Bio Mechanics and Kinetics

Task 1. Produce a labelled diagram showing the forces involved on the skeletal structure of a human arm as a weight is being held at a particular angle. Your report should use the principle of moments to fully describe all of the forces that are being applied at various points along the forearm through the muscle and tendons. (Assessment criteria 1.1, 1.2, 1.3) Skeletal muscles consist of many parallel fibres wrapped in a flexible sheath along a bone, narrowing at both ends into tendons. Some of the muscles thin into two or three tendons which are known as “biceps” and “triceps”. Muscle fibres contract after receiving an electrical signal from the nerve ending attached to them; hence the muscles shorten and a pulling force is sensed by the two bones. Primarily, the main function of the muscles is to pull and not to push.

As part of this task, I am going to demonstrate the forces involved on the skeletal structure of the human arm when a weight is being held at a particular angle. The image below shows a free-body diagram illustrating the forces exerted on the forearm bar. According to the Laws of Statics such as Newton’s Law, the net force on the immobile bar must be zero, and the total torque (which will be discussed later) is also zero.

Figure 1

Hence the forces acting on the forearm are its weight (W), the weight of the hand (H), force from the bicep muscle (B, which pulls upward the forearm at an angle α) and the force from the humerus bone (A).

The muscular system within the arm generates linear force. Linear force refers to the force that acts in straight line between the origin and the insertion. However the linear force is manifested by the rotational moment which is generated at the joint centre. This is due to the geometrical relationship between the lines of action of the muscles and the joint centre.

The maximum force a muscle is able to exert is equivalent to its cross-sectional area, i.e. the legs are capable of lifting heavier load due to having greater cross-sectional area compared to the arms. Therefore the estimated maximum force a muscle can apply is about 7×10⁶ dyn/cm² = 7 x 10⁵ Pa = 102lb/in². The formula to calculate the moment of force is:

For example, if an arm (weighing 7kg) lifts a load of 5kg by 1cm, what is the moment of force applied on the arm?

Firstly I will need to find the force of both objects, by using this formula:

Where acceleration is 9.8m/s (Earths gravitational field, since it is constant).

The force of the object = 5kg X 9.8m/s = 49N

The force of the arm = 7kg X 9.8m/s = 68.6N

Therefore the moment of a force =49N X 0.01m = 0.49Nm

The moment of force of 0.49Nm is applied on the arm.

The various joints in the body are known as levers which causes rotations about a fulcrum (axis rotation). This is used to figure out the forces exerted by the muscles such as lifting loads and transfer movement from one point to another. For a lever, the force F required to balance a load of weight (W) is:

Where d₁ and d₂ are the lengths of the lever arms (illustrated in figure 2)

If d₁ is 5cm and d₂ is 35cm, find out the force required to balance the weight of 5kg. Using the above formula:

Therefore, a force of 0.71Nm is required to balance the weight of 5kg on the arm.

If the load is close to the fulcrum, the mechanical advantage is greater (d₁2