MNB.7 Skeletal Lever System PDF

Musculoskeletal System, Nervous System & Bioelectricity MNB.7 Skeletal Lever System D r A ndy Ma Learning outcomes At the end of this lecture, the learner will be able to Define Centre of Gravity (CofG) and understand its significance in biomechanical calculations of force and in determining...

Musculoskeletal System, Nervous System & Bioelectricity MNB.7 Skeletal Lever System D r A ndy Ma Learning outcomes At the end of this lecture, the learner will be able to Define Centre of Gravity (CofG) and understand its significance in biomechanical calculations of force and in determining an object’s stability. Draw a diagram showing the rate of force development for human muscle and outline what factors determine the maximum tension achievable by human muscle. Define moment, torque, and lever. Outline the conditions required to achieve static equilibrium and recognise these conditions in anatomical examples. Demonstrate the ability to calculate biomechanical forces using the principle of moments. Define and calculate mechanical advantage. Differentiate between Class I, II, and III levers using anatomical examples. Discuss the risk of back injury associated with poor lifting technique. MN B.7 Ske le ta l L eve r Syst em 2 Centre of Gravity The centre of gravity of a body is the point around which its mass or weight is evenly distributed and through which the force of gravity acts. The centre of gravity is the single point that moves in accordance with Newton’s simple laws of motion. Weight Weight For uniformly dense and symmetric objects, the centre of gravity is located at their geometric centres. MN B.7 Ske le ta l L eve r Syst em 3 Centre of Gravity When investigating levers, we will need to establish the location from which the force acts. In other words, where does the centre of gravity lie for a particular body part or object. We usually assume symmetry and homogeneity. Therefore, the centre of gravity will typically lie at the midpoint, e.g. halfway along the length of the arm. Weight of Lower arm Weight of Shot put MN B.7 Ske le ta l L eve r Syst em 4 Centre of Gravity We can divide the human body into various segments each with its own mass and centre of gravity. Consider a body of mass M (70kg) Segment Mass Torso and Head 0.593 x M Upper arms 0.053 x M Forearms and hands 0.043 x M Upper legs (thighs) 0.193 x M Lower legs and feet 0.118 x M Mass of torso and head ~ 0.593 x 70 kg, or 41.5 kg. MN B.7 Ske le ta l L eve r Syst em 5 Centre of Gravity and Stability An object is stable when its centre of gravity lies above the base of support (part of the body that is in contact with the supporting surface). Stable: Centre of gravity lies within the Centre of supports Gravity Support Unstable: Centre of gravity lies outside the supports MN B.7 Ske le ta l L eve r Syst em 6 Centre of Gravity and Stability Think about a rugby Increasing the area of player who stands with base support their feet well apart if increases stability. they are expecting to be tackled. The same rugby Lowering the center player will also of gravity increases maintain a low body stability. position. MN B.7 Ske le ta l L eve r Syst em 7 Force Development MN B.7 Ske le ta l L eve r Syst em 8 Rate of Force Development The rate of force development is a measure of explosive strength, or simply how fast you can develop force in the initial contraction phase. It determines the maximum force that can be generated in the early phase of muscle contraction (0–200 ms) and it has important functional consequences (think sprinter vs long distance runner!). MN B.7 Ske le ta l L eve r Syst em 9 Force Development Muscles have different architectures depending on their purpose. Some muscles may be designed for fine control e.g. muscles in eyelids, while others may be designed for force and power output e.g. quadriceps. Each sarcomere is a single contractile unit which can generate a MN B.7 Ske le ta l L eve r Syst em 10 force, F. Force Development In parallel muscles, the sarcomeres are in series which result in fibres orientated parallel to the tendon. In pennate muscles the sarcomeres are parallel which results in increased force production but this arrangement has a reduce range of motion. Parallel muscle Pennate muscles 𝐼𝑛 𝑆𝑒𝑟𝑖𝑒𝑠 𝐼𝑛 𝑃𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝐹𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 𝐹1 = 𝐹2 𝐹𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 𝐹1 + 𝐹2 + … ML: muscle length FL: fibre length MN B.7 Ske le ta l L eve r Syst em 11  Force Development Pennate muscles can generate more force than parallel fibered muscles of the same size because the fiber length is shorter. Parallel muscle Pennate muscles 𝒎 𝒙 𝑪𝒐𝒔α 𝑭𝒐𝒓𝒄𝒆 ∝ 𝒍𝒙ρ l: fibre length m: muscle mass α: fibre angle ρ: muscle density ML: M N Bmuscle length.7 Ske le tal Lever Syst em FL: fibre length 12 Force Development Anatomical CSA is The maximum tension (specific tension) a muscle can produce is the area of the cross- proportional to the physiological cross-sectional area of the muscle at its section of a muscle thickest part. perpendicular to its longitudinal axis. In pennate muscles the PCSA is larger than in parallel muscles, which results in increased force production. Physiological CSA is the area of the cross section of a muscle perpendicular to its MNB.7 Ske le ta l L eve r Syst em fibers. 13 Force Development The tension (force) also depends on the length of the muscle when it is stimulated. Maximum tension of muscles is achieved when the muscle is stimulated in its relaxed state, i.e. at its normal resting length. MN B.7 Ske le ta l L eve r Syst em 14 Levers MN B.7 Ske le ta l L eve r Syst em 15 Levers Simple movement and motion is possible by the articulation of joints using muscles and tendons. The skeletal and muscle systems consist of a series of levers which operate through the application of moments and torque. A lever consists of a beam or rigid rod (e.g. a bone) pivoted at a fixed hinge or fulcrum (e.g. the elbow) that is capable of rotating due to the action of forces. The turning effect of the force on the body on which it is acting is measured by the moment of a force. The moment of a force depends on the magnitude of the force and the distance from the axis of rotation. The rotational force that causes an object to rotate about its axis is known as torque. MN B.7 Ske le ta l L eve r Syst em 16 Levers Levers are examples of simple machines where a force is applied to lift or balance a load. The load (resistance) is the object you are trying to move. The effort (applied force) is the force applied to move the load. The fulcrum (or pivot) is the point where the load is pivoted. A lever can be used to magnify the size of a force or change its direction. MN B.7 Ske le ta l L eve r Syst em 17 Moment of a Force The magnitude of the torque generated is the product of the force, and the perpendicular distance of the line of action of the force from the centre of the rotation at the pivot. d1 d2 F1 F2 The perpendicular distance (d) is the F1 acts a distance of d1 from the pivot and shortest distance between the pivot and the causes an anticlockwise moment. line of action of the force (F). F2 acts a distance of d2 from the pivot and causes a clockwise moment. Assign a direction to the moment, e.g. MN B.7 Ske le ta l L eve r Syst em clockwise is + and anticlockwise is -. 18 Calculate Torque In the human body, muscle attachments mean that the force is very rarely acting perpendicular to the distance (i.e. at 90°). Therefore, the calculation of the moment, or torque τ, must take this into consideration. τ = 𝐹𝑠𝑖𝑛𝜃 𝑑 Where θ is measured between How much of F acts the force and the distance. perpendicular to x? A force of 445 N is generated in the biceps and acts at an angle of 82°. 𝐹⟂ = 𝐹 𝑠𝑖𝑛𝜃 𝐹⟂ = 445 sin(82) 𝐹⟂ = 440.7 𝑁 𝐹𝑦 = 𝐹 𝑠𝑖𝑛𝜃 τ = 440.7 0.003 MN B.7 Ske le ta l L eve r Syst em = 1.32 Nm 19 Calculate Torque When you hold your arm outstretched as shown below, it's supported primarily by the deltoid muscle. The deltoid exerts a 67 N force at 15° as shown below. If the force-application point is 18 cm from the shoulder joint, calculate the torque due to the deltoid. τ = 𝐹𝑠𝑖𝑛𝜃 𝑑 τ = 67 sin 15 0.18 τ = 3.12 𝑁𝑚 MN B.7 Ske le ta l L eve r Syst em 20 Equilibrium The load and weight of the lower leg produce a clockwise torque about the knee. The lower leg will rotate in a clockwise direction. If the hamstring muscle at the back of the upper leg contracts with a strong force, it produces an anticlockwise torque that holds the leg stationary. When held stationary, the leg is in equilibrium. 1. Net Torque = Zero 2. Net Force = Zero MN B.7 Ske le ta l L eve r Syst em 21 Equilibrium The image shows a human arm holding a weight in equilibrium. (a) Using the information shown, calculate the force generated in the biceps muscle. 𝑁𝑒𝑡 𝑇𝑜𝑟𝑞𝑢𝑒 = 0 𝐶𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 𝑇𝑜𝑟𝑞𝑢𝑒 + 𝐴𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 𝑇𝑜𝑟𝑞𝑢𝑒 = 0 (30 0.46 + 18 0.15 ) + −𝐹𝐵 0.04 = 0 30 0.46 + 18 0.15 − 𝐹𝐵 0.04 = 0 30 0.46 + 18 0.15 = 𝐹𝐵 0.04 𝐹𝐵 = 412.5 𝑁 (b) Calculate the force at the elbow joint. 𝑁𝑒𝑡 𝐹𝑜𝑟𝑐𝑒 = 0 𝐷𝑜𝑤𝑛𝑤𝑎𝑟𝑑 𝐹𝑜𝑟𝑐𝑒 + 𝑈𝑝𝑤𝑎𝑟𝑑 𝐹𝑜𝑟𝑐𝑒 = 0 30 + 18 + 𝐹𝐵 + 𝐹𝐸 = 0 30 + 18 − 412.5 + 𝐹𝐸 = 0 𝐹𝐸 = 412.5 − (48) 𝐹𝐸 = 364.5 𝑁 MN B.7 Ske le ta l L eve r Syst em 22 + value therefore acting downward Mechanical Advantage The Mechanical Advantage (M.A.) is defined as the ratio of the load force to the applied force. It indicates whether you need to generate a larger applied force than the load force you are attempting to move. 𝐹𝑙 Fl : Resistance load (weight) 𝑀𝐴 = 𝐹𝑎 Fa : Force applied by muscles The mechanical advantage can be >1, Fl → MA >1 → MA 1. Always efficient. Designed for strength. Load force: weight of the body Applied force: generated in the calf muscles Fulcrum: ball of the foot MN B.7 Ske le ta l L eve r Syst em 25 Class III Fa is between the fulcrum and Fl MA is always < 1. Always inefficient. Designed for speed and range of motion. Load force: weight of the arm Applied force: generated in the bicep muscle Fulcrum: elbow joint MN B.7 Ske le ta l L eve r Syst em 26 Back Injury MN B.7 Ske le ta l L eve r Syst em 27 Back Injury Normal load (100%) occurs in the back for a healthy person standing upright (shaded box). The stress (load) on the lumbar vertebrae can increase significantly depending on the activity of the body at a given time. Lifting heavy objects increases loading on the back muscles. Poor lifting technique can result in back injury due to the requirement to generate large forces in the back muscles. MN B.7 Ske le ta l L eve r Syst em 28 Back Injury 𝑇𝑜𝑟𝑞𝑢𝑒 𝑎𝑟𝑜𝑢𝑛𝑑 𝐿5 𝐶𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 𝑇𝑜𝑟𝑞𝑢𝑒 = 𝐴𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 𝑇𝑜𝑟𝑞𝑢𝑒 111 0.4 + 58 0.25 + 81 0.2 + 328 0.1 = 𝐹𝐵 0.06 𝐹𝐵 = 1798 𝑁 𝑀𝑜𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑏𝑜𝑥 10 𝑐𝑚 𝑓𝑢𝑟𝑡ℎ𝑒𝑟 𝑎𝑤𝑎𝑦 𝐹𝐵 = 2762 𝑁 MN B.7 Ske le ta l L eve r Syst em 29 Back Injury o Knees bent, back straight: uses muscles in legs and arms. o Keep mass close to body: reduces torque. o Feet slightly apart: increase stability. o Use stomach muscles to help: reduces forces generated in back muscles. o Maintain natural curve of spine: reduces pressure between vertebral discs. MN B.7 Ske le ta l L eve r Syst em 30 Learning outcomes At the end of this lecture, the learner will be able to Define Centre of Gravity (CofG) and understand its significance in biomechanical calculations of force and in determining an object’s stability. Draw a diagram showing the rate of force development for human muscle and outline what factors determine the maximum tension achievable by human muscle. Define moment, torque, and lever. Outline the conditions required to achieve static equilibrium and recognise these conditions in anatomical examples. Demonstrate the ability to calculate biomechanical forces using the principle of moments. Define and calculate mechanical advantage. Differentiate between Class I, II, and III levers using anatomical examples. Discuss the risk of back injury associated with poor lifting technique. MN B.7 Ske le ta l L eve r Syst em 31 Thank you F O R M O R E I N F O R M AT I O N P L E A S E C O N TA N T D r. A nd y M a E MA IL: ama@rc si.c om 32