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Lesson 5 · 45 min

Equilibrium, simple machines, pressure and fluids

Connect force balance with moments, mechanical advantage, fluid pressure and upthrust.

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, simple machines, and fluids

What complete requires

An object in static has neither translational nor angular . In an elementary planar problem, resultant horizontal , resultant vertical and resultant moment about any chosen point are all zero. A beam can have no net vertical yet still turn if forces form a couple. Start with a free-body diagram showing weight, loads and support reactions. Choose a pivot that removes an inconvenient unknown from the moment equation. Include the beam's own weight unless its mass is explicitly negligible. For a uniform beam, weight acts through its midpoint in a uniform gravitational field.

Centre of gravity and stability

The centre of gravity is the point through which resultant weight may be considered to act. A freely suspended object settles with its centre of gravity vertically below the suspension point. Intersecting plumb lines from different suspension positions can locate it for an irregular lamina. A body is stable when a small produces a restoring effect, unstable when increases and neutral when its centre of gravity stays at the same height in the ideal model. A broad base and low centre of gravity usually improve to overturning. The weight line must remain within the support base to avoid an unsupported tipping tendency.

Key definition

Centre of gravity

The point through which resultant weight may be considered to act.

Exam keywords:resultantweightconsidered

Mechanical advantage and ratio

A machine can change the magnitude or direction of an effort . Mechanical advantage, MA, is load divided by effort in matching units. ratio, VR, is effort distance divided by load distance over the same interval. Efficiency is MA/VR as a decimal or100MA/VR as percentage. An ideal loss-free machine has. Real machines use input work in and deformation, so MA is normally smaller than VR. A large mechanical advantage does not create : the effort generally moves through a greater distance than the load rises. Show both displacements when explaining the exchange.

Key equation
MA=VR

Levers, pulleys and inclined planes

For a lever in balance, effort ×effort arm equals load ×load arm when lever weight and are negligible. Arms are perpendicular pivot-to--line distances. In an ideal pulley system count rope sections actually supporting the moving load, not every visible rope length. A single fixed pulley mainly changes direction and has ideal VR =1. For an ideal inclined plane, rise; effort decreases while its travel distance increases. Wheel-and-axle, thread and gear arrangements also exchange or torque for under conservation. Their ratios must follow the actual geometry rather than an arbitrary count of parts.

Key equation
VR=slopelengthvertical

and contact area

is normal per area: in pascals, where1. A parallel to a surface is not used directly as normal . The same gives greater average on a smaller area, explaining cutting edges and broad foundations. Area units need, so20. differs from total . Equal pressures on unequal piston areas produce different forces. In a fluid at rest acts normal to a surface, and the on an extended surface may depend on how varies over it.

Key equation
p=FA
Pa=1Nm²
care:1cm²=10⁻⁴m²
cm²=0.002m²

Hydrostatic and atmospheric measurements

For constant- liquid, increase with depth h isρgh. Absolute includes surface ; gauge is a difference from a reference such as atmosphere. At the same depth within one connected resting liquid is equal regardless of container shape. A manometer compares through a liquid-column height difference; identify which side is greater before setting signs. A mercury barometer balances atmospheric against a column under the relevant assumptions. Atmospheric generally decreases with altitude, allowing calibrated barometric altitude estimates, though weather affects readings. An aneroid barometer uses deformation of a sealed flexible chamber instead of a liquid column.

Pascal's principle and hydraulics

A change in an enclosed fluid is transmitted throughout the elementary incompressible system. For two comparable-level pistons. The larger piston can therefore exert greater . Volume conservation requires A₁d₁ when leakage and compressibility are negligible. The small piston travels farther, preserving ideal input and output work. A hydraulic press multiplies but not . Real equipment has losses, structural limits and operating precautions. When a question supplies piston diameters rather than areas, square their ratio because area is proportional to diameter squared; a tenfold diameter ratio does not only a tenfold ratio.

Key equation
F₁A₁=F₂A₂
=A₂d₂

Upthrust and floating conditions

A fluid's greater on a submerged body's lower surface produces an upward resultant. Archimedes' principle identifies upthrust as displaced fluid weight. A freely floating body at rest has upthrust equal to weight. Its submerged fraction is object /fluid for a uniform body under the simple model. Apparent weight in a fluid is weight minus upthrust when the support arrangement matches that definition. A ship's average includes enclosed spaces, so a steel hull can float despite solid steel being denser than water. A hydrometer floats to a depth depending on fluid and can be calibrated for a specified comparison.

Surface effects and terminal

Surface tension describes the tendency of an interface to resist increasing area through molecular interactions. Cohesion concerns attraction within one substance and adhesion between different materials; their balance affects wetting and capillary behaviour. Viscosity describes to relative flow between fluid layers. A falling body can reach terminal when weight equals upthrust plus drag. At that point is zero while remains nonzero. The drag relationship must match conditions, and it is not always constant or linear in . A separate -and-flow unit treats stress, strain and the conditions under which Stokes' drag model applies.

Self-check and reasoning

A machine lifts600 N with200 N effort and VR =4. Its MA is3 and efficiency75%. A hydraulic input area2 cm² receives50 N; output area20 cm² gives ideal500 N while moving one tenth as far. A block displacing0.003 m³ water receives upthrust1,000 ×10 ×0.003 =30 N at g =10 m/s². In a beam question balance moments and forces separately. A300 N downward load may require unequal support reactions depending on position even though their sum is300 N. Each result follows a defined , or fluid principle; identify the arrangement before selecting a convenient equation.

Topic revision bank

Key rules & principles

  • needs and moment balance.
  • Efficiency is MA/VR.
  • Upthrust equals displaced fluid weight.

Worked examples

A uniform4 m beam weighs100 N and carries200 N at1 m from the left. Find end reactions.
  1. Moments about left:4Rright =100(2) +200(1) =400.
  2. Rright =100 N.
  3. Vertical balance gives Rleft +100 =300.

Answer: Left200 N; right100 N.

Find upthrust on0.002 m³ fully submerged in water, withρ =1,000 and g =10.
  1. Displaced fluid volume equals submerged volume.
  2. Upthrust =ρgV.
  3. Use1,000 ×10 ×0.002.

Answer: 20 N upward.

Related videos and resources

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Statics: Crash Course Physics #13

CrashCourse

Static equilibrium; combine with the separate fluid resource and written machines explanations.

Fluids at Rest: Crash Course Physics #14

CrashCourse

Hydrostatic fluid concepts, not the complete machines unit.

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