Free body diagram: how to do it, examples, exercise
A free – body diagram, isolated – body diagram, or force diagram is a diagram in which the forces acting on a body are represented by arrows.
Be sure to include in the diagram all the forces acting on the object, and since it is a vector quantity, the arrow is responsible for pointing out its direction and direction, while the length of the arrow gives an idea of the magnitude or intensity.
In Figure 1, we have an example of a free-body diagram that we are going to look at.
The situation is as follows: a traffic light hanging from cables (figure 1a). Two forces act on it, one is exerted by the Earth, which is the weight. In the diagram it is indicated as F g and acts vertically downwards.
The other force is the tension in the vertical rope, called T 3, which runs in the vertical upward direction, holding the traffic light and preventing it from resting on the ground.
When a problem has more than one object, it is necessary to draw a diagram for each one separately.
The knot between the slanted ropes and the one holding the semaphore is considered a point object and its free-body diagram is shown in figure 1c. Note that for the node the voltage T 3 is directed downwards.
It is important to emphasize that in the free-body diagram, the forces that the object exerts on other bodies should not appear, but only those that act on it .
Free Body Diagram Examples
The free-body diagram allows the application of Newton’s laws and, with them, determines the state of motion or the rest of the object on which the forces act. In the case of the semaphore shown, we can determine the value of the voltages in the cables that support the semaphore, known the weight of this.
Once this data is known, suitable cables are selected to hang on the traffic light and fulfill its function without breaking.
Free-body diagrams are used to describe various everyday situations, such as these:
A person pulling a trunk or container
It is very common for people to have to move heavy objects such as the picture container. To do this, they must exert a force F on the container , which in this example is horizontal and to the right, which is the direction of movement.
But this is not the only force acting on it, there is also the normal n , exerted by the flat surface of the wheeled platform. And finally, there’s his weight: F g , directed vertically down.
Normal is a force that arises whenever two surfaces are in contact and is always perpendicular to the surface exerting it. In this case, the platform with wheels exerts a normal on the container.
A block that slides on an inclined plane
Some tables have a slightly slanted table for easy note-taking and reading. It also has a pencil compartment, but we’ve all already placed the pencil on the table outside the compartment and watched it slide across the table.
What forces act on the pencil?
The same ones that act in the block shown in the following free-body diagram:
The normal F N is the force that the table surface exerts on the pencil or support pad. Unlike the previous example, the normal is not vertical but slanted. Remember that normal is the force exerted by the table on the block and is perpendicular to it. As the table is tilted, so is the normal one.
As always, the weight F g is vertical, regardless of the system’s inclination.
And finally we have a new operating force, which is the kinetic friction F fr between the table and the pencil or pad. Friction is also a contact force, but unlike normal, it is a tangential (parallel) force to the surface. Also note that it is always directed in the opposite direction of the movement.
The Atwood machine is a simple machine consisting of a light, frictionless pulley on the rail, through which a light, inextensible rope is also passed.
Two objects of mass 1 in 2 are hanging from it. When one of the objects goes up, the other goes down, as shown in figure 4a:
Since there are two objects, a free-body diagram is drawn for each one separately. For both objects, there are only two forces: the tension on the string T and the respective weights.
In the figure, each weight is expressed directly as the product of the mass per acceleration. In turn, the tension is always directed vertically along the tensioned rope.
Apply Newton’s laws to determine the acceleration with which the Atwood machine masses shown in the previous section move.
Newton’s second law states that the sum of forces is equal to the product of mass times acceleration.
The sign convention in each mass can be different, so we will take as positive the movement, as indicated in the graph, the first mass goes up and the second goes down.
In some problems, the statement does not provide information, so the signs must be assigned arbitrarily, and if the acceleration result is negative, the mass system will move in the opposite direction from the one initially assumed.
-For mass 1 (increase):
T – m 1 g = m 1 a
-For mass 2 (low):
-T + m 2 g = m 2 a
Both equations form a system of linear equations of two unknowns, since the voltage appears with a different sign in each equation, we simply add them term by term and the voltage is canceled:
m 2 g – m 1 g = m 1 a + m 2 a
a = m 2 g – m 1 g / (m 1 + m 2 )