Applied physics

Chapter Three Applied physics

Levers, beams and moments

The lever is perhaps the simplest of all machines; its existence pre-dates the wheel and is testified to by the existence of the Pyramids and Stonehenge. “Give me a fulcrum and a lever long enough and I will move the earth,” boasted Archimedes in the 3rd century BCE. Stone Age technology was less prosaic, but – together with rollers and ramps – enabled the movement and erection of blocks of stone weighing over 10 tonnes at sites such as Stonehenge, Avebury and Carnac.

It is important for us to understand levers, and the mathematics associated with them: the biomechanics of the human body is fundamentally little more than a series of interconnected levers.

Three elements make up a lever system: the load (the thing that is being moved), the fulcrum (the point around which the lever turns) and the effort (the force used to move the lever).

Two factors define a lever. The first of these is the mechanical advantage (MA), also known as the force ratio. A lever is a magnifying force – the amount by which it magnifies the force is given by the MA:


For example, if an object weighs 30 N but the lever allows a person to move it using just 5 N (Fig. 3.1):


Note: the MA does not have any units; it is, in effect, a ratio. Of course, it is not possible to do this without a trade-off, otherwise we would be getting energy for free and violating pretty much every law of physics. The payback comes with the fact that, although we don’t have to push so hard, we do have to push further. The ratio of the distance moved by the effort (SE) to the distance moved by the load (SL) is called the distance ratio (DR), also known as the velocity ratio:


If a system is 100% efficient, which, of course, is only ever hypothetically possible, then these two factors will balance each other out: moving the load using the lever in Figure 3.1 will require only 1/6 of the effort but the load will only move 1/6 of the distance compared with the effort.

Levers come in three flavours, depending on the relative position of the elements that make up a lever system:

Muscle action

Not all muscles act directly on the bones they move to articulate a given joint. Although most do – called direct action – there are muscles that precipitate movement indirectly, called indirect action. Examples of direct action are seen all over the body: think of the action of biceps brachialis (Fig. 3.4) as a classic example. Some muscles, however, do not insert directly in to the two bones that they articulate. Perhaps the best – and most easily understood – example of this is the action of the four quadriceps muscles, which, in this circumstance, can be considered as a functional whole.

The quadriceps (Fig. 3.5) arise from the femur and act to extend the knee. The muscle faces a problem, however: if the knee is flexed, it has to somehow ‘get round the corner’. It achieves this by inserting into the superior pole of the patella, a large sesamoid bone that acts as a pulley, its reciprocally curved deep surface sliding between the two condyles of the femur. A tendon then runs from the inferior pole of the patella to the tibial tuberosity, so that contraction of the muscle will extend the knee: the quadriceps move the patella; the patella moves the tibia.

Moment of inertia

The direction of pull of a muscle – called its line of action – is an important consideration when considering the turning force created in a joint by a muscle. This turning effect is known as a moment of inertia (moment for short) and applied to any object that is able to rotate about an axis such as the forearm, which rotates about the elbow or, in the first instance, a door, which rotates about its hinge (Fig. 3.6).

In Figure 3.6, we see four ways in which a force could be applied to the door and we know intuitively from our daily experience what will happen if we push the door in each of the instances A–D. Force A will not move the door as it is not generating a turning force, merely pushing the door into its hinges. Force C will move the door with all of the force directed in the right direction but we all know the door will move much more easily if we push further away from the hinge, exactly as is the case with a lever and its fulcrum, as is the instance with Forces B and D.

So, from this we can deduce that there are two factors that govern the size of the turning force or moment. The first is the distance from the pivot to the point where the force is being applied; the second is the direction in which the force is being applied, the line of action. The formula from which we can calculate a moment (I) is obtained by multiplying the force (F) by the perpendicular distance (d) from the line of action of the force to the axis:


NOTE: Remember, however, that force is a vector quantity and, therefore, only the component of force perpendicular to the door need be considered.

This is best understood by reconsidering the four examples on an individual basis. If we consider example C (Fig. 3.7) first, we do not need to resolve the vector of forces because all of the force is being directed at right angles to the door. The distance, s, from the line of action of the force (F = 10 N) is 20 cm; therefore, the moment is:


In example D (Fig. 3.8), the force is again acting perpendicularly to the door but this time the perpendicular distance from the line of action of the force to the axis of rotation is 80 cm.


When we consider example C (Fig. 3.9), we have a more complex situation as the calculation of the perpendicular distance from the line of action of the force to the pivot now requires some consideration and a little bit of trigonometry. If we look at the situation in (A), we can see the distance that we need to calculate, d. In order to do this, we need to construct a short series of right-angled triangles so that we can use basic trigonometry to calculate the angles and lengths required.

If we draw a vertical line from the intersection of d and the extension of the axis and a perpendicular to this, which intersects F, there are now three right-angled triangles, I, II and III.

As triangles I and II are exactly the same shape, their internal angles must be the same and so θ must be 30°. As there are 180° on a straight line, this means λ must be 180°− 30° = 150°. The same principle shows that γ = 180° − 150° = 30°. This means that we now have the ability to calculate all the dimensions of triangle III: we know it is a right-angled triangle with hypotenuse of 80 cm and internal angles 90°, 30° (and, by rule of internal angles, 60°). Therefore:


Therefore the moment:


In the final example, A (Fig. 3.6), the line of action is perpendicular to the axis; therefore, the distance is zero:


From the standpoint of the manual physician, doors have only a limited interest: they can be closed to allow privacy and opened to admit a patient. Let us therefore take an example that instead relates to the human body.

Example 1

In Figure 3.10, the biceps brachialis muscle is holding the arm stationary against two forces, a 20 N weight held in the hand and the weight of the forearm itself, which is regarded as operating through its centre of gravity. We wish to know what force the muscle needs to generate in order to maintain this posture.

All the moments acting on a body are summative (that is, they can be added and subtracted); however, it is necessary to realize that those acting in a clockwise direction (in this case the action of the muscle) will counter those acting in an anticlockwise direction. It is convenient to regard those moments acting in a clockwise direction as being positive and those acting in an anticlockwise direction as negative. This is by no means essential, so long as you make one direction positive and the other negative, the maths will come out right. Because this is a purely voluntary notion, it is always a good idea to state your convention at the outset, so, in this case: clockwise = +ve.

If the arm is stationary, then the sum of the positive moments must be equal to the sum of the negative moments:


which can be written as:


Calculating the moments created by the mass and the forearm weight is quite straightforward as the perpendicular distance from the lines of actions of the forces to the axis is simply the distance from the elbow at which the forces act. In the case of the forearm’s centre of gravity, this is 12 cm; in the case of the weight held in the hand, 20 cm (remember that these quantities will need to be converted into metres). As the weight is given in newtons rather than kilogrammes, it will not need to be converted; therefore:




So, the anticlockwise (−ve) moment being generated by the action of the biceps brachialis muscle is −5.2 Nm (assuming, of course, for the sake of simplicity, that this is the only muscle involved in flexion of the elbow). We can construct a second diagram (B) showing the dimensions pertinent to the biceps brachialis muscle. We know that:


As it is the force that we are trying to calculate, we can rearrange this as:


Apr 4, 2017 | Posted by in GENERAL & FAMILY MEDICINE | Comments Off on Applied physics
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