Course Aim:

To equip cadets with fundamental physics knowledge essential for successful performance in maritime training and onboard operations.

Learning Objectives

By the end of this module, cadets will be able to:

1. Define fundamental physical quantities (length, mass, time, temperature) and their standard SI units.

2. Explain the importance and application of the International System of Units (SI) in maritime contexts.

3. Perform conversions between different measurement units (meters to nautical miles, km/h to knots, etc.).

4. Differentiate between precision and accuracy in measurements, providing practical maritime examples.

5. Calculate absolute and relative errors from measurement data and interpret their significance.

6. Demonstrate correct usage of measurement instruments relevant to maritime operations (e.g., measuring tapes, thermometers, gauges).

7. Apply correct rounding and significant figures rules to report measurement results clearly and accurately.

Key Concepts:

1. What is Measurement?

• Measurement is the process of obtaining the magnitude of a quantity relative to a defined standard or unit.

• In maritime operations, measurements provide vital data for navigation, stability, cargo management, maintenance, and engineering.

2. Importance of Accurate Measurement:

• Safety (e.g., accurate depth measurements prevent grounding)

• Operational efficiency (precise fuel measurements enhance performance)

• Compliance with regulations (maintaining proper records of measurements for inspection)

3. Basic Physical Quantities in Maritime Operations:

• Length and Distance: Used for navigation, ship dimensions, and cargo handling.

  • Units: meters (m), nautical miles (NM), kilometers (km)

• Mass: Essential for cargo loading, stability calculations, and ship structure analysis.

  • Units: kilograms (kg), tonnes (t)

• Time: Important for navigation calculations, voyage planning, maintenance schedules.

  • Units: seconds (s), minutes (min), hours (h)

• Temperature: Relevant for machinery, cargo storage, environmental control.

  • Units: degrees Celsius (°C), Kelvin (K)

• Pressure: Crucial for engine room operations, tank inspections, cargo safety.

  • Units: pascal (Pa), bar, psi (pounds per square inch)

4. Common Maritime Measurement Instruments:

• Measuring Tape: For length, width, and cargo measurements.

• Thermometer: For measuring temperature (engine coolant, cargo holds).

• Pressure Gauge: For monitoring tank and machinery pressures.

• Echo Sounder: For determining water depth below the keel.

Practical Maritime Examples:

• Measuring water depth to ensure safe navigation through shallow waters.

• Recording accurate cargo weights to maintain ship stability.

• Monitoring engine room pressures to prevent mechanical failures.

Summary of Key Points:

• Measurement is vital to safe, efficient maritime operations.

• Familiarity with common units and their applications onboard enhances operational competence.

• Accurate measurements lead to better decisions and safer voyages.

Quick Review Questions:

1. What is the definition of measurement?

2. List three physical quantities commonly measured onboard ships.

3. Why is accurate measurement important for maritime safety?

4. Name two units used for measuring pressure onboard.

Key Concepts:

1. Definition and Importance of SI Units:

• SI Units (International System of Units) provide a consistent global standard for measurements.

• Ensures uniformity, reducing confusion and errors in international maritime communication and operations.

3. Maritime Relevance of SI Units:

• Distance:

  • Meters (m): For short distances onboard (length of ropes, height of masts, cargo hold dimensions).

  • Nautical Miles (NM): Specifically for navigation and chart plotting (1 NM ≈ 1852 meters).

• Speed:

  • Knots: Maritime speed measurement equal to one nautical mile per hour (1 knot ≈ 1.852 km/h).

• Mass:

  • Kilograms (kg) and Tonnes (t): Essential for cargo weights, stability calculations, and fuel load monitoring.

• Temperature:

  • Celsius (°C): Commonly used onboard for practical temperature monitoring (engine cooling water, cargo conditions).

  • Kelvin (K): Used in specific scientific and engineering calculations onboard.

Example Conversions:

• Distance:

  • 1 nautical mile (NM) = 1852 meters (m)

• Speed:

  • 1 knot = 1.852 kilometers/hour (km/h) ≈ 0.514 m/s

• Temperature:

  • °C = K – 273.15

Practical Maritime Examples:

• Bridge officers use nautical miles to measure distances on navigational charts.

• Engine cadets use Celsius to monitor machinery temperatures and detect abnormalities.

• Deck crew measure ropes and mooring lines in meters for precision during operations.

Summary of Key Points:

• SI units provide global consistency, crucial for accurate maritime communication and operations.

• Understanding base and derived units simplifies maritime tasks and enhances onboard safety and efficiency.

• Familiarity with nautical-specific units (nautical miles, knots) is vital for professional maritime competency.

Quick Review Questions:

1. What is the purpose of using SI units in maritime operations?

2. List the four basic SI units commonly used onboard ships.

3. Convert 5 nautical miles to meters.

4. How many kilometers per hour is a ship traveling at 15 knots?

Key Concepts:

1. Why Unit Conversion Matters in Maritime Work:

• Ships use both SI units and traditional nautical units.

• Ensures consistency in voyage planning, cargo handling, maintenance, and reporting.

• Prevents operational errors and enhances safety.

3. Practical Shipboard Examples:

• Navigation: Converting chart distances in meters to nautical miles.

• Speed Monitoring: Converting GPS speed in km/h to knots for bridge reporting.

• Cargo Handling: Converting cargo mass from kg to tons for manifest entries.

• Engineering: Using °C and K in engine temperature monitoring and calculations.

Summary of Key Points:

• Unit conversions are essential for safety, reporting, and operational consistency.

• Mastery of standard conversion formulas prevents errors in voyage and cargo management.

• Regular practice with real-world shipboard scenarios builds competency.

Quick Review Questions:

1. Convert 9256 meters to nautical miles.

2. A ship travels at 10 knots. What is its speed in km/h?

3. Convert 4.5 tons to kilograms.

4. Convert 310 K to degrees Celsius.

Homework / Practice Exercise Examples:

1. Distance Conversion Practice:

• A buoy is 7.408 km from your current position. How many nautical miles is that?

2. Speed Conversion Practice:

• Your vessel is cruising at 12.5 knots. What is this in m/s?

3. Mass Conversion Practice:

• You are loading 3,200 kg of supplies. How many tons is this?

3. Error Calculations:

Absolute Error (AE):

The difference between the measured value and the true value.
Formula:

AE = |Measured Value – True Value|

Relative Error (RE):

Shows how significant the error is relative to the true value.
Formula:

RE = (Absolute Error ÷ True Value) × 100%

Examples:

1. Tank Level Measurement:

   • True value: 6.00 m

   • Measured value: 5.85 m

   • AE = |5.85 – 6.00| = 0.15 m

   • RE = (0.15 ÷ 6.00) × 100% = 2.5%

2. Pressure Reading Error:

   • True pressure: 3.0 bar

   • Reading: 2.7 bar

   • AE = 0.3 bar

   • RE = 10%

4. Practical Onboard Scenarios:

• Tank Sounding Error:
  Manual soundings may vary due to rough seas or incorrect tape alignment. Cadets must repeat measurements and average values for better accuracy.

• Engine Gauge Inaccuracy:
  Worn-out gauges may consistently display incorrect readings. Systematic error can be reduced by periodic calibration.

Weather Impact:

Random errors may arise from wind or wave effects during deck measurements or cargo checks. 

Summary of Key Points:

• Accuracy refers to closeness to the correct value, while precision refers to repeatability.

• Systematic errors are predictable and often due to instrument faults, while random errors vary unpredictably.

• Use absolute and relative error formulas to quantify measurement quality.

• Reducing errors improves safety, efficiency, and trust in shipboard operations.

Quick Review Questions:

1. What is the difference between accuracy and precision?

2. Name one example of systematic error on a ship.

3. Calculate the relative error if the true value is 20°C and the measured value is 18.5°C.

4. Why is it important to understand measurement errors in tank sounding?

Key Concepts:

1. Common Measurement Instruments Onboard:

Practical Onboard Examples:

• Measuring engine room temperature every 4 hours to detect overheating trends.

• Using measuring tape to check freeboard height before departure.

• Reading a pressure gauge on a compressed air bottle for maintenance checks.

• Echo sounder helping the bridge team avoid grounding in shallow areas.

Summary of Key Points:

• Instruments must be used correctly and checked regularly.

• Proper calibration ensures safety and compliance with international standards.

• Understanding how to read and interpret data improves decision-making and avoids costly errors.

Quick Review Questions:

1. Name four instruments commonly used onboard ships.

2. Why is calibration important for pressure gauges?

3. What precaution must be taken while reading a thermometer?

4. How does an echo sounder help during navigation?

This module introduces the fundamental concepts of motion (kinematics) and the forces that cause motion (dynamics). Understanding how objects move and interact with forces is crucial for navigating a vessel, managing loads, and maintaining ship stability.

Learning Objectives:

By the end of this module, cadets will be able to:

1. Differentiate between scalar and vector quantities with examples relevant to ship operations.

2. Define and distinguish between distance and displacement, speed and velocity.

3. Calculate average speed, velocity, and acceleration using kinematic formulas.

4. Explain uniform and non-uniform motion and their practical implications.

5. Apply Newton’s three laws of motion to shipboard scenarios.

6. Define momentum and explain the law of conservation of momentum.

7. Use Newton’s Second Law (F = ma) to solve motion-related problems.

8. Use vector addition to determine the resultant of two or more forces.

9. Apply dynamic principles to situations such as tug operations, anchoring, and mooring.

5. Kinematic Formulas

• Speed (v) = Distance (d) ÷ Time (t)
  v = d ÷ t

• Velocity (v) = Displacement (s) ÷ Time (t)
  v = s ÷ t

• Acceleration (a) = Change in velocity (Δv) ÷ Time (t)
  a = (v - u) ÷ t
  Where:

  • v = final velocity

  • u = initial velocity

  • t = time

Maritime Application Examples:

• Calculating the time for a ship to travel 100 nautical miles at 10 knots.
  → Time = Distance ÷ Speed = 100 ÷ 10 = 10 hours

• Estimating ship acceleration when increasing speed from 5 knots to 10 knots in 2 minutes.
  → a = (10 – 5) ÷ 2 = 2.5 knots/minute

Quick Review Questions:

1. What is the main difference between scalar and vector quantities?

2. If a ship sails 6 km north and then 8 km south, what is the total distance and displacement?

3. A ship travels 30 NM in 3 hours. What is its average speed?

4. A vessel speeds up from 8 knots to 12 knots in 4 minutes. What is the acceleration?

🔢 Kinematic Concepts – Problems & Answers

✅ Formula Recap:

1. Speed (v) = Distance ÷ Time

2. Velocity (v) = Displacement ÷ Time

3. Acceleration (a) = (Final Velocity – Initial Velocity) ÷ Time

🧮 Calculation Problems

Problem 1: Calculating Speed

A crew member walks 120 meters in 3 minutes.
What is the average speed in m/s?

Convert 3 minutes = 180 seconds
Speed = 120 ÷ 180 = 0.67 m/s

✅ Answer: 0.67 m/s

Problem 2: Calculating Velocity

A lifeboat is launched and travels 400 meters east in 200 seconds.
What is its average velocity?

Velocity = Displacement ÷ Time = 400 ÷ 200 = 2 m/s east

✅ Answer: 2 m/s east

Problem 3: Acceleration from Speed Change

A ship increases speed from 5 m/s to 15 m/s in 20 seconds.
What is the acceleration?

a = (v₂ - v₁) ÷ t = (15 - 5) ÷ 20 = 10 ÷ 20 = 0.5 m/s²

✅ Answer: 0.5 m/s²

Problem 4: Time to Accelerate

A small craft accelerates from 0 to 10 m/s with an acceleration of 2 m/s².
How long does it take?

t = (v₂ - v₁) ÷ a = (10 - 0) ÷ 2 = 5 seconds

✅ Answer: 5 seconds

Problem 5: Distance Traveled at Constant Speed

A vessel moves at a constant speed of 12 knots for 2 hours.
How far does it travel?

1 knot = 1 nautical mile/hour
Distance = 12 × 2 = 24 nautical miles

✅ Answer: 24 nautical miles

Problem 6: Displacement from Velocity

A ship travels due north at 7 m/s for 600 seconds.
What is the displacement?

Displacement = v × t = 7 × 600 = 4,200 meters north

✅ Answer: 4,200 meters north

Problem 7: Average Speed with Two Segments

A ship sails 30 km at 20 km/h, then 40 km at 10 km/h.
What is the average speed for the total trip?

Time₁ = 30 ÷ 20 = 1.5 h
Time₂ = 40 ÷ 10 = 4 h
Total Distance = 70 km
Total Time = 1.5 + 4 = 5.5 h
Average speed = 70 ÷ 5.5 ≈ 12.73 km/h

✅ Answer: Approximately 12.73 km/h

Key Concepts:

1. Distance-Time Graphs

• Y-axis: Distance

• X-axis: Time

🛳️ Ship Example: A constant upward sloping distance-time graph indicates a ship cruising at steady speed.

From a Distance-Time Graph:

• Speed = Slope of the line
  Speed = Change in Distance ÷ Change in Time

2. Speed-Time Graphs

• Y-axis: Speed or velocity

• X-axis: Time

🛳️ Ship Example: If a ship accelerates from rest to full speed, the graph shows an upward slope.

From a Speed-Time Graph:

• Acceleration = Slope of the line
  Acceleration = Change in Speed ÷ Change in Time

• Distance Traveled = Area under the speed-time graph

Example:
A ship accelerates from 0 to 10 knots in 5 minutes (speed-time graph = triangle).

• Acceleration = (10 – 0) ÷ 5 = 2 knots/min

• Distance = ½ × 5 × 10 = 25 NM (approx, if converted properly)

Summary of Key Points:

• Graphs are visual tools that simplify motion analysis.

• Slopes represent speed or acceleration; areas under curves represent distance.

• Recognizing graph patterns helps understand and predict ship behavior during operations.

Quick Review Questions:

1. What does a straight, sloped line on a distance-time graph represent?

2. How can you calculate acceleration from a speed-time graph?

3. What does the area under a speed-time graph indicate?

4. A ship maintains 12 knots for 30 minutes. How far has it traveled?

Key Concepts:

1. Newton’s First Law – The Law of Inertia

"An object remains at rest or in uniform motion unless acted upon by an external force."

• Inertia: The tendency of an object to resist changes in its state of motion.

Ship Application:

• A ship at rest stays still unless pushed by wind, tug, or engine thrust.

• A moving vessel continues unless resisted by drag, waves, or propeller reversal.

• Large ships have high inertia due to their mass — stopping them requires time and force.

🛳️ Example:
When the engine stops, the ship doesn’t stop immediately—it continues moving due to inertia.

2. Newton’s Second Law – Force and Acceleration

"The force acting on an object is equal to its mass multiplied by its acceleration."
Formula: F = m × a

• F = Force (in newtons, N)

• m = Mass (in kilograms, kg)

• a = Acceleration (in m/s²)

Ship Applications:

• Used to calculate the force needed to accelerate a vessel or machinery.

• Important in mooring, launching lifeboats, and moving heavy ship components.

🛳️ Example:
A 500 kg lifeboat accelerates at 2 m/s².
→ F = 500 × 2 = 1000 N force needed.

🛳️ Example:
Calculating the tension on a mooring rope during strong winds.

3. Newton’s Third Law – Action and Reaction

"For every action, there is an equal and opposite reaction."

• Forces always come in pairs.

• When one body exerts a force on another, the second body exerts an equal force in the opposite direction.

Ship Applications:

• Propellers push water backward → water pushes ship forward.

• Tugboat pushes ship → ship pushes back on the tug.

• Anchor chain resists ship drift → seabed exerts reactive force on anchor.

🛳️ Example:
Launching a lifeboat – the downward force results in an upward tension on the davit.

Quick Review Questions:

1. What does Newton’s first law say about a moving ship with no forces acting on it?

2. A 300 kg winch motor accelerates at 1.5 m/s². What force is applied?

3. Give an example of Newton’s third law onboard a ship.

4. Why does a larger ship require more time to stop?

🔢 Newton’s Laws of Motion – Problems & Answers

✅ Formula Recap:

• Newton’s First Law (Inertia):

An object remains at rest or moves with constant velocity unless acted upon by a net external force.

• Newton’s Second Law (F = m × a):

Force is the product of mass and acceleration.

• Newton’s Third Law (Action = Reaction):

For every action, there is an equal and opposite reaction.

🧮 Calculation Problems

Problem 1: Force Required to Accelerate a Lifeboat

A 2,000 kg lifeboat is accelerated at 2.5 m/s² by a towing force.
What is the force applied?

Formula: F = m × a

Solution:
F = 2,000 × 2.5 = 5,000 N

✅ Answer: 5,000 newtons

Problem 2: Acceleration of a Cargo Container

A 1,200 kg container is pushed with a force of 3,600 N.
What is the acceleration?

Formula: a = F ÷ m

Solution:
a = 3,600 ÷ 1,200 = 3 m/s²

✅ Answer: 3 m/s²

Problem 3: Mass from Force and Acceleration

A net force of 800 N causes an object to accelerate at 4 m/s².
What is the object’s mass?

Formula: m = F ÷ a

Solution:
m = 800 ÷ 4 = 200 kg

✅ Answer: 200 kilograms 

Problem 4: Newton’s First Law – Ship at Constant Speed

A ship is sailing at a constant speed of 15 knots with no net force acting on it.
What is its acceleration?

Explanation:
According to Newton’s First Law, if the speed is constant and no net force acts, acceleration is zero.

✅ Answer: 0 m/s² 

Problem 5: Newton’s Third Law – Mooring Force

A tugboat pulls a ship with a force of 10,000 N.
What force does the ship exert on the tugboat?

Action = Reaction → Equal and opposite

✅ Answer: 10,000 N (in opposite direction) 

Problem 6: Friction Force Slowing Down a Crate

A 500 kg crate is sliding across the deck and slows down due to a friction force of 1,000 N.
What is the deceleration?

a = F ÷ m = 1,000 ÷ 500 = 2 m/s²

✅ Answer: 2 m/s² (deceleration)

Problem 7: Combined Forces – Net Acceleration

Two tugs pull a barge (mass = 10,000 kg) with forces of 6,000 N and 4,000 N in the same direction.
What is the resulting acceleration?

Total Force = 6,000 + 4,000 = 10,000 N
a = F ÷ m = 10,000 ÷ 10,000 = 1 m/s²

✅ Answer: 1 m/s²

Key Concepts:

1. Momentum

Momentum (p) = Mass (m) × Velocity (v)
Unit: kg·m/s

• A measure of how much motion an object has.

• Heavier and faster ships have greater momentum and are harder to stop.

🛳️ Example:
A 20,000-ton ship moving at 3 m/s:

p = 20,000,000 kg × 3 m/s = 60,000,000 kg·m/s

2. Conservation of Momentum

In a closed system with no external forces, total momentum before = total momentum after.

• Applies to ship collisions and barge transfers.

• Helps analyze direction and speed changes after impact.

🛳️ Example:
If a ship collides with a floating dock, the combined system’s momentum remains the same before and after the collision (though velocities change).

3. Impulse

Impulse = Force (F) × Time (t)
Impulse = Change in momentum
Unit: N·s (same as kg·m/s)

• Impulse is the effect of applying a force over time to change an object’s momentum.

• Longer contact time = smaller force needed (important in fender design).

🛳️ Example:
During berthing, fenders increase contact time to reduce impact force on the hull.

Berthing Force Calculation (Simplified):

• If a ship of mass m reduces speed from u to 0 in time t:

  • Impulse = m × (0 – u) = –m × u

  • Force = Impulse ÷ t = –(m × u) ÷ t

The negative sign shows direction reversal (deceleration). 

Safety Note:

• The greater the ship's momentum, the more critical the need for early stopping, tugs, and fendering systems.

• Crew must understand how to calculate and manage momentum to prevent damage or accidents.

Summary of Key Points:

• Momentum = mass × velocity; more mass or speed = more momentum.

• Conservation of momentum applies to collisions; total motion is maintained.

• Impulse is the change in momentum over time; longer time reduces required force.

• Practical understanding of these concepts improves ship handling and safety during maneuvers.

Quick Review Questions:

1. What is the formula for momentum?

2. What does the conservation of momentum mean?

3. A 5000 kg vessel slows from 4 m/s to 0 m/s in 5 seconds. What is the average force applied?

4. Why are fenders important during berthing?

🔢 Momentum and Impulse – Problems & Answers

✅ Formula Recap:

1. Momentum (p):

p = m × v

1. • p = momentum (kg·m/s)

   • m = mass (kg)

   • v = velocity (m/s)

2. Impulse (J):

J = F × t = Δp = m × (v₂ - v₁)

2. • J = impulse (N·s)

   • F = force (N)

   • t = time (s)

Δp = change in momentum 

Problem 1: Momentum of a Moving Boat

A small boat with a mass of 3,000 kg is moving at a velocity of 6 m/s.
What is its momentum?

Solution:
p = m × v = 3,000 × 6 = 18,000 kg·m/s

✅ Answer: 18,000 kg·m/s

Problem 2: Impulse to Stop a Moving Object

A 500 kg lifeboat moving at 4 m/s is brought to rest in 10 seconds.
What is the impulse?

Δp = m × (v₂ - v₁) = 500 × (0 - 4) = –2,000 N·s

✅ Answer: –2,000 N·s (negative indicates direction of force is opposite to motion)

Problem 3: Force from Impulse

A force acts on a 1,200 kg container for 5 seconds, changing its velocity from 0 m/s to 3 m/s.
What is the applied force?

Step 1: Δp = m × (v₂ - v₁) = 1,200 × (3 - 0) = 3,600 N·s
Step 2: F = Δp ÷ t = 3,600 ÷ 5 = 720 N

✅ Answer: 720 newtons 

Problem 4: Change in Momentum

A 2,500 kg ship changes speed from 8 m/s to 6 m/s due to current resistance.
What is the change in momentum?

Solution:
Δp = 2,500 × (6 - 8) = 2,500 × (–2) = –5,000 kg·m/s

✅ Answer: –5,000 kg·m/s 

Problem 5: Impulse from Force and Time

A mooring line applies a 1,500 N force to a drifting lifeboat for 8 seconds.
Calculate the impulse imparted to the boat.

Solution:
J = F × t = 1,500 × 8 = 12,000 N·s

✅ Answer: 12,000 newton-seconds

Problem 6: Final Velocity from Impulse

A 300 kg life raft receives an impulse of 900 N·s.
What is its final velocity if it was initially at rest?

J = Δp = m × v → v = J ÷ m = 900 ÷ 300 = 3 m/s

✅ Answer: 3 m/s

Problem 7: Ship Collision Scenario

Ship A (mass = 5,000 kg) traveling at 3 m/s collides and stops.

• What is the momentum before the collision?

• What impulse must be absorbed to stop it?

Solution:
p = 5,000 × 3 = 15,000 kg·m/s
Impulse needed = –15,000 N·s (opposite direction)

✅ Answer: Momentum = 15,000 kg·m/s; Impulse = –15,000 N·s

This module covers the fundamental principles of forces, motion, equilibrium, and resistance as applied to real-life situations on board ships. Cadets will explore how different forces act on vessels and equipment, and how to manage them for safe and efficient operations.

Learning Objectives:

By the end of this module, cadets will be able to:

1. Define and differentiate between scalar and vector quantities in maritime operations.

2. Explain the concept of resultant force and how it affects motion and stability.

3. Identify conditions for equilibrium and assess whether forces acting on an object are balanced.

4. Use vector addition to calculate resultant forces acting on ship structures or during anchoring.

5. Apply force analysis to practical tasks like towing, lifting, mooring, and anchoring.

6. Describe the roles of friction and resistance in ship movement and equipment operation.

7. Identify common sources of resistance and techniques to reduce it (e.g., lubrication, hull cleaning).

8. Solve basic problems involving forces, tension, friction, and resistance in marine contexts.

Key Concepts:

1. Representing Forces with Arrows (Vectors)

• Vector: A quantity with both magnitude (size) and direction.

• Represented by arrows:

  • Length of arrow = magnitude of the force

  • Arrowhead = direction of the force

🛳️ Example:

• A 10 N wind force from the east is drawn as a 10-unit arrow pointing west to east.

2. Vector Addition (Graphical Method)

Two main graphical methods:

a) Triangle (Head-to-Tail) Method

• Place the tail of the second vector at the head of the first.

• The resultant vector is drawn from the start of the first to the end of the second.

b) Parallelogram Method

• Place both vectors with tails at the same point.

• Complete the parallelogram using the two vectors as adjacent sides.

• The diagonal represents the resultant vector.

3. Resultant Force

The resultant force is a single force that has the same effect as all the individual forces combined.

• Vector addition determines how forces like wind, current, and engine thrust combine.

• Useful for understanding actual ship direction and speed (resultant velocity).

 Example:

• A vessel moves at 5 knots due north (engine thrust), but a current flows east at 2 knots.

• Resultant direction is northeast — calculated via vector addition.

5. Estimating Resultant Speed and Direction (Optional Extension)

For more advanced cadets:

• Use the Pythagorean theorem for right-angle vectors:

Resultant speed = √(A² + B²)
Where A and B are magnitudes of perpendicular vectors.

Summary of Key Points:

• Vectors represent physical quantities with both direction and magnitude.

• Vector addition (graphically or mathematically) is used to find resultant forces.

• Navigation is influenced by multiple forces — understanding their combination is essential for control and safety.

• Graphical methods help visualize and predict ship movement.

Quick Review Questions:

1. What does the length and direction of a force arrow represent?

2. Name two graphical methods used to add vectors.

3. If a vessel is moving at 4 knots east and a current flows 3 knots north, what is the approximate direction of the vessel?

4. Why is it important for officers to understand vector addition during navigation?

Key Concepts:

1. Conditions for Equilibrium

Static Equilibrium

• The object is at rest.

• All forces acting on the object cancel out.

Condition:
Σ Forces = 0 and Σ Moments = 0

🛳️ Example:
A stationary gangway is in static equilibrium when the support forces at both ends balance the weight of the gangway and the person walking on it.

Dynamic Equilibrium

• The object is moving at constant velocity.

• No change in speed or direction (i.e., no net acceleration).

Condition:
Σ Forces = 0 but object is in motion

🛳️ Example:
A ship sailing steadily at 10 knots on calm seas is in dynamic equilibrium.

2. Balanced vs. Unbalanced Forces

Balanced Forces

• Forces are equal in size and opposite in direction

• No movement or constant velocity

Unbalanced Forces

• Net force is not zero

• Object accelerates or changes motion

 Examples:

• Balanced: A moored ship held in place by equal tension in bow and stern lines.

• Unbalanced: Wind pressure increases on the starboard side, causing the ship to drift toward port.

3. Real-Life Maritime Applications

a) Gangway Support

• Weight of the gangway and user is balanced by the support forces at the ship and shore ends.

• Improper balance = tipping or collapse risk.

b) Mooring Tension

• Balanced mooring = steady ship position.

• Unbalanced tension = ship surges or shifts.

c) Crane Stability

• Lifting a load shifts the center of gravity.

• Balance must be maintained to prevent tipping or boom failure.

4. Equilibrium Diagrams (Free-Body Diagrams)

• A free-body diagram shows all the forces acting on an object.

• Use arrows to represent forces (magnitude and direction).

• Add vectors to check for net force (ΣF = 0 → equilibrium).

Summary of Key Points:

• Equilibrium occurs when all forces and moments are balanced.

• Static = at rest; Dynamic = moving at constant velocity.

• Balanced forces → stability; Unbalanced forces → movement or danger.

• Mooring, gangways, and cranes all require force balance for safety.

• Free-body diagrams help visualize and analyze force balance.

Quick Review Questions:

1. What are the conditions for static equilibrium?

2. How can you tell if a system is in dynamic equilibrium?

3. Give one example of a balanced and one of an unbalanced force system onboard.

4. Why is force balance important in crane operations?

Key Concepts:

1. Friction Between Moving Surfaces

Friction is the resistance that one surface or object encounters when moving over another.

2. Water Resistance (Hydrodynamic Drag)

Drag is the resistance a ship faces while moving through water.

• Increases with speed, hull roughness, and wave interaction.

• Major factor in determining fuel efficiency and engine load.

🛳️ Impact:

• Slows the vessel down.

• Increases fuel consumption.

Affects voyage planning and cost efficiency. 

3. Methods to Reduce Friction and Resistance

a) Lubrication

• Applying oil or grease to reduce surface contact and wear.

• Used in:

  • Engine parts (pistons, bearings)

  • Winch systems

  • Hinges and moving joints

b) Streamlining

• Designing hulls with smooth, tapered shapes to reduce drag.

• Bow shapes, bulbous bows, and fairings aid in minimizing resistance.

c) Hull Maintenance

• Regular cleaning to remove marine growth (biofouling).

• Use of anti-fouling paint to prevent barnacle and algae buildup.

🛳️ Results:

• Improved speed

• Reduced engine strain

• Lower fuel costs and emissions

Summary of Key Points:

• Friction occurs wherever two surfaces move against each other onboard.

• Water resistance is a major external force affecting ship performance.

• Reducing friction and drag is critical for safety, efficiency, and cost-effectiveness.

• Proper maintenance and design reduce wear and optimize ship performance.

Quick Review Questions:

1. What is friction and where does it commonly occur onboard ships?

2. How does water resistance affect ship speed and fuel efficiency?

3. What are two common methods to reduce mechanical friction?

4. Why is hull cleaning important for performance?

This module introduces the essential physics concepts of work, energy, and power, and how they relate to ship operations. Cadets will learn how to calculate these quantities, understand energy conservation and efficiency, and apply their knowledge to improve performance and safety onboard.

Learning Objectives:

By the end of this module, cadets will be able to:

1. Define work, energy, and power in physical terms.

2. Calculate work done and power output in mechanical systems.

3. Identify different types of energy: kinetic, potential, and mechanical energy.

4. Explain the law of conservation of energy and how it applies to shipboard operations.

5. Perform energy efficiency calculations.

6. Apply energy principles to real-life maritime scenarios, such as engine performance and machinery operation.

Key Concepts:

1. Work

Work is done when a force moves an object in the direction of the force.

• Formula:
  W = F × d

  • W = Work (joules, J)

  • F = Force (newtons, N)

  • d = Distance (meters, m)

🛳️ Example:
If a winch pulls a mooring rope with a force of 500 N over 10 meters:
W = 500 × 10 = 5,000 J

• Note: If there is no movement or the force is perpendicular to motion, no work is done.

2. Power

Power is the rate at which work is done or energy is transferred.

• Formula:
  P = W ÷ t or P = F × d ÷ t

  • P = Power (watts, W)

  • t = Time (seconds, s)

• Common Units:

  • Watt (W), Kilowatt (kW): 1 kW = 1000 W

  • Horsepower (HP): 1 HP ≈ 746 W

🛳️ Example:
If a crane does 10,000 J of work in 20 seconds:
P = 10,000 ÷ 20 = 500 W

🛳️ Example:
A ship’s engine delivers 300 kW of power. This output is used to calculate fuel consumption and propeller efficiency.

Summary of Key Points:

• Work is force applied over distance in the direction of movement.

• Power is the rate at which work is performed.

• Shipboard equipment must be designed with appropriate work and power ratings.

• Calculations of work and power support planning, efficiency, and safety.

Quick Review Questions:

1. What is the formula for calculating work?

2. A 1,500 N force moves an object 5 meters. How much work is done?

3. What is power, and what unit is it measured in?

4. If 12,000 J of work is done in 30 seconds, what is the power?

🔢 Work and Power – Problems & Answers

✅ Formula Recap:

1. Work:

W = F × d
Where:

1. • W = Work (Joules, J)

   • F = Force (Newtons, N)

   • d = Distance (meters, m)

2. Power:

P = W ÷ t or P = F × d ÷ t
Where:

2. • P = Power (Watts, W)

   • t = Time (seconds, s)

Problem 1: Calculating Work

A crane lifts a 2,000 N container 12 meters high.
How much work is done?

Solution:
W = F × d = 2,000 × 12 = 24,000 J

✅ Answer: 24,000 joules 

Problem 2: Calculating Power

A mooring winch does 36,000 J of work in 30 seconds.
What is the power output?

Solution:
P = W ÷ t = 36,000 ÷ 30 = 1,200 W

✅ Answer: 1,200 watts (or 1.2 kW)

Problem 3: Work with Vertical Lift

A lifeboat (mass = 500 kg) is raised 8 meters.
How much work is done against gravity?

Use F = m × g, then W = F × d
g = 9.81 m/s²

Solution:
F = 500 × 9.81 = 4,905 N
W = 4,905 × 8 = 39,240 J

✅ Answer: 39,240 joules

Problem 4: Power of Engine Room Pump

An engine room pump moves a 1,000 N fluid load 10 meters in 20 seconds.
Calculate the power.

Solution:
P = (F × d) ÷ t = (1,000 × 10) ÷ 20 = 10,000 ÷ 20 = 500 W

✅ Answer: 500 watts

Problem 5: Time to Complete Work

A deck winch does 90,000 J of work at a rate of 1.5 kW.
How long does it take?

t = W ÷ P

Solution:
t = 90,000 ÷ 1,500 = 60 seconds

✅ Answer: 60 seconds

Problem 6: Power in Horsepower (HP)

A motor delivers 3,000 W of power. Convert this to horsepower (HP).

1 HP = 746 W

Solution:
HP = 3,000 ÷ 746 ≈ 4.02 HP

✅ Answer: Approximately 4.02 horsepower

Problem 7: Efficiency Check (Advanced)

A motor uses 5,000 J of energy and performs 4,000 J of useful work.
What is its efficiency?

η = (Output ÷ Input) × 100

Solution:
η = (4,000 ÷ 5,000) × 100 = 80%

✅ Answer: 80% efficient

Key Concepts:

1. Kinetic Energy (KE)

The energy a body possesses due to its motion.

• Formula:
  KE = ½ × m × v²

  • m = mass (kg)

  • v = velocity (m/s)

• Unit: Joules (J)

🛳️ Example:
A 10,000 kg ship moving at 2 m/s:
KE = ½ × 10,000 × (2)² = 20,000 J

2. Potential Energy (PE)

The stored energy due to an object’s position or height.

• Formula:
  PE = m × g × h

  • m = mass (kg)

  • g = gravity (9.81 m/s²)

  • h = height (m)

🛳️ Example:

A 500 kg container lifted 8 meters:

PE = 500 × 9.81 × 8 = 39,240 J 

3. Mechanical Energy

The sum of kinetic and potential energy in a system.

• Formula:
  ME = KE + PE

🛳️ Application:
A ship crane lifting a moving load involves both kinetic and potential energy.

Summary of Key Points:

• Kinetic energy relates to motion; potential energy relates to height or position.

• Mechanical energy is the total of kinetic and potential energy.

• Ships use various energy types, often transforming one form into another.

• Efficient energy use improves performance, fuel economy, and safety.

Quick Review Questions:

1. What is the formula for kinetic energy?

2. What happens to the potential energy of a lifeboat as it is lowered into the water?

3. Give an example of an energy transformation in an electric motor.

4. What two energy forms make up mechanical energy?

🔢 Types of Energy – Problems & Answers

✅ Formula Recap:

1. Kinetic Energy (KE):

KE = ½ × m × v²

1. • m = mass (kg)

   • v = velocity (m/s)

   • KE = energy in joules (J)

2. Potential Energy (PE):

PE = m × g × h

2. • m = mass (kg)

   • g = 9.81 m/s² (gravity)

   • h = height (m)

   • PE = energy in joules (J)

3. Mechanical Energy (ME):

ME = KE + PE

Problem 1: Kinetic Energy of a Moving Boat

A small boat (mass = 1,000 kg) is moving at 5 m/s.
Calculate its kinetic energy.

Solution:
KE = ½ × 1,000 × 5² = 0.5 × 1,000 × 25 = 12,500 J

✅ Answer: 12,500 joules 

Problem 2: Potential Energy of a Lifeboat

A 750 kg lifeboat is suspended 10 meters above the waterline.
Calculate its potential energy.

Solution:
PE = 750 × 9.81 × 10 = 73,575 J

✅ Answer: 73,575 joules

Problem 3: Mechanical Energy of a Crane Load

A 500 kg load is 6 meters above the deck and swinging at 2 m/s.
What is its total mechanical energy?

Step 1: PE = m × g × h = 500 × 9.81 × 6 = 29,430 J
Step 2: KE = ½ × 500 × 2² = 0.5 × 500 × 4 = 1,000 J
ME = PE + KE = 29,430 + 1,000 = 30,430 J

✅ Answer: 30,430 joules

Problem 4: Velocity from Kinetic Energy

A 200 kg object has 1,800 J of kinetic energy.
What is its speed?

Rearranged KE formula: v = √(2 × KE ÷ m)

Solution:
v = √(2 × 1,800 ÷ 200) = √(18) ≈ 4.24 m/s

✅ Answer: Approximately 4.24 m/s

Problem 5: Height from Potential Energy

A 600 kg object has 29,430 J of potential energy.
What is its height?

Rearranged formula: h = PE ÷ (m × g)

Solution:
h = 29,430 ÷ (600 × 9.81) = 29,430 ÷ 5,886 ≈ 5 m

✅ Answer: 5 meters

Problem 6: Mass from Kinetic Energy

An object moving at 3 m/s has 1,350 J of kinetic energy.
What is its mass?

Rearranged formula: m = 2 × KE ÷ v²

Solution:
m = 2 × 1,350 ÷ 9 = 2,700 ÷ 9 = 300 kg

✅ Answer: 300 kilograms

Problem 7: Crane Safety Scenario

A 2,000 kg cargo is lifted 12 meters and is moving at 1 m/s.
What is the total energy the crane is handling?

• PE = 2,000 × 9.81 × 12 = 235,440 J

• KE = 0.5 × 2,000 × 1² = 1,000 J

• ME = 235,440 + 1,000 = 236,440 J

✅ Answer: 236,440 joules of mechanical energy 

Key Concepts:

1. Law of Conservation of Energy

“Energy cannot be created or destroyed — it can only be converted from one form to another.”

• Total energy in a closed system remains constant.

• In practice, energy is often transformed from one form to another, but some is lost as heat, sound, or vibration.

🛳️ Example:
A ship’s diesel engine transforms chemical energy from fuel into thermal energy, then mechanical energy — with some lost as heat and sound.

2. Energy Efficiency

Efficiency is a measure of how much useful energy is obtained from the total input energy.

• Formula:
  Efficiency (η) = (Useful Output Energy ÷ Input Energy) × 100%

• Expressed as a percentage.

🛳️ Example:
If a generator receives 1000 kJ of fuel energy and produces 400 kJ of electrical energy:
η = (400 ÷ 1000) × 100 = 40%

3. Causes of Energy Loss

Heat ➡️ Friction in engines, bearings, or turbines

Sound and Vibration ➡️ Pump noise, engine block resonance

Electrical Resistance ➡️ Loss in cables and circuits

🛳️ Note: Total energy remains — but some becomes non-useful forms like heat and noise.

4. Improving Efficiency Onboard

Regular maintenance of engines ➡️ Reduces friction and energy loss

Using high-efficiency motors ➡️ Converts more input into mechanical work

Proper lubrication ➡️ Minimizes energy loss due to friction

Monitoring energy usage ➡️ Identifies and corrects inefficiencies

Summary of Key Points:

• Energy is conserved; it only changes form.

• Efficiency tells how much input energy is turned into useful output.

• Some energy is always lost as heat, friction, or noise.

• Efficient equipment saves fuel, reduces costs, and improves performance.

Quick Review Questions:

1. What does the law of conservation of energy state?

2. How is efficiency calculated?

3. If a generator is 60% efficient, how much energy is lost if it receives 1000 kJ?

4. Name two ways to improve energy efficiency on a ship.

Key Concepts:

1. Main Engine and Propulsion System

• Converts chemical energy (fuel) → thermal energy → mechanical energy

• Drives the propeller for vessel movement

• Work and power output measured in kW or HP

🛳️ Example:
Main engine rated at 10,000 kW — power used to overcome resistance and move the vessel at cruising speed.

2. Cargo Handling Systems (Cranes, Winches)

• Use electrical energy → mechanical energy to lift and move cargo

• Work done = Force × Distance lifted

• Power needed depends on weight and speed of lifting

🛳️ Example:
A deck crane lifts a 2,000 kg container 15 meters:
Work = m × g × h = 2000 × 9.81 × 15 = 294,300 J

3. Pumping Systems

• Fuel, ballast, and bilge systems use pumps

• Energy input = electrical or mechanical power to move fluids

• Efficiency matters for minimizing energy loss and avoiding delays

🛳️ Example: A ballast pump rated at 5 kW transfers 10 m³ of water every 10 minutes.

4. Refrigeration and HVAC Systems

• Use electrical energy to drive compressors

• Transfers heat from cold storage spaces (reverse work process)

• Affects food preservation and crew comfort

🛳️ Efficiency Tip:
Well-maintained insulation reduces workload on compressors, saving energy.

5. Lighting and Electrical Systems

• Energy usage measured in kWh

• LEDs offer higher efficiency than traditional bulbs

• Monitoring consumption helps with fuel budgeting and generator load management

🛳️ Example:
Lighting load: 2 kW operating for 12 hours/day → 2 × 12 = 24 kWh/day

Summary of Key Points:

• Ship operations rely on converting energy into useful work (lifting, moving, heating, etc.)

• Calculating work and power helps size equipment and optimize performance

• Energy efficiency directly impacts fuel use, emissions, and system reliability

• Regular maintenance and smart operation improve energy performance across all systems

Quick Review Questions:

1. How is energy used in a ship’s propulsion system?

2. What energy conversion occurs in a cargo crane during lifting?

3. Why is pump efficiency important during ballast operations?

4. How can lighting systems onboard be made more energy-efficient?

This module introduces cadets to the principles of fluid mechanics and buoyancy, which are essential for understanding ship stability, ballast operations, pump systems, and underwater pressure effects. Cadets will learn to apply these concepts in both deck and engine department tasks.

Learning Objectives:

By the end of this module, cadets will be able to:

1. Describe key fluid properties: density, pressure, and viscosity.

2. Explain Pascal’s law and identify its applications on ships.

3. Apply Archimedes' principle to understand buoyancy and ship stability basics.

4. Calculate pressure in fluids at different depths.

5. Compute flow rates and relate them to pump operations and ballast control.

6. Apply these principles to practical tasks such as cargo tank filling, ballast exchange, and floatation assessment.

Key Concepts:

1. Density (ρ)

Density is the mass of a fluid per unit volume.
Formula: ρ = m ÷ V

• Unit: kg/m³

• Affects buoyancy and fuel/payload calculations.

🛳️ Ship Application:

• Fuel oil and ballast water have different densities, affecting stability and trim.

2. Pressure (P)

Pressure is the force exerted by a fluid per unit area.
Formula: P = F ÷ A or P = ρgh (for static fluids)

• Unit: Pascal (Pa), bar

• Increases with fluid depth and density.

🛳️ Example:

• Pressure at the bottom of a 5 m tank filled with seawater (ρ ≈ 1025 kg/m³):
  P = 1025 × 9.81 × 5 ≈ 50,306 Pa

🛳️ Ship Application:

• Pressure gauges monitor tank levels and pump discharge.

3. Viscosity (μ)

Viscosity is a fluid’s resistance to flow.

• High viscosity = flows slowly (e.g., heavy fuel oil)

• Low viscosity = flows easily (e.g., water)

🛳️ Ship Application:

• Affects pump selection, filter performance, and heating requirements for fuel oil.

4. Real-Life Maritime Applications

Density ➡️ Ballast/fuel tank calculations and draft control

Pressure ➡️ Pumping systems, hydraulic controls, tank levels

Viscosity ➡️ Fuel system design, lube oil flow, cargo handling

Summary of Key Points:

• Fluids onboard include liquids and gases (e.g., water, oil, air).

• Density influences buoyancy, load distribution, and fluid mass.

• Pressure increases with depth and is essential for tank and pipeline management.

• Viscosity affects fluid flow rate and energy required for pumping.

Quick Review Questions:

1. What is the formula for fluid density?

2. How does pressure change with fluid depth?

3. Why is viscosity important when handling heavy fuel oil?

4. Give one onboard system where pressure measurement is essential. 

Key Concepts:

1. Pascal’s Law

“When pressure is applied to a confined fluid, the pressure change is transmitted equally in all directions throughout the fluid.”

• Formula:
  P = F ÷ A
  Where:

  • P = Pressure (Pa)

  • F = Force (N)

  • A = Area (m²)

🛳️ Example:
If a force of 500 N is applied to a piston with an area of 0.01 m²:
P = 500 ÷ 0.01 = 50,000 Pa

2. Hydraulic Force Multiplication

• In hydraulic systems, a small force applied on a small-area piston can generate a larger force on a larger-area piston.

Output Force (F₂) = P × A₂
Input Force (F₁) = P × A₁
→ Since pressure is the same:
F₂ ÷ A₂ = F₁ ÷ A₁

🛳️ Example:
A 100 N force on a small piston (area = 0.01 m²) creates pressure = 10,000 Pa.
This pressure on a larger piston (0.1 m²) gives output force =
P × A = 10,000 × 0.1 = 1,000 N

4. Advantages of Hydraulic Systems

• Transmit large forces through small input

• Smooth and precise control

• Compact and reliable

• Ideal for confined ship spaces

Summary of Key Points:

• Pascal’s Law states that pressure in a confined fluid is transmitted equally in all directions.

• Hydraulic systems use this principle to multiply force and control heavy machinery.

• Understanding how force and pressure interact helps in operating and maintaining marine hydraulic equipment safely and effectively. 

Quick Review Questions:

1. What does Pascal’s Law state?

2. A piston with area 0.05 m² receives 500 N of force. What is the pressure?

3. Name two shipboard systems that use hydraulic pressure.

4. Why is hydraulic power preferred in marine machinery?

🔢 Pascal’s Law and Applications – Problems & Answers

Problem 1: Pressure Calculation

A force of 500 N is applied to a piston with an area of 0.01 m².
What is the pressure applied to the fluid?

Solution:
P = F ÷ A = 500 ÷ 0.01 = 50,000 Pa

✅ Answer: 50,000 Pascals (Pa)

Problem 2: Force Output on Larger Piston

In a hydraulic system, the small piston (area = 0.005 m²) is pushed with 400 N of force.
If the large piston has an area of 0.05 m², what is the output force?

Pressure is the same in both pistons:
F₁ ÷ A₁ = F₂ ÷ A₂
→ F₂ = (F₁ × A₂) ÷ A₁

Solution:
F₂ = (400 × 0.05) ÷ 0.005 = 20 ÷ 0.005 = 4,000 N

✅ Answer: 4,000 N

Problem 3: Area Calculation

A hydraulic jack produces a pressure of 25,000 Pa and supports a force of 1,000 N.
What is the area of the piston?

A = F ÷ P

Solution:
A = 1,000 ÷ 25,000 = 0.04 m²

✅ Answer: 0.04 square meters 

Problem 4: Pressure Transmission – Deck Crane

A hydraulic crane uses a cylinder with an area of 0.03 m².
If the pressure in the fluid is 60,000 Pa, what is the lifting force generated?

F = P × A

Solution:
F = 60,000 × 0.03 = 1,800 N

✅ Answer: 1,800 newtons 

Problem 5: Comparing Pistons

Two pistons in a hydraulic system have areas of 0.01 m² and 0.08 m².
If a 200 N force is applied to the small piston, how much force is transmitted to the large piston?

Solution:
F₂ = (200 × 0.08) ÷ 0.01 = 16 ÷ 0.01 = 1,600 N

✅ Answer: 1,600 N

Problem 6: Shipboard Scenario

During mooring, a hydraulic winch applies 300 N of force through a piston area of 0.006 m².
What is the fluid pressure in the system?

Solution:
P = F ÷ A = 300 ÷ 0.006 = 50,000 Pa

✅ Answer: 50,000 Pascals

Problem 7: Reverse Calculation – Small Piston Force Needed

To lift a 5,000 N load with a large piston area of 0.1 m², how much force is needed on the small piston of area 0.01 m²?

F₁ = (F₂ × A₁) ÷ A₂
F₁ = (5,000 × 0.01) ÷ 0.1 = 50 ÷ 0.1 = 500 N

✅ Answer: 500 N

Key Concepts:

1. Archimedes’ Principle

“A body wholly or partially submerged in a fluid experiences an upward (buoyant) force equal to the weight of the fluid it displaces.”

• Explains why objects float or sink.

• Buoyant force is the upward force exerted by a fluid that opposes the weight of an object.

2. Buoyant Force (FB)

Formula:
FB = ρ × V × g
Where:

• ρ = Density of fluid (kg/m³)

• V = Volume of fluid displaced (m³)

• g = Acceleration due to gravity (9.81 m/s²)

🛳️ Example:
A steel block displaces 0.5 m³ of seawater (ρ = 1025 kg/m³):
FB = 1025 × 0.5 × 9.81 ≈ 5,028 N

4. Buoyancy and Ship Stability

• As cargo is loaded or ballast is adjusted, the displacement and draft of the ship change.

• Center of buoyancy moves as the underwater shape of the hull changes.

• Stability is maintained by ensuring the center of gravity remains below the center of buoyancy.

🛳️ Application:

• Ballast tanks are used to control the ship’s trim and stability.

• Overloading can reduce freeboard and compromise safety.

Summary of Key Points:

• Archimedes’ Principle explains floating by comparing object weight to displaced fluid weight.

• Buoyant force equals the weight of displaced fluid.

• Ship stability, draft, and safety depend on managing buoyancy through load and ballast control.

• Understanding buoyancy is essential for safe cargo operations and emergency equipment use.

Quick Review Questions:

1. What is Archimedes’ Principle?

2. What is the formula to calculate buoyant force?

3. What happens when a ship’s weight exceeds the buoyant force?

4. How does ballast water affect a ship’s buoyancy?

🔢 Archimedes’ Principle and Buoyancy – Problems & Answers

Problem 1: Buoyant Force of a Floating Object

A wooden block displaces 0.6 m³ of seawater (ρ = 1025 kg/m³).
Calculate the buoyant force acting on the block.

Solution:
FB = 1025 × 0.6 × 9.81 = 6,030.15 N

✅ Answer: 6,030.15 newtons (N)

Problem 2: Volume of Displaced Water

A floating object experiences a buoyant force of 4,905 N in freshwater (ρ = 1000 kg/m³).
What is the volume of water displaced?

Rearranged: V = FB ÷ (ρ × g)

Solution:
V = 4,905 ÷ (1000 × 9.81) = 4,905 ÷ 9,810 = 0.5 m³

✅ Answer: 0.5 cubic meters

Problem 3: Will the Object Float?

A steel object weighs 2,000 N and displaces 0.15 m³ of seawater.
Will it float or sink?

Step 1: Calculate FB = 1025 × 0.15 × 9.81 ≈ 1,507.24 N
Step 2: Compare FB to object weight (2,000 N)

✅ Answer: It will sink (because 1,507.24 N < 2,000 N)

Problem 4: Required Displacement to Float

What volume of seawater must be displaced for a 3,500 N lifeboat to float?

Solution:
V = FB ÷ (ρ × g) = 3,500 ÷ (1025 × 9.81) ≈ 3,500 ÷ 10,055.25 ≈ 0.348 m³

✅ Answer: Approximately 0.348 cubic meters

Problem 5: Apparent Weight in Water

A metal block weighs 500 N in air and experiences a buoyant force of 120 N in water.
What is its apparent weight in water?

Apparent weight = Actual weight – Buoyant force
500 – 120 = 380 N

✅ Answer: 380 newtons

Problem 6: Stability and Draft

A barge displaces 200 m³ of seawater. What is the total buoyant force acting on it?

Solution:
FB = 1025 × 200 × 9.81 = 2,009,250 N

✅ Answer: 2,009,250 newtons