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School Physics Notes: Forces 5.1 Turning forces and skeletal muscles

GCSE level physics exam revision notes on Forces 5

Turning forces: 5.7 Turning force calculations applied to the physics of the human skeleton and muscle system

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[KEY POINTS and learning objectives for this page, after initial notes]

Index of physics notes: 5. Turning forces, calculating moments, problem solving

This page contains online questions only.

Jot down your answers and check them against the worked out answers at the end of the page.

5.7 Turning forces and the physics of the human skeleton and muscle system

skeleton1.jpg

1. The knee joint

Cartilage of joints in your skeletal-muscle system, has many functions, including the ability to resist compressive forces, enhance bone resilience, and provide support on bony areas where there is a need for flexibility e.g.

The cartilage is a softer material between the bones of a joint and acts as a shock absorber to minimise the chance of bone fracture damage.

Cartilage being a softer material between the bones of a joint, reduces friction that would cause pain and wear of the bones.

Cartilage connective tissue also acts as a binding agent to hold the bones in place, but not too rigidly, the bone-cartilage system must be flexible.

 

2. The arm joint

The comments on cartilage in 1. apply here too.

The arm joint is a lever system operated by the expansion and contraction of muscles e.g.

to bend and raise your forearm the biceps contract and the triceps relax, and,

to lower or straighten your forearm the biceps relax and the triceps contract

 

3. The pelvic hip joint

The comments on cartilage in 1. apply here too.

Weight lifted and force calculations based on the action of muscles

Repeat of calculations in biology notes from Part 3 of

INDEX of biology notes on the skeleton and muscles

force and moment calculations of arm elbow joint force generated by biceps and triceps calculating force needed to lift raise a weight

Some example questions (ANSWERS)

ALL based on the diagram above for the antagonistic biceps and triceps muscles of the elbow joint

Assume the gravitational field constant is 10 N/kg

Question 1

 If d1 = 0.40 m, d2 = 5 cm, using the above diagram for help, and applying the principle of moments, calculate the minimum force needed by the biceps contraction to lift a weight of 20 N.

 

Question 2

 If d1 = 0.36 m, d2 = 0.06 m, using the above diagram for help, and applying the principle of moments, calculate the minimum force needed by the biceps contraction to lift a mass of 3 kg.

 

Question 3

 If d1 = 45 cm, d2 = 5 cm, using the above diagram for help, and applying the principle of moments, calculate the minimum force needed by the biceps contraction to lift a weight of 15 N.

 

Question 4

 If d1 = 48 cm, d2 = 6 cm, using the above diagram for help, and applying the principle of moments, calculate the minimum force needed by the biceps contraction to lift a weight of 2 kg.

 

Question 5

 If d1 = 0.50 m, d2 = 0.05 m, using the above diagram for help, and applying the principle of moments, if the maximum strength of a person's biceps muscle contraction creates a force of 150 N, calculate the maximum weight the person's arm can raise and hold.

 

Question 6

 If d1 = 0.60 m, d2 = 0.06 m, using the above diagram for help, and applying the principle of moments, if the maximum strength of a person's biceps muscle contraction creates a force of 180 N, calculate the maximum mass the person's arm can raise and hold.

 

Question 7

 If d1 = 0.50 m, d2 = 0.05 m, using the above diagram for help, and applying the principle of moments, if the maximum strength of a person's biceps muscle contraction creates a force of 200 N, calculate the maximum weight the person's arm can raise and hold.

 

Question 8

 If d1 = 0.40 m, d2 = 0.06 m, using the above diagram for help, and applying the principle of moments, if the maximum strength of a person's biceps muscle contraction creates a force of 240 N, calculate the maximum mass the person's arm can raise and hold.

 

WORKED out answers

Key points about turning forces - turning forces and muscle action

Information sources for Doc Brown's key points: IGCSE-GCSE physics are based on textbooks and syllabus-specifications for students taking the UK AQA, Edexcel, OCR 21st Century Science, OCR Gateway science suite, WJEC, CCEA and CIE GCSE physics 9-1 level science examinations

A structured set of summary revision notes on moment forces and human muscle action, including biomedical applications, tailored to the GCSE/IGCSE physics specifications across major UK exam boards: WJEC, CCEA, CIE, AQA, Edexcel, and OCR.

Moment Forces and Human Muscle Action – Summary Revision Notes

Core Concepts

  • Moment of a Force
    Moment = Force x Perpendicular Distance from Pivot
    • Measured in Newton-metres (Nm)
    • Force in Newtons (N), distance in metres (m)
  • Pivot (Fulcrum)
    • Fixed point about which rotation occurs
    • In the human body, joints act as pivots
  • Muscle Action as Force
    • Muscles apply force via tendons to bones
    • The distance from the joint (pivot) determines the moment produced
  • Antagonistic Muscle Pairs
    • Muscles work in pairs: one contracts while the other relaxes
    • Example: biceps and triceps control elbow movement

Human Anatomy and Biomechanics

Component Role in Moment Generation
Bones Act as rigid levers
Joints Serve as pivots
Muscles Apply force to create rotation
Tendons Transmit force from muscle to bone
  • Example:
    • Biceps apply force to lift the forearm
    • Elbow joint = pivot
    • Forearm = lever
    • Moment = force × distance from elbow

Biomedical Applications

  • Prosthetics Design
    • Engineers calculate required moments to mimic natural limb movement
    • Adjust lever lengths and force application points
  • Joint Replacement Surgery
    • Artificial joints must replicate natural pivot points and moment ranges
  • Rehabilitation Devices
    • Use controlled moments to assist recovery (e.g. resistance bands, robotic arms)
  • Ergonomics and Injury Prevention
    • Understanding moments helps design safer tools and workspaces

Typical Exam Board Specification Content

Key Focus Areas
Emphasises practical applications of moments in biomechanics and lever systems
Covers muscle action, centre of gravity, and stability in human movement
Includes human levers, torque, and biomedical relevance
Focuses on static equilibrium, moment calculations, and biological examples
Applies moment principles to human anatomy and prosthetic design
Explores forces in action, couples, and muscle-generated moments

Student Exam Tips

  • Use labelled diagrams of limbs to identify pivot, force, and distance
  • Practice calculating moments using real anatomical examples
  • Understand lever classes (first, second, third) and relate them to body parts
  • Revise antagonistic muscle pairs and their roles in movement
  • Explore case studies in prosthetics or sports science for applied understanding
  • Convert units carefully (e.g. cm to m) in moment calculations

Keywords, phrases and learning objectives for turning forces

Be able to describe and explain the application of turning forces and the physics of the human skeleton and muscle system e.g. knee joints, arm joints and pelvic hip joints, and where necessary be able to do calculations on the forces involved.

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INDEX for physics notes on forces 5 TURNING FORCES

INDEX of all my physics notes on FORCES

INDEX of all my physics notes on FORCES and MOTION

INDEX of all my PHYSICS NOTES

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Index Forces 5. Turning forces, calculating moments, problem solving
Worked out ANSWERS to the moment calculations

Remember 1 kg mass is equivalent to 10 N on the Earth's surface.

Q1 F1 x d1 = F2 x d2, convert cm to m, 20 x 0.40 = F2 x 0.05, F2 = 20 x 0.40 / 0.05 = 160 N

Q2 3 kg = 3 x 10 = 30 N, F1 x d1 = F2 x d2, convert cm to m, 30 x 0.36 = F2 x 0.06 x F2, F2 = 30 x 0.36 / 0.06 = 180 N

Q3 F1 x d1 = F2 x d2, convert cm to m, 15 x 0.45 = F2 x 0.05, F2 = 15 x 0.45 / 0.05 = 135 N

Q4 2 kg = 2 x 10 = 20 N, F1 x d1 = F2 x d2, convert cm to m, 20 x 0.48 = F2 x 0.06, F2 = 20 x 0.48 / 0.06 = 160 N

Q5 F1 x d1 = F2 x d2, F1 x 0.50 = 150 x 0.05, F1 = 150 x 0.05 / 0.50 = 15 N

Q6 F1 x d1 = F2 x d2, F1 x 0.60 = 180 x 0.06, F1 = 180 x 0.06 / 0.60 = 18 N, mass = 18 / 10 = 1.8 kg

Q7 F1 x d1 = F2 x d2, F1 x 0.50 = 200 x 0.05, F1 = 200 x 0.05 / 0.50 = 20 N

Q8 F1 x d1 = F2 x d2, F1 x 0.40 = 240 x 0.06, F1 = 240 x 0.06 / 0.40 = 36 N, mass = 36 /10 = 3.6 kg

 

Index Forces 5. Turning forces, calculating moments, problem solving

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