por We explain what the principle of transmissibility is, with examples and solved exercises
What is the principle of transferability?
He transferability principle It is applied to solid objects and states that a force applied to 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 effect of translational and rotational motion on the object, as long as 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 straight dotted line that contains them is called line of action of force.
The principle of transmissibility is very useful, since it allows the forces acting on the object to be conveniently slipped, in order to facilitate analysis.
Explanation of transferability principle
The principle of transmissibility is based on the fact that two forces F1 and F2 are equivalent, provided they have the same magnitude and the same direction.
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 the 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 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 it is an excellent approximation, if the deformation 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.
Observe that in both figures the body (rigid or not) is in equilibrium, since the forces have the same magnitude and direction and opposite directions. In addition, 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 transferability principle
Example 1
Suppose you have a heavy trunk on a level ground. The effect of pushing it on the left side is the same as pulling on a horizontal rope on the right side, as long as both forces are applied along the green horizontal line shown. In such a case, the movement of the trunk on the ground is the same.
Example 2
You have a long plank as a shelf. To install it, it is equivalent to hold it from the ceiling by means of ropes at its ends, than to place props below, 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 that there is a force F applied at a point A, the moment that causes this force about point O shown in the figure is:
mOR = rA × F
Well, the principle of transferability ensures that Facting on any point along its line of action, for example points B, C and more, causes the same moment with respect to point O. Therefore it is valid to affirm that:
mOR = rA × F = rB × F = rC × F
solved exercises
Exercise 1
A homogeneous sphere has mass M = 5 kg and is resting at rest on a frictionless horizontal surface.
a) Draw in a diagram the force exerted by the surface on the sphere. b) Build 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
Graph a) shows the force exerted by the surface on the sphere, called normal No., since 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) and the line of action is the vertical that passes through the geometric center of the sphere.
Graph b) shows the free body diagram of the sphere, where apart from the normal, the weight is shown, which is applied to the center of gravity, denoted with the yellow point.
Thanks to the principle of transmissibility, the normal force No. it can be moved 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.
Since the sphere is in equilibrium, taking the vertical upwards as positive and the vertical downward as negative, 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/s2 = 49 N, directed vertically upwards.
Exercise 2
Indicate whether the transferability principle is fulfilled in the following cases:
A force of 20 N applied horizontally on a rigid body is replaced by another force of 15 N applied at another point on the body, although both are applied in the same direction.
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 essential conditions of the principle of transferability does not exist.
A force of 20 N applied horizontally on a rigid body is replaced by a force of 20 N applied vertically at another point on the body.
On this occasion, the principle of transmissibility is not fulfilled since, although the two forces have the same module, they are not applied in the same direction. Again, one of the essential conditions of the principle of transferability does not exist. It can be said that the two forces are equivalent.
A force of 10 N applied horizontally on a rigid body is exchanged for another force of 10 N also applied at another point on the body, but in the same direction and direction.
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 sense. All the necessary conditions of the principle of transferability are fulfilled. It can be said that the two forces are equivalent.
A force slides in the direction of its line of action.
In this case, the transmissibility principle is fulfilled since, being the same force, the magnitude of the applied force does not vary and it slides along its line of action. Once again, all the necessary conditions of the principle of transferability 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 sense. If the magnitude of the first is 15 N and that of the second is 25 N, what conditions must a third external force meet that replaces the resultant of the two previous ones in order to comply with the principle of transmissibility?
On the one hand, the value of the resulting force has to be 40 N, which is the result of adding the magnitude of the two forces.
On the other hand, the resultant force must act at any point on the straight line that joins the two points of application of the two forces.
References
Bedford, 2000. A. Engineering Mechanics: Statics. Addison Wesley.
Beer, F. 2010. Vector Mechanics for Engineers. McGraw Hill. 5th. Edition.
Giancoli, D. 2006. Physics: Principles with Applications. 6th. Ed Prentice Hall.
Hibbeler, R. 2004. Engineering Mechanics: Statics. Prentice Hall. Meriam, JL 2012. Engineering Mechanics: Statics. 7th edition. Wiley&Sons.