6.4: Free-Body Diagrams

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As Figure 6.2.1 shows, trying to draw every single force acting on every single object can very quickly become pretty messy. And anyway, this is not usually what we need: what we need is to separate cleanly all the forces acting on any given object, one object at a time, so we can apply Newton’s second law, \(F_{net} = ma\), to each object individually.

In order to accomplish this, we use what are known as free-body diagrams. In a free-body diagram, a potentially very complicated object is replaced symbolically by a dot or a small circle, and all the forces acting on the object are drawn (approximately to scale and properly labeled) as acting on the dot. Regardless of whether a force is a pulling or pushing force, the convention is to always draw it as a vector that originates at the dot. If the system is accelerating, it is also a good idea to indicate the acceleration’s direction also somewhere on the diagram.

The figure below shows, as an example, a free-body diagram for block 1 in Figure \(\PageIndex{1}\), in the presence of both a nonzero acceleration and a kinetic friction force. The diagram includes all the forces, even gravity and the normal force, which were left out of the picture in Figure \(\PageIndex{1}\).

Figure \(\PageIndex{1}\), with the friction force adjusted so as to be compatible with a nonzero acceleration to the right.

Note that I have drawn \(F^n\) and the force of gravity \(F^G_{E,1}\) as having the same magnitude, since there is no vertical acceleration for that block. If I know the value of \(\mu_k\), I should also try to draw \(F_k = \mu_k F^n\) approximately to scale with the other two forces. Then, since I know that there is an acceleration to the right, I need to draw \(F^t\) greater than \(F^k\), since the net force on the block must be to the right as well. And, if I were drawing a free-body diagram for block 2, I would have to make sure that I drew its weight, \(F^G_{E,2}\), as being greater in magnitude than \(F^t\) , since the net force on that block needs to be downwards.