WJEC A-Level Physics 1420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Solids Under Stress
Using Radiation to Investigate Stars
Particles and Nuclear Structure
Conduction of Electricity
Resistance
D.C. Circuits
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Nuclear Decay
Nuclear Energy
Capacitance
Electrostatic and Gravitational Fields of Force
Orbits and the Wider Universe
Magnetic Fields
Electromagnetic Induction
Option A: Alternating CurrentsOption A
Option B: Medical PhysicsOption B
Option C: The Physics of SportsOption C
Option D: Energy and the EnvironmentOption D
Mathematical Requirements (Appendix C)
Option C: The Physics of Sports
Moments in the muscles and joints
- 3 things to know
- A2 Unit 4 Section B
Overview
Recall questions on Moments in the muscles and joints.
What you need to know
Answer each one before you reveal it.
Why can a limb be treated as a lever when muscle forces are analysed?
The elbow is an example of a third-class lever. What is gained in exchange for the large muscle force it demands?
Questions from this subtopic
Each one with the model answer underneath.
- 1
Why can a limb be treated as a lever when muscle forces are analysed?
Model answer
A bone is a rigid rod that turns about the joint at one end, so the joint acts as the pivot. The muscle force and the load then produce opposing moments about that pivot, exactly as they would on a lever.
- 2
About which point are moments taken when the force in a muscle across a joint is calculated?
Model answer
Moments are taken about the joint itself. The force the joint exerts on the bone acts through that point, so it has no moment about it and drops out of the equation.
- 3
The biceps attaches to the forearm 4.0 cm from the elbow joint. A weight of 50 N is held in the hand 32 cm from the joint with the forearm horizontal. Calculate the muscle force, ignoring the weight of the arm.
Model answer
Taking moments about the elbow, F × 0.040 = 50 × 0.32 = 16 N m, so F = 16 ÷ 0.040 = 400 N. Substituting back, 400 × 0.040 = 16 N m, matching the moment of the load.
- 4
Repeat the elbow calculation including the forearm's own weight of 16 N acting 15 cm from the joint, with the biceps 4.0 cm from the joint and a 50 N load 32 cm from it.
Model answer
The total clockwise moment is 50 × 0.32 + 16 × 0.15 = 16 + 2.4 = 18.4 N m, so F = 18.4 ÷ 0.040 = 460 N. Substituting back, 460 × 0.040 = 18.4 N m.
- 5
A biceps force of 460 N supports a forearm of weight 16 N holding a 50 N load. Calculate the force the humerus exerts on the forearm at the elbow.
Model answer
For vertical equilibrium the upward and downward forces must balance, so the joint force is 460 − 16 − 50 = 394 N. Since the muscle pull exceeds the two weights, this 394 N acts downwards on the forearm. Substituting back, 460 = 16 + 50 + 394.
- 6
Explain why the biceps must pull with a force many times larger than the weight being held.
Model answer
The muscle attaches very close to the joint while the load is held out near the end of the forearm, so the muscle's perpendicular distance from the pivot is a small fraction of the load's. To match the load's moment with such a short lever arm it must supply a proportionately larger force.
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