Principle of transmittance of forces

The principle of transmissibility is applied to solid objects and affirms that a force applied on some point of the body is equivalent to another force of equal magnitude and direction, as long as said force is applied in the same line that contains the original force.

Therefore, any force of the same magnitude and direction will cause the same translational and rotational motion effect on the object, provided that its point of application is located on the same line, as shown in the following figure.

The forces shown F and F ‘are said to be equivalent forces and the dotted straight line that contains them is called the line of action of the force .

The principle of transmissibility is very useful, since it allows the forces acting on the object to slide as conveniently as possible, in order to facilitate the analysis.

In addition, they must produce the same moment with respect to any point O, which is guaranteed by having the same line of action and because the moment is the product of force times the distance from O to said line.

Note that the principle applies only to a rigid body, that is, an object in which the relative distances between its parts do not change, because the internal forces that hold it together are strong enough. Therefore, the object does not change its shape, whether or not external forces act on it.

On the other hand, if the object is not rigid, modifying the point of application of the forces would produce variations in terms of the tension or compression applied to the body, which would lead to changes in its shape.

Of course, assuming that a body is rigid is nothing more than an idealization, since in reality all objects are deformable to a greater or lesser extent. However, in many cases this is an excellent approximation, if the strain is small enough to be considered negligible.

Limitations

The principle of transmissibility has, as indicated, a limitation regarding the internal effects of rolling or sliding forces. The following figure shows an object, with the forces F and F ‘applied at different points on the same line of action.

Note that in both figures the body (rigid or not) is in equilibrium, since the forces have the same magnitude and direction and opposite directions. Furthermore, the forces are, as has been said, on the same line of action, but in the left figure the effect on the body is of tension while in the right the effect is of compression.

Therefore, although the body remains at rest, the internal effects are different and become apparent if the object is not totally rigid. In the case of the left the forces tend to lengthen the body, while in the right they tend to shorten it.

Examples of the principle of transmissibility

Example 1

Suppose you have a heavy trunk on a horizontal floor. The effect of being pushed from the left side is the same as being pulled by a horizontal rope from the right side, while both forces are applied along the green horizontal line shown. In this case, the movement of the trunk on the ground is the same.

Example 2

There is a long plank as a shelf. To install it, it is equivalent to fasten it to the ceiling by means of ropes at its ends, than to place struts underneath, also at the same ends.

In both cases, the forces that balance the plank will have the same magnitude and direction, acting on the same lines of action, but are being applied at different points.

The principle of transmissibility and moments

Suppose we have a force F applied at a point A, the moment that originates this force around the point O shown in the figure is:

M _O = r _A × F

Well, the principle of transmissibility ensures that F , acting on any point along its line of action, for example points B, C and more, originates the same moment with respect to point O. Therefore it is valid to affirm what:

M _O = r _A × F  = r _B × F = r _C × F

Solved exercises

Exercise 1

A homogeneous sphere has mass M = 5 kg and is supported at rest on a horizontal surface without friction.

• • a) Draw in a diagram the force exerted by the surface on the sphere.
  • b) Construct the free-body diagram of the sphere
  • c) Calculate the value of the normal force exerted by the surface on the sphere.

• Solutions a and b

In the graph a) the force exerted on the sphere surface shown, normal call N , as it is perpendicular to the surface. The point of application of the force coincides with the point of support of the sphere on the surface (point in green color) and the line of action is the vertical one that passes through the geometric center of the sphere.

In graph b) there is the free-body diagram of the sphere, where apart from the normal, the weight is shown , which is applied at the center of gravity, denoted by the point in yellow.

Thanks to the principle of transmissibility, the normal force N can be transferred to this point, without changing its effects on the sphere. These effects are none other than keeping the sphere resting on the table in balance.

• Solution c

Since the sphere is in equilibrium, taking vertical upwards as a positive direction and negative vertical downwards, Newton’s second law results in:

N – P = 0

That is, the weight and the normal balance, therefore they are equal in magnitude:

N = P = Mg = 5kg × 9.8 m / s ² = 49 N, directed vertically upwards.

Exercise 2

Indicate whether the principle of transferability is met in the following cases:

• First case

A 20 N force applied horizontally on a rigid body is replaced by another 15 N force applied at another point on the body, although both are applied in the same direction.

• • Solution

In this case, the principle of transmissibility will not be fulfilled since, although the two forces are applied in the same direction, the second force does not have the same magnitude as the first. Therefore, one of the indispensable conditions of the principle of transferability does not exist.

• Second case

A 20 N force applied horizontally on a rigid body is replaced by another 20 N force, applied at another point on the body and vertically.

• • Solution

On this occasion, the principle of transmissibility is not fulfilled since, although the two forces have the same modulus, they are not applied in the same direction. Again, one of the indispensable conditions of the principle of transmissibility does not exist. It can be said that the two forces are equivalent.

• Third case

A 10 N force applied horizontally on a rigid body is exchanged for another 10 N applied at another point on the body, but in the same direction and direction.

• • Solution

In this case, the transmissibility principle is fulfilled, since the two forces are of the same magnitude and are applied in the same direction and direction. All the necessary conditions of the principle of transmissibility are fulfilled. It can be said that the two forces are equivalent.

• Fourth case

A force slides in the direction of your action line.

• • Solution

In this case, the principle of transmissibility is fulfilled since, being the same force, the magnitude of the applied force does not vary and it slides along its line of action. Again all the necessary conditions of the principle of transmissibility are fulfilled.

Exercise 3

Two external forces are applied to a rigid body. The two forces are applied in the same direction and in the same direction. If the modulus of the first is 15 N and that of the second is 25 N, what conditions must a third external force meet to replace the resultant of the two previous ones to fulfill the principle of transmissibility?

• Solution

On the one hand, the value of the resultant force has to be 40 N, which is the result of adding the modulus of the two forces.

On the other hand, the resultant force must act at any point on the straight line joining the two points of application of the two forces.