Moment of a force

Moment

Moment the turning effect of a force about a pivot.

Equation

Equation moment = Fd, in N m, where d is the perpendicular distance from the pivot to the line of action of the force.
If the force is at an angle, use the perpendicular component of the force or distance.
A door handle far from the hinge gives a larger moment, so less force is needed.
Carry extra significant figures until the final answer to avoid rounding errors.

Section 4 of 15Couples and torque

Couple

Couple a pair of forces equal in size, opposite in direction and acting along different lines, which produces rotation only.
The resultant force is zero, so the object does not accelerate linearly.

Torque of a couple

τ = Fd, where F is one of the forces and d is the perpendicular distance between them.
The forces of a couple cannot share a line of action.

Section 5 of 15Principle of moments

Statement

Statement for a system in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about the same point.

Method

Method apply the principle in these steps.
Choose a pivot and mark every force and its perpendicular distance.
Treat the weight of a uniform beam as acting at its centre.
Equate clockwise and anticlockwise moments and solve for the unknown.

Section 6 of 15Conditions for equilibrium

Equilibrium

Equilibrium there is no resultant force and no resultant torque, so the object stays at rest or moves with constant velocity and does not rotate.
A pivot can be chosen at the point of one force to remove its moment (d = 0).
Coplanar forces in equilibrium form a closed vector triangle (tip to tail) when drawn to scale.

Common forces

Common forces weight, normal reaction, tension and friction.

To draw the triangle

To draw the triangle make a free-body diagram, then place the arrows tip to tail keeping lengths and directions;
- the path must close.

Section 7 of 15Density

Density

Density mass per unit volume: ρ = m/V, in kg m⁻³.
Gases are less dense than liquids, which are less dense than solids, because there are fewer particles per unit volume.

Volumes

Volumes cuboid lwh; cylinder πr2h; sphere 4/3 πr3.

Unit conversions

Unit conversions larger to smaller unit: multiply;
- smaller to larger: divide;
- cube the factor for volumes (1 cm³ = 10⁻⁶ m3, 1 mm³ = 10⁻⁹ m3).

Section 8 of 15Pressure

Pressure
- force per unit area: p = F/A, in pascals (1 Pa = 1 N m⁻²).
  What the symbols mean: p = pressure (Pa = N m⁻²) · F = force acting at right angles to the surface (N) · A = area (m²)
- It is a scalar;
  - a force spread over a small area gives a large pressure (a high-heeled shoe) and over a large area gives a small pressure.
- Use the cross-sectional area in contact with the surface.

Section 9 of 15Hydrostatic pressure: Δp = ρgΔh

Consider a column of fluid of area A and height h above a point; its weight is W = mg.

Using ρ = m/V with V = Ah gives m = ρAh, so W = ρAhg.

Pressure = W/A = ρhg, so the change in pressure with depth is Δp = ρgΔh.

The pressure acts in all directions at a point in a fluid at rest and increases with depth, density and g.

Total pressure = hydrostatic pressure + atmospheric pressure (101 325 Pa).

Section 10 of 15U-tube manometer

A U-tube manometer measures pressure differences using the heights of liquid in the two arms.

With equal pressure on both sides the levels are equal.

What the symbols mean:

If the pressure on one side rises, that level falls and the other rises; the height difference gives the pressure difference (Δp = ρgΔh).

What the symbols mean: p = pressure due to the liquid (Pa) · ρ = density of the liquid (kg m⁻³) · g = 9.81 N kg⁻¹ · h = depth (m)

With two different liquids, the pressure at a point equals atmospheric pressure plus the weight of liquid above it, which lets densities be compared.

Section 11 of 15Upthrust

Upthrust

Upthrust the upward force on an object immersed in a fluid, caused by the difference in hydrostatic pressure between the bottom and top of the object.

Calculation

F = (pbottom − ptop)A, where A is the cross-sectional area of the object.
Upthrust is much greater in liquids than in gases because liquids are denser.
It is why objects seem lighter in water;
- the object rises if upthrust exceeds its weight, and floats if its density is less than the fluid's.

Section 12 of 15Archimedes' principle

Statement

Statement the upthrust on an object in a fluid equals the weight of the fluid displaced.

Equation

F = ρgV, where V is the volume of fluid displaced and ρ is the fluid density.
- (the immersed volume)

Floating

Floating the object floats when the upthrust equals its own weight.

Tip

Tip you need the immersed volume only to find the mass (m = ρV) and then the weight (mg) of the displaced fluid.

Section 13 of 15More from the specification

Section 14 of 15Key terms

Term	Meaning
Centre of gravity	The point where the weight of an object may be considered to act.
Moment	The turning effect of a force, equal to force × perpendicular distance from the pivot.
Couple	A pair of equal, opposite, parallel forces that produces rotation only.
Torque	The moment of a couple, τ = Fd.
Principle of moments	In equilibrium the total clockwise moment equals the total anticlockwise moment.
Density	Mass per unit volume.
Pressure	Force per unit area.
Upthrust	The upward force on an object in a fluid, equal to the weight of displaced fluid.

Section 15 of 15Exam technique

Using the distance along the beam instead of the perpendicular distance from the pivot to the line of action.

Treating the two forces of a couple as acting along the same line, or using the distance between a force and the pivot for the torque.

Forgetting atmospheric pressure when asked for total pressure at depth.

Believing upthrust depends on the density of the object — it depends on the displaced fluid (ρfluidgV).

Failing to cube or square unit conversions — 1 cm³ = 10-6 m³, not 10-2.