Strand 3 Study Notes: Mechanical Systems (Machines, Power Transmission, Mechanisms, and Fluid Power)

Mechanical systems thinking: forces, moments, power, and efficiency

A mechanical system is any arrangement of components that transmits and transforms motion, forces, and energy to do useful work—think bicycles, gearboxes, robots, conveyors, presses, and elevators. When you study mechanical systems, you’re mostly asking a few recurring questions:

• What forces and moments exist at each component?
• How do speed and torque change as motion is transmitted?
• Where does energy go—useful output vs losses?
• Which component limits the performance (strength, traction, overheating, vibration)?

This section builds the “language” you’ll use everywhere else: force balance, torque, power, and efficiency.

Force, moment, and torque (the core ideas)

A force is a push or pull that can accelerate an object. A moment (or torque) describes the turning effect of a force about a point or axis. In mechanical systems, torque is often the most important quantity because shafts, gears, pulleys, and motors primarily transmit rotation.

For a force $F$ acting at a perpendicular distance (moment arm) $r$ from a pivot, the magnitude of the moment is:

$M = F\,r$

• $M$ is moment/torque in $\text{N\,m}$
• $F$ is force in $\text{N}$
• $r$ is perpendicular distance in $\text{m}$

Why this matters: many “simple machines” (levers, wrenches, cranks) work by increasing $r$ so the same force produces a larger torque.

Common conceptual trap: using the wrong distance. It must be the perpendicular distance from the axis to the line of action of the force—not just any length you see in a diagram.

Free-body diagrams and equilibrium

A free-body diagram (FBD) isolates one component and shows all external forces and moments acting on it. FBDs are the fastest way to avoid guesswork in mechanical systems.

If a component is in static equilibrium (not accelerating), then:

$\sum F_x = 0$
$\sum F_y = 0$
$\sum M = 0$

Even in moving machines, parts may be treated as quasi-static if accelerations are small compared to loads (for example, slow presses or steady conveyors). When accelerations matter, you add inertia terms (dynamics), but the FBD mindset stays the same.

Work, energy, and power in machines

Work is energy transferred by a force acting through a distance. In rotation, torque does work through an angular displacement.

Translational work:

$W = F\,d$

Rotational work:

$W = T\,\theta$

• $T$ is torque
• $\theta$ is angular displacement in radians

Power is the rate of doing work. It’s the most useful “bridge” between motors and loads.

Translational power:

$P = F\,v$

Rotational power:

$P = T\,\omega$

• $P$ in $\text{W}$
• $v$ in $\text{m/s}$
• $\omega$ in $\text{rad/s}$

A key idea you’ll use repeatedly: ideal transmissions trade speed for torque without changing power. Real transmissions lose some power to friction, heat, fluid shear, and deformation.

Mechanical advantage and speed ratio

A mechanical advantage (MA) compares output force (or torque) to input force (or torque). For a lever, the MA comes from geometry; for gears, it comes from gear ratio.

A related idea is velocity ratio (or speed ratio): how output speed compares to input speed. In an ideal machine:

• Higher torque output implies lower speed output.
• Higher speed output implies lower torque output.

Efficiency: where power goes

Real systems are never perfectly efficient. Efficiency is defined as:

$\eta = \frac{P_{\text{out}}}{P_{\text{in}}}$

Efficiency matters because it tells you:

• how large a motor must be,
• how much heat will be generated,
• whether a design is practical or will overheat/wear quickly.

Misconception to avoid: “Gear ratios increase power.” They don’t. Gear ratios repackage power as different combinations of torque and speed; efficiency determines how much power survives the transmission.

Worked example: power, torque, and speed

A motor delivers $P = 2.0\,\text{kW}$ at $n = 1500\,\text{rpm}$. Find torque.

1) Convert rotational speed to angular speed:

$\omega = 2\pi\,\frac{n}{60} = 2\pi\,\frac{1500}{60} = 50\pi\,\text{rad/s}$

2) Use $P = T\,\omega$:

$T = \frac{P}{\omega} = \frac{2000}{50\pi} \approx 12.7\,\text{N\,m}$

Interpretation: at this speed, the motor torque is modest; if you need much higher torque at the load, you’ll typically use a gearbox or belt reduction.

Exam Focus

• Typical question patterns:
  • Given speed and power, compute torque (or vice versa) and interpret what changes through a transmission.
  • Use equilibrium (sum of moments) to find reaction forces at supports or required input force at a lever/crank.
  • Compute efficiency and determine required input power for a desired output.
• Common mistakes:
  • Mixing $\text{rpm}$ with $\text{rad/s}$ in $P = T\,\omega$ (always convert).
  • Using the wrong moment arm (not perpendicular).
  • Treating torque as conserved through gears (power is the better conservation idea, then apply efficiency).

Shafts, bearings, and couplings: carrying loads and transmitting torque

Many mechanical systems are built around shafts (to transmit torque) supported by bearings (to carry loads while allowing rotation) and connected using couplings (to join rotating members). Failures here can be catastrophic because these components often run continuously and are sensitive to misalignment, lubrication, and fatigue.

Shafts: what they do and what can go wrong

A shaft is a rotating element designed to transmit torque and sometimes support gears, pulleys, or impellers. A shaft experiences:

• torsion from transmitted torque,
• bending from transverse loads (gear forces, belt tensions),
• axial loads in some cases (helical gears, thrust loads),
• fatigue because loads repeat each revolution.

Even if a shaft is sized for strength, it can still fail by excessive deflection (misalignment of gears/bearings), critical speed vibration, or fatigue at stress concentrators (keyways, shoulders, threads).

Torsion basics (strength and stiffness)

For a solid circular shaft of diameter $d$ transmitting torque $T$, the maximum shear stress at the surface is:

$\tau_{\max} = \frac{T\,c}{J}$

For a solid round shaft:

$c = \frac{d}{2}$
$J = \frac{\pi d^4}{32}$

So:

$\tau_{\max} = \frac{16T}{\pi d^3}$

Why this matters: it shows how strongly diameter affects torsional stress. Small increases in $d$ greatly reduce stress.

Torsional angle of twist over length $L$ is:

$\theta = \frac{T\,L}{J\,G}$

• $G$ is shear modulus

Stiffness matters when timing/position accuracy is important (robot joints, machine tools) or when twist causes uneven load sharing.

Common misconception: “If stress is OK, the shaft is OK.” In many systems, deflection and fatigue govern the design long before static strength.

Keys and splines: attaching hubs to shafts

When a gear or pulley is mounted on a shaft, you must transmit torque between the shaft and hub.

• A key is a removable piece that locks the hub to the shaft.
• A spline is a set of ridges/grooves that distributes torque around the circumference.

Keys are simple and common but introduce stress concentrations (keyway corners). Splines can handle higher torque with better concentricity but cost more to manufacture.

Bearings: reducing friction and carrying load

A bearing supports a rotating shaft while allowing relative motion with minimal friction. Two major categories:

• Rolling-element bearings (ball/roller): low friction, easy installation, limited by fatigue life and lubrication cleanliness.
• Plain bearings (journal bearings): shaft rides on a lubricating film; excellent for high speed and heavy loads when properly lubricated.

Bearings must be matched to load type:

• Radial loads: perpendicular to shaft axis (belt pull, gear forces).
• Thrust loads: along shaft axis (helical gear thrust, propellers).

Why lubrication matters: frictional heat and wear rise sharply when lubrication film breaks down. Contamination is a common real-world cause of premature bearing failure.

Couplings: connecting shafts (and managing misalignment)

A coupling connects two shafts to transmit torque. Couplings are selected based on:

• torque capacity,
• allowable misalignment (angular, parallel, axial),
• torsional stiffness (important for vibration),
• need for shock absorption.

Rigid couplings demand good alignment; flexible couplings tolerate some misalignment and reduce transmitted vibration.

Worked example: sizing a shaft in torsion (conceptual)

A solid shaft must transmit $T = 200\,\text{N\,m}$. Choose the minimum diameter so that $\tau_{\max} \le 40\,\text{MPa}$.

Use:

$\tau_{\max} = \frac{16T}{\pi d^3}$

Rearrange:

$d^3 = \frac{16T}{\pi\tau_{\max}}$

Substitute $T = 200\,\text{N\,m} = 200\times 10^3\,\text{N\,mm}$ and $\tau_{\max} = 40\,\text{MPa} = 40\,\text{N/mm}^2$:

$d^3 = \frac{16\times 200\times 10^3}{\pi\times 40} \approx 25465\,\text{mm}^3$

$d \approx 29.4\,\text{mm}$

In practice you’d round up and then check bending, deflection, keyway stress concentration, and fatigue.

Exam Focus

• Typical question patterns:
  • Compute shaft torque from power and speed, then compute torsional stress or angle of twist.
  • Identify appropriate bearing type for radial vs thrust loads and explain the reasoning.
  • Diagnose failures (misalignment, poor lubrication, fatigue at keyway) from symptoms.
• Common mistakes:
  • Unit slips between $\text{N\,m}$ and $\text{N\,mm}$ (be consistent).
  • Forgetting that mounted components create bending loads, not just torsion.
  • Assuming bearings only “reduce friction” and ignoring load direction and lubrication.

Gear systems: ratios, torque flow, and gear trains

Gears are used when you need compact, positive (no-slip) transmission with precise speed ratios—common in gearboxes, robots, clocks, and machine tools.

What gear ratio really means

For two meshing gears (gear 1 driving gear 2), the speed ratio is set by tooth counts:

$\frac{\omega_1}{\omega_2} = \frac{N_2}{N_1}$

• $\omega$ is angular speed
• $N$ is number of teeth

If $N_2 > N_1$, the driven gear turns slower and torque increases.

In an ideal (lossless) mesh:

$P_1 = P_2$

So:

$T_1\,\omega_1 = T_2\,\omega_2$

which implies:

$\frac{T_2}{T_1} = \frac{\omega_1}{\omega_2}$

In real gears, include efficiency $\eta$:

$P_2 = \eta\,P_1$

Spur, helical, bevel, and worm gears (how they differ)

• Spur gears: teeth parallel to axis; simplest; can be noisy at high speed due to abrupt tooth engagement.
• Helical gears: angled teeth; smoother and quieter; generate axial thrust that bearings must carry.
• Bevel gears: transmit between intersecting shafts (often 90°).
• Worm gears: high reduction in one stage; can be less efficient due to sliding contact; sometimes self-locking depending on geometry and friction.

Why this matters: gear type affects efficiency, noise, load direction (thrust), and packaging.

Gear tooth forces (the “hidden loads” on shafts and bearings)

Even if you only care about torque, gears create forces at the mesh that load shafts and bearings.

A helpful concept is tangential force at the pitch circle:

$F_t = \frac{T}{r}$

• $r$ is pitch radius

This tangential force is what transmits torque. Many gear types also create radial (and for helical gears, axial) components, which increase bearing load and shaft bending.

Gear trains: combining ratios

A gear train uses multiple gears to achieve larger reductions, reverse direction, or fit geometry.

• In a simple gear train, intermediate idler gears change direction/spacing but do not change the overall ratio (ratio depends on first and last gears).
• In a compound gear train, gears share shafts so ratios multiply.

For compound stages, overall speed ratio is the product of stage ratios.

Worked example: two-stage compound gear reduction

A motor drives gear A (20 teeth) which meshes with gear B (60 teeth). Gear B is on the same shaft as gear C (15 teeth), which drives gear D (45 teeth). Motor speed is $n_{\text{in}} = 1800\,\text{rpm}$, motor torque is $T_{\text{in}} = 5\,\text{N\,m}$. Assume overall efficiency $\eta = 0.90$. Find output speed and output torque.

1) Stage 1 ratio:

$\frac{\omega_A}{\omega_B} = \frac{N_B}{N_A} = \frac{60}{20} = 3$

So $\omega_B = \frac{\omega_A}{3}$.

2) Stage 2 ratio:

$\frac{\omega_C}{\omega_D} = \frac{N_D}{N_C} = \frac{45}{15} = 3$

Since B and C share a shaft, $\omega_C = \omega_B$. So overall reduction is $3\times 3 = 9$.

3) Output speed:

$n_{\text{out}} = \frac{1800}{9} = 200\,\text{rpm}$

4) Power in:

Convert input speed to angular speed:

$\omega_{\text{in}} = 2\pi\,\frac{1800}{60} = 60\pi\,\text{rad/s}$

$P_{\text{in}} = T_{\text{in}}\,\omega_{\text{in}} = 5\times 60\pi = 300\pi\,\text{W}$

5) Power out:

$P_{\text{out}} = \eta\,P_{\text{in}} = 0.90\times 300\pi = 270\pi\,\text{W}$

6) Output angular speed:

$\omega_{\text{out}} = 2\pi\,\frac{200}{60} = \frac{20\pi}{3}\,\text{rad/s}$

7) Output torque:

$T_{\text{out}} = \frac{P_{\text{out}}}{\omega_{\text{out}}} = \frac{270\pi}{20\pi/3} = 40.5\,\text{N\,m}$

Interpretation: about a 9x torque increase would be ideal (45 N·m), reduced here to 40.5 N·m by losses.

Exam Focus

• Typical question patterns:
  • Compute speed/torque changes through a gear pair or multi-stage gear train, including efficiency.
  • Determine direction of rotation (especially with idlers) and interpret what an idler does to ratio.
  • Use pitch radius to find tangential force at the gear mesh.
• Common mistakes:
  • Inverting the tooth ratio (mixing up driver vs driven).
  • Multiplying torque ratios without accounting for efficiency.
  • Ignoring that gear forces create bending loads on shafts and increase bearing reactions.

Belt, rope, and chain drives: flexible power transmission

Flexible drives are everywhere because they are cheap, quiet, and tolerant of misalignment. Compared with gears, they usually have lower cost and easier layout—but belts can slip and chains require lubrication and tensioning.

Belts vs chains: choosing the right system

• Belt drives (flat, V-belt, timing belt):
  • Can be quiet and absorb shock.
  • V-belts rely on friction and can slip under overload (sometimes a safety feature).
  • Timing belts have teeth and do not slip (more like gears).
• Chain drives:
  • Positive engagement with sprockets (no slip under normal operation).
  • Good for dirty environments compared to some belts, but still need lubrication.
  • Can be noisy and exhibit polygonal speed variation at high speed.

Speed ratio for belt and chain drives

If there is no slip (or for chains/timing belts), the belt speed is the same on both pulleys. With pulley diameters $D_1$ and $D_2$:

$\frac{\omega_1}{\omega_2} = \frac{D_2}{D_1}$

This mirrors gears: a larger driven pulley reduces speed and increases torque.

Slip warning: with friction belts, real ratios can deviate due to slip/creep, especially under high load or poor tension.

Tension difference creates torque

A belt transmits torque because the tight side tension exceeds the slack side tension.

Let $T_1$ be tight-side belt tension and $T_2$ be slack-side belt tension. The net driving force is:

$F = T_1 - T_2$

Torque on a pulley of radius $r$ is:

$T = (T_1 - T_2)\,r$

Power can be written as:

$P = (T_1 - T_2)\,v$

where $v$ is belt speed.

Friction limit (capstan equation)

For a friction belt on a pulley, the maximum tension ratio before slip is approximately:

$\frac{T_1}{T_2} = e^{\mu\theta}$

• $\mu$ is coefficient of friction
• $\theta$ is wrap angle in radians

Why this matters: increasing wrap angle (using an idler) or using a V-belt (wedging effect) increases transmissible torque without slip.

Worked example: belt drive power capacity

A belt runs over a pulley of radius $r = 0.10\,\text{m}$ at belt speed $v = 8.0\,\text{m/s}$. Tight-side tension is $T_1 = 600\,\text{N}$ and slack-side tension is $T_2 = 200\,\text{N}$.

1) Net driving force:

$F = T_1 - T_2 = 400\,\text{N}$

2) Torque on pulley:

$T = F\,r = 400\times 0.10 = 40\,\text{N\,m}$

3) Power transmitted:

$P = F\,v = 400\times 8.0 = 3200\,\text{W} = 3.2\,\text{kW}$

Interpretation: if the load demands more than 3.2 kW at that speed, either tensions must increase (within belt limits), pulley size must change, or you need a different transmission.

Exam Focus

• Typical question patterns:
  • Given pulley sizes and speeds, determine speed ratio and resulting torque change.
  • Use $P = (T_1 - T_2)\,v$ to compute power, or solve for required belt tensions.
  • Apply the capstan equation to determine whether slip occurs.
• Common mistakes:
  • Confusing torque $T$ with belt tension $T_1,\,T_2$ (same symbol used in many texts).
  • Using diameter when the formula needs radius (or vice versa).
  • Forgetting wrap angle must be in radians in $e^{\mu\theta}$.

Mechanisms and kinematics: linkages, cams, and degrees of freedom

A mechanism is a set of rigid bodies connected by joints that constrains motion in a useful way. Unlike transmissions (gears/belts) that mainly trade torque and speed, mechanisms often convert one motion type to another—rotation to oscillation, rotation to translation, intermittent motion, and so on.

Kinematics vs dynamics (what you are solving for)

• Kinematics describes motion (positions, velocities, accelerations) without considering forces.
• Dynamics connects motion to forces and torques.

Many mechanical-systems questions start kinematically (what motion results?) and then add dynamics (what torque/force is required?). Keeping the two steps separate prevents confusion.

Common joints and links

A link is a rigid body in a mechanism. A joint (kinematic pair) connects links and constrains relative motion.

Common planar joints:

• Revolute (pin) joint: allows rotation, one degree of freedom.
• Prismatic (slider) joint: allows translation, one degree of freedom.

Degrees of freedom and mobility (how constrained is the mechanism?)

In planar mechanisms, a widely used mobility estimate is:

$M = 3(n - 1) - 2j_1 - j_2$

• $M$ is mobility (degrees of freedom)
• $n$ is number of links including the ground
• $j_1$ is number of 1-DOF joints (revolute/prismatic)
• $j_2$ is number of 2-DOF joints (less common in simple mechanisms)

Why this matters: if $M = 1$, a single input (like a motor) can fully determine the motion. If $M > 1$, you may need multiple actuators or additional constraints. If $M \le 0$, the mechanism may be overconstrained or immobile.

Caution: mobility formulas are a guide, not a proof. Special geometries (parallel joints, redundant constraints) can make the real behavior different.

Four-bar linkages (the “workhorse” mechanism)

A four-bar linkage consists of four links forming a loop with four revolute joints. One link is fixed (ground). By choosing link lengths, you can create:

• crank-rocker motion (continuous rotation input produces oscillating output),
• double-crank motion,
• double-rocker motion.

Why it matters: four-bars are used in windshield wipers, suspension linkages, packaging machines, and many folding/closing systems.

Crank-slider (rotation to translation)

A crank-slider converts rotation into approximately sinusoidal translation—used in piston engines, pumps, and reciprocating compressors.

Key idea: the slider’s velocity and acceleration vary through the stroke, so inertia forces (and required torque) vary even under steady rotation.

Cams and followers (custom motion profiles)

A cam is a shaped rotating element that drives a follower according to a designed displacement profile. Cams are used when you need:

• specific timing (dwell, rise, return),
• non-sinusoidal motion,
• compact packaging.

Real-world constraint: aggressive cam profiles can cause high acceleration (jerk), leading to vibration, noise, and follower separation.

Worked example: mobility of a crank-slider

A planar crank-slider has:

• ground link (frame),
• crank, connecting rod, slider.

So $n = 4$ links.

Joints:

• revolute between ground and crank,
• revolute between crank and connecting rod,
• revolute between connecting rod and slider,
• prismatic between slider and ground.

All are 1-DOF joints, so $j_1 = 4$ and $j_2 = 0$.

$M = 3(4 - 1) - 2(4) = 9 - 8 = 1$

Interpretation: one input (crank rotation) determines the entire motion.

Exam Focus

• Typical question patterns:
  • Identify mechanism type from a diagram and describe what motion conversion it performs.
  • Compute mobility $M$ and interpret whether one actuator is sufficient.
  • Reason qualitatively about velocity/acceleration variation in crank-slider and cam systems.
• Common mistakes:
  • Miscounting links (forgetting the ground link is counted in $n$).
  • Treating mobility as always exact (ignoring special constraints).
  • Assuming slider speed is constant during crank rotation (it is not).

Fasteners and joints: bolts, threads, and load paths

Mechanical systems are assemblies. Even a perfectly sized gear or shaft can fail if the joints are poorly designed—loose bolts, slipping clamps, fatigue in threaded regions, or crushing of connected parts.

Threads and why bolts work

A bolt is essentially a spring: when tightened, it stretches and creates preload (clamping force) that holds joint members together by friction.

Why preload matters:

• It prevents separation under external load.
• It helps the joint resist shear by friction rather than by bolt shank bearing.
• It reduces fatigue by keeping bolt load variation smaller (when properly designed).

A common misconception is that the bolt “holds parts together by its shear strength.” In many well-designed joints, the goal is that the parts are held by clamping friction, and the bolt mostly stays in tension.

Torque–preload relationship (practical, not exact)

In practice, tightening torque is often related to preload by an approximate relation:

$T \approx K\,F\,d$

• $T$ is tightening torque
• $F$ is desired preload
• $d$ is nominal bolt diameter
• $K$ is a factor capturing thread and under-head friction

This relation is useful conceptually, but it is not highly accurate because friction varies widely. Critical joints often use direct tension measurement, angle tightening, or controlled lubrication.

Shear, bearing, and tear-out in joints

When loads act parallel to the joint interface, several failure modes are possible:

• Shear of bolt: bolt shank shears across a plane.
• Bearing failure: bolt presses into the plate material, causing crushing/ovalization.
• Net-section tension/tear-out: the plate tears from the hole to the edge.
• Slip: friction is exceeded and the joint slides (problem for alignment and fatigue).

Good joint design is about load path clarity: you want to know whether shear is intended to be carried by friction (preloaded joint) or by direct bolt bearing (non-preloaded, clearance holes).

Worked example: clamp friction vs direct shear (conceptual)

Suppose two plates are clamped by a bolt with preload $F = 10\,000\,\text{N}$ and the friction coefficient at the interface is $\mu = 0.3$. The maximum shear force resisted without slipping is approximately:

$F_{\text{shear,max}} = \mu\,F = 0.3\times 10\,000 = 3000\,\text{N}$

If the applied shear exceeds this, the joint will slip unless there are dowels, fitted bolts, or more preload/rougher surfaces.

Exam Focus

• Typical question patterns:
  • Explain how preload helps prevent joint separation and reduces slip.
  • Determine whether a joint will slip given preload and friction.
  • Identify likely joint failure mode from loading direction and geometry.
• Common mistakes:
  • Assuming bolt strength alone prevents slip (ignoring friction and preload).
  • Treating tightening torque as an exact indicator of preload.
  • Ignoring edge distance and plate tear-out when placing holes.

Fluid power systems: hydraulics and pneumatics for actuators

Fluid power uses pressurized fluids to transmit power. Hydraulics (liquid) are used for high force and stiffness (excavators, presses), while pneumatics (gas) are used for fast, clean actuation with simpler infrastructure (factory automation).

Pressure and Pascal’s principle

Pressure is force per area:

$P = \frac{F}{A}$

In a confined fluid, applied pressure is transmitted throughout the fluid (Pascal’s principle). This is why a small input force on a small piston can create a large output force on a larger piston.

For a cylinder with pressure $P$ and piston area $A$, the actuator force is approximately:

$F = P\,A$

Why it matters: fluid power gives you a direct, scalable way to generate large forces without huge mechanical leverage.

Flow rate controls speed

While pressure is tied to force, flow rate is tied to actuator speed.

Volumetric flow rate:

$Q = A\,v$

• $Q$ in $\text{m}^3/\text{s}$
• $A$ piston area
• $v$ piston velocity

This is the key design trade:

• high force needs high pressure and/or large area,
• high speed needs high flow.

Hydraulic vs pneumatic behavior

Hydraulic fluids are nearly incompressible, so hydraulics are stiff (good for precise force control and holding loads). Air is compressible, so pneumatics are springy—good for rapid motion, but less ideal for precision positioning or constant force under varying load.

Valves and circuits (how control actually happens)

In real systems, actuators don’t connect directly to a pump/compressor. Valves manage:

• direction (extend/retract),
• flow (speed control),
• pressure (force limiting and safety).

Even without memorizing every symbol, you should understand the logic: the pump provides flow; the relief valve protects against overpressure; the directional valve routes flow; the throttle/flow-control valve sets speed.

Worked example: cylinder sizing from force and pressure

A hydraulic cylinder must produce $F = 50\,000\,\text{N}$ at available pressure $P = 10\,\text{MPa}$. Find required piston area and diameter.

1) Area from $F = P\,A$:

$A = \frac{F}{P} = \frac{50\,000}{10\times 10^6} = 5.0\times 10^{-3}\,\text{m}^2$

2) Diameter from $A = \frac{\pi d^2}{4}$:

$d = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4\times 5.0\times 10^{-3}}{\pi}} \approx 0.0798\,\text{m}$

$d \approx 80\,\text{mm}$

Interpretation: if you need faster motion, you must also check whether the pump can supply enough $Q$ for the desired speed.

Exam Focus

• Typical question patterns:
  • Compute cylinder force from pressure and area, or size a cylinder for required force.
  • Compute actuator speed from flow rate using $Q = A\,v$.
  • Explain practical differences between hydraulics and pneumatics (stiffness, force density, cleanliness, control).
• Common mistakes:
  • Confusing pressure (force capability) with flow (speed capability).
  • Mixing gauge and absolute pressure without stating assumptions (especially in pneumatics).
  • Forgetting unit conversions for $\text{MPa}$ to $\text{Pa}$.

Brakes and clutches: friction torque and heat

Brakes and clutches both rely on friction to transmit torque. A clutch connects/disconnects power flow; a brake removes kinetic energy to slow/stop motion. In mechanical systems, these are often limited not by peak torque but by heat (energy that must be dissipated) and by wear.

Friction fundamentals

Dry friction force is often modeled as:

$F_f = \mu\,N$

• $\mu$ coefficient of friction
• $N$ normal force

For torque transmission, friction acts over a radius. A simple torque estimate is:

$T \approx \mu\,N\,R$

where $R$ is an effective radius. Real clutches/brakes often require integrating pressure distribution across a contact area; many curricula use “effective radius” or provide a formula for the specific geometry.

Why heat matters more than peak torque

When you stop a rotating system, you remove its kinetic energy. For a rotating inertia $I$ with angular speed $\omega$, the rotational kinetic energy is:

$E = \frac{1}{2}I\,\omega^2$

A brake must convert most of this energy into heat. If heat cannot be absorbed and rejected fast enough, you get fade (reduced friction), material damage, or fluid boil (in hydraulic brakes).

Worked example: braking energy (rotational)

A flywheel with $I = 2.5\,\text{kg\,m}^2$ is spinning at $n = 600\,\text{rpm}$. How much energy must the brake dissipate to stop it?

1) Convert speed:

$\omega = 2\pi\,\frac{600}{60} = 20\pi\,\text{rad/s}$

2) Energy:

$E = \frac{1}{2}I\,\omega^2 = 0.5\times 2.5\times (20\pi)^2$

$E \approx 4935\,\text{J}$

Interpretation: if stopping happens repeatedly, average power dissipation can be large and drives thermal design.

Exam Focus

• Typical question patterns:
  • Compute braking energy from inertia and speed, then relate it to heat load.
  • Estimate friction torque capacity from normal force and effective radius.
  • Compare clutch vs brake function and explain why overheating causes fade.
• Common mistakes:
  • Treating friction coefficient as constant under all temperatures and speeds.
  • Forgetting to convert $\text{rpm}$ to $\text{rad/s}$ in energy calculations.
  • Focusing only on torque capacity and ignoring energy/thermal limits.

Springs and vibration basics: stiffness, natural frequency, and resonance

Mechanical systems often include elasticity—sometimes intentionally (springs, compliant couplings), sometimes as an unavoidable property of materials. Elasticity plus mass creates vibration. If you ignore vibration, you can end up with noisy, inaccurate, or even failing machines.

Springs: force–deflection and stored energy

A linear spring follows Hooke’s law:

$F = k\,x$

• $k$ is stiffness in $\text{N/m}$
• $x$ is deflection

The energy stored in a linear spring is:

$E = \frac{1}{2}k\,x^2$

Why this matters: springs store and release energy, which can smooth motion (suspension), maintain contact (cam followers), or create return forces (valves). But spring energy can also drive oscillations.

Mass–spring natural frequency (why resonance happens)

For a simple mass–spring system (mass $m$, stiffness $k$), the undamped natural angular frequency is:

$\omega_n = \sqrt{\frac{k}{m}}$

and natural frequency in Hz is:

$f_n = \frac{\omega_n}{2\pi}$

If a system is excited near $f_n$, oscillation amplitude can grow large—this is resonance.

Damping: why vibrations don’t grow forever

Real systems have damping (energy dissipation through friction, material hysteresis, fluid drag). Damping reduces resonance peaks and helps systems settle.

A practical design mindset:

• avoid operating continuously at resonant speeds,
• increase stiffness (raise $f_n$) or mass (lower $f_n$) depending on constraints,
• add damping when possible.

Worked example: natural frequency

A component of mass $m = 5.0\,\text{kg}$ is mounted on an isolator of stiffness $k = 20\,000\,\text{N/m}$.

1) Natural angular frequency:

$\omega_n = \sqrt{\frac{20\,000}{5.0}} = \sqrt{4000} \approx 63.2\,\text{rad/s}$

2) Frequency in Hz:

$f_n = \frac{63.2}{2\pi} \approx 10.1\,\text{Hz}$

Interpretation: periodic forcing near 10 Hz (or rotating speed near 606 rpm if it maps 1:1 to excitation frequency) may produce large vibrations.

Exam Focus

• Typical question patterns:
  • Use $F = k\,x$ to find stiffness, deflection, or force.
  • Compute natural frequency $\omega_n$ or $f_n$ and discuss resonance risk.
  • Relate vibration issues to stiffness, mass, and damping changes.
• Common mistakes:
  • Confusing $\omega_n$ (rad/s) with $f_n$ (Hz).
  • Treating stiffness as a force (stiffness is force per displacement).
  • Ignoring that adding flexibility can introduce new resonances and alignment issues.

Integrating mechanical systems: reading load paths and matching components

Real machines are systems: changing one element changes the loads and performance of others. A reliable way to “think like a mechanical systems designer” is to trace load paths and power paths.

Power path vs load path

• Power path: where energy flows (motor → transmission → load). You track $P$, $\omega$, $T$, efficiency.
• Load path: where forces travel (gear mesh forces → shaft bending → bearings → frame). You track forces, moments, reactions, deflections.

A gearbox problem, for example, is never just “compute gear ratio.” You also need to consider:

• gear tangential force increases with transmitted torque,
• higher gear forces increase shaft bending,
• increased bearing loads reduce bearing life and increase friction,
• misalignment worsens gear contact and noise.

A systematic analysis workflow (how to avoid getting lost)

When you’re given a mechanical system to analyze, a consistent workflow prevents errors:

1) Define the function: what motion/force is needed at the output?
2) Choose a transmission/mechanism: gear, belt, linkage, hydraulic cylinder, etc.
3) Compute power and torque from output back to input (include $\eta$).
4) Compute forces at interfaces (gear mesh forces, belt tensions, cylinder loads).
5) Draw FBDs for shafts/supports to find reactions.
6) Check limiting factors: stress, deflection, slip, overheating, vibration.

Mini worked integration example: motor + gearbox + drum

A winch drum of radius $r = 0.15\,\text{m}$ must pull $F = 2000\,\text{N}$ at rope speed $v = 0.30\,\text{m/s}$. A gearbox with reduction 20:1 and efficiency $\eta = 0.85$ is used.

1) Output torque required at drum:

$T_{\text{out}} = F\,r = 2000\times 0.15 = 300\,\text{N\,m}$

2) Output power at drum:

$P_{\text{out}} = F\,v = 2000\times 0.30 = 600\,\text{W}$

3) Input power needed:

$P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{600}{0.85} \approx 706\,\text{W}$

4) Drum speed:

$\omega_{\text{out}} = \frac{v}{r} = \frac{0.30}{0.15} = 2.0\,\text{rad/s}$

5) Motor speed:

$\omega_{\text{in}} = 20\times 2.0 = 40\,\text{rad/s}$

6) Motor torque:

$T_{\text{in}} = \frac{P_{\text{in}}}{\omega_{\text{in}}} = \frac{706}{40} \approx 17.7\,\text{N\,m}$

Interpretation: the motor torque is much smaller than drum torque because the gearbox multiplies torque; the motor power must be higher than the drum power because of losses.

Exam Focus

• Typical question patterns:
  • Multi-step system problems that combine torque-speed-power with efficiency and an actuator geometry (drum, pulley, gear).
  • “Identify the limiting component” questions based on slip, maximum stress, or overheating.
  • Explain design choices qualitatively (why gears vs belts vs hydraulics for a given task).
• Common mistakes:
  • Jumping straight to equations without defining the power path and load path.
  • Applying the gear ratio to power instead of speed/torque (power changes mainly due to efficiency).
  • Forgetting that interface forces (gear mesh, belt tension) feed into shaft and bearing loading.