The concept of a free body diagram (FBD) might seem intimidating at first but it stems from a fairly simple idea. Let’s start with something familiar, like money. Every weekend, I go to my favorite café and enjoy a cup of coffee while working on my computer. Afterward, I purchase my weekly airtime and buy groceries on my way back home. Below is a simple cash flow for my typical Saturday, you may note that it looks very similar to a free-body diagram. The major difference is that the arrows on a cash flow are arbitrary and numbers represent money while the arrows on a free body diagram are precise and the numbers represent magnitudes of force vectors.
Introduction to free-body diagram
For starters, you can think of a free-body diagram as some kind of a cash flow-esque diagram between a body and its environment – except instead of cash, bodies “trade” forces with their environment. Think of a light bulb hanging from a ceiling, for example, it is constantly trading forces with its environment. Here is how
The Earth pulls on the bulb with a force of gravity that is equal to its weight. In turn, the bulb pulls on the cord attached to it. The cord attached to the bulb then pulls on the ceiling. The ceiling pulls on the roofing of the house, which in turn pulls on the walls, which in turn pulls on the Earth. This might sound like a lot of things going on. But what if there was a way to simplify this seemingly never-ending trade-off of forces?
Why do we need a free-body diagram?
What a free-body-diagram does is to “isolate” a single body (i.e. it “frees” it from other bodies) and represents all the forces acting on the body without showing other bodies in the environment. A free-body diagram comprises a single body and the interactions (forces) on the said body from other bodies in its environment. This is why we say the body is “free”.
It is a common practice to represent bodies as particles or sometimes, little black squares when dealing with free-body diagrams. This is only done for simplification purposes similar to how mountain peaks are represented as little black triangles on geographical maps.
When dealing with analytical problems in classical physics, the internal structures and motions of the body can be ignored if the body in general moves as a unit. For example, we can represent a rocket as a particle when analyzing its motion in space without considering the internal combustion movements because the whole thing moves as a unit.
Example of a free-body diagram
For our first example, let’s create a free-body diagram of a hanging light bulb. Since we are analyzing forces that act on the bulb as a whole and not just certain parts of it, we’ll represent the bulb as a dot or box. This is the first step: to recognize that the body in question can rightly be simplified to a dot.
The next step is to select a frame of reference and form a coordinate system. This will help us to consistently represent forces or other vectors in the right direction. In our case, we can select the frame of reference to be the room that houses the bulb – i.e. we are formulating a free-body diagram from the point of view of an observer in the room. We can then select the orientation of our coordinate axes. In our case, the positive direction may be taken to be upward.
Next, we identify all forces involved with the bulb. Some of the most common forces encountered in mechanics are friction, gravity, tension, buoyancy, normal, and applied force. In our case, the bulb experiences the force of gravity due to the gravitational pull of the Earth. This force acts vertically downwards and will be represented as such. Another force is the tension in the cord, it acts vertically upwards along the cord. This is appropriately represented by arrows.
Using a free-body diagram to solve an analytical problem
Free-body diagrams are a powerful tool when analyzing the motion of an object. A well-drawn free-body diagram can help you understand the forces causing a body to accelerate, decelerate or remain at rest. It is a like a map, it points you in the right direction of where you want to go.
Worked Example
A 77-kg person is parachuting and experiencing a downward acceleration of 2.5 m/s2 shortly after opening the parachute. The mass of the parachute is 5.2 kg. (a) Find the upward force exerted on the parachute by the air. (b) Calculate the downward force exerted by the person on the parachute.
Let’s start with the first part of the question, evaluating the upward force on the parachute by the air.
Identifying the body in question can rightly be approximated to a particle
Although the parachute is an extended body, the forces of gravity and buoyancy act equally on the body making it move as a unit. It can therefore be regarded as a particle for this problem.
Select a frame of reference and form a coordinate system
We select the Earth frame of reference. This means from our point of view; the parachute is accelerating towards the ground at a rate of 2.5m/s2. We also, out of convenience, take the direction of increasing y to be upwards.
Identify all forces acting on the body
We identify three forces acting on the parachute:
- The gravitational pull of the Earth, this is its weight. Directed vertically downwards
- Tension in the cords. This is the applied force from the parachutist. Directed vertically downwards
- Buoyancy due to the air resistance. Directed vertically upwards.
The internal forces within the parachute do not affect the motion. We also neglect any sideways force during the fall.
We now start with the free-body diagram of the parachute, which is shown in (b) above.
\begin{equation}
\sum F_y=m a_y
\end{equation}
The negative sign indicates that the net force is acting downwards – i.e. in the direction of decreasing “y”.
\begin{equation}
\sum F_y=-m a_y
\end{equation}
\begin{equation}
D-m g-T=-m a_y
\end{equation}
\begin{equation}
D-T=m\left(g-a_y\right)
\end{equation}
\begin{equation}
D-T=2.5(9.8-2.5)=18.25
\end{equation}
\begin{equation}
D=T+18.25 \ldots \ldots(e q 01)
\end{equation}
Next, we turn our attention to the parachutist whose free-body diagram is shown in (c) in the figure above.
\begin{equation}
\sum F_y=m a_y
\end{equation}
\begin{equation}
T-m g=-m a_y
\end{equation}
$$T\;=\;m(g\;-\;a)\;$$
$$T\;=\;77(9.8\;-\;2.5)$$
$$T\;=\;562.1N\;….\;(eq\;02)$$
This is the magnitude of the tension in the cords of the parachute. Note that its value is positive indicating that its direction is vertically upwards in the positive “y” direction.
Next, we evaluate the drag force, D, by substituting equation 2 into 1.
$$D\;=\;T\;+18.25$$
$$D\;=\;562.1\;+18.25$$
$$D\;=\;580.35N$$
Thus the upward force exerted on the parachute by the air is 580.35N and the downward force exerted by the person on the parachute is equal to the tension in the cords which is 562.1N.
