Forces Affecting Musculoskeletal System PDF

Forces a ecting musculoskeletal system: BONES There are elements in the MSK: - Bones - Ligaments - Tendons - Muscles Which can be modified due to external forces. They are deformable. They will each respond differently in response to external forces. Most simple effect = bones, most complicated = mu...

Forces a ecting musculoskeletal system: BONES There are elements in the MSK: - Bones - Ligaments - Tendons - Muscles Which can be modified due to external forces. They are deformable. They will each respond differently in response to external forces. Most simple effect = bones, most complicated = muscles. We will first focus on bones. BONES There are different classes of joints. The main difference between each is the degree of movement / freedom: one displacement or rotation possibility for each one of these joints - Displacement - Rotation Depending on the joint, we will have a max of 3 degrees of rotation and 3 degrees of displacement. But in a normal joint this is not the usual situation, typically in a joint we will find a limited degree of freedom (simplified version): - No displacement - 1 to 2 rotations NOT ALLOWED We are going to analyze joints in which there is only a single rotation axis and no displacements. Typical example: elbow, focusing only on the flexion and extension and eliminating pronation and supination. All the consequences we obtain from this simplified version can also be applied to a more general system. FORCES AND EQUILIBRIUM Example: in this elbow we find: bones in the forearm; humerus in the arm; axis of rotation in which the 2 sides of the joints can modify their orientation. The angle alpha can be modified. This angle will be changes due to a set of different forces. Initially, 2 forces: - The load (object in the hand, mass we are lifting), external force due to the presence of this load (FL) - Force exerted by the biceps: internal force which balances the action of the load. All these muscles can only produce a pulling force, a force of contraction, in the direction of its fibers. The only force which can be done is contraction force, pulling from one end of the tendon to the other. If we are looking for an equilibrium (system at rest), the acceleration has to be 0. The Ftotal will also be 0 = equilibrium. Is it possible to have a 0 total force with these 2 forces? Yes, because we can obtain: - total force in X direction = 0 - total force in Y direction = 0 Vectors: the direction / orientation of the forces plays a role, as the load can be applied in different directions. In our diagram, the orientation of the force is tilted, not diagonal: there is an X component and a Y component. X is pulling left and Y is pulling the load upwards (Fm) and down (FL). Joint force (Fa) From this diagram we can obtain that the muscular force is pointing upwards (Y component +), and load is pointing downwards (Y component -), so each force can be balanced in the Y axis. Fy = 0 → equilibrium. But there is also an X component in the muscular force, which we cannot balance by the action of any of the forces we have considered up till now. If we are looking for an equilibrium, the only possibility for it to be at rest is that the X component of the muscular force be 0: there is something missing to compensate for Fx. This would be the JOINT FORCE (Fa) which compensates muscular Fx. “For any joint, the minimum set of forces to reach equilibrium consists of 3 forces”: - Load on the system (FL) - Muscular force (FM) - Force that appears at the joint (Fa) The joint force is the reaction due to interaction in the interface, meaning everytime we find a place where one of these bones is in contact with something, an action-reaction force will appear (Fa). BEHAVIOUR OF THE 3 FORCES If the reaction force Fa is large, in response to a large X component, the ligament can be stressed and torn (sprain). In the same way, Fm can also be very large in response to a heavy load (Y axis), tearing muscles and tendons (strain). Lastly, FL can also break bones if very prominent. What determines how high the reaction force will have to be: the load, the distance to the joint and orientation. E ect of distance. “Shopping bag situation” 2 situations: similar forces, produced by similar loads, pointing in the same orientation. FL1 and FL2 are similar. The point is that the distance to the joint is different. In each situation, we will need a different Fm. FM1 will have to be larger than FM2 (FB1 >>> FB2), as the distance to the joint is larger. The load is similar in both situations but something changes making FM1 larger than FM2. The only component that changes is the distance. The distance between where the force is applied and the axis of rotation are different in both cases: distance will be a parameter when considering the equilibrium. Why? We can understand it by using the example of a lever. A lever is a rod which can rotate around a single point: a fulcrum or hinge. This is a fixed point, it cannot be displaced; it's only possibility is to rotate around the hinge. When applying forces F1 and F2, perpendicular to the axis of the rod, to obtain equilibrium: F1 ·d1 = F2 · d2 In any lever, in order to reach equilibrium, this law has to be fulfilled. Model that can be applied in some joints of the body (which are fixed and only rotate): ○ 1st class lever: the different forces act at the same distance from the joint in the same orientation, to reach equilibrium. ○ 2nd class lever: FL is acting close to the joint / hinge, and Fm is far from the joint ○ 3rd class lever: The Fm acts close to the hinge, and FL acts far from it. The latter is the situation for the elbow. It’s a 3rd class lever. E ect of orientation. Flexion or extension Effect of holding the load by flexing it or extending it: we establish another 2 different limit situations. We have the same load being applied in both. In which situation will the biceps need to apply a stronger force? - Flexed (close to 90º): less force needed (Fm) - Extended (close to 180º): more force (Fm) needed to pull and contract In order to reach an equilibrium, we need to fulfill the formula between the distances and the forces which are perpendicular to each other. - In the extended situation, Fb (muscle, Y axis) is much smaller: the Y component is reduced as the force is distributed in the X and Y axis. Therefore, the muscle force Fm will have to be very big, as for there to be enough Y component to compensate for FL. - The force needed in flexion to balance out FL is almost completely distributed in the Y axis. Therefore, for a given load, in the flexed position we will need the minimum force to reach equilibrium, as it will all or almost all be present in the Y axis. In these 2 situations, the load and distances remain the same, but orientation changes: FB · sen (α) · dB = FL · dL Parameters to consider when seeking equilibrium of joints: - The magnitude of the force - The distance between the axis of the joint and where the force is being applied - The orientation of the force