Free body diagrams

A free-body diagram (abbreviated as FBD, also called force diagram) is a diagram used to show the magnitude and direction of all applied forces, moments, and reaction and constraint forces acting on a body. They are important and necessary in solving complex problems in mechanics.

What is and is not included in a free-body diagram is important. Every free-body diagram should have the following:

  • The body represented as a dot if it is a point mass, and the body itself if it is a rigid body.
  • The external forces/moments. The force vector should indicate: relative magnitude, point of application, and the direction.
  • A properly defined coordinate system.

A free-body diagram should not include the following:

  • Bodies other than the body we are interested in.
  • Forces applied by the body.
  • Internal forces depending on the chosen system. For example, a free-body diagram on a truss should not include the forces between individual truss members.
  • Kinematic quantities (velocity and acceleration).

Always assume the direction of forces/moments to be positive according to the appropriate coordinate system. The calculations from Newton/Euler equations will provide you with the correct direction of those forces/moments. Things that should not follow this are:

  • Gravity
  • Tension
  • Friction if the velocity $\vec{v}$ is provided

If forces/moments are present, always begin with a free-body diagram. Do not write down equations before drawing the FBD as those are often simple kinematic equations, or Newton/Euler equations.


Drawing free-body diagrams on point-masses/particles is straightforward, as all the forces are applied at a single point, and the point-mass cannot rotate about its own axis due to the absence of a moment of inertia. In some instances, Newton's 2nd Law in rotation is not necessary at all. We will look at one such example.

Example Problem: Point-mass on an incline

An object of mass $m$ is at rest on an incline at an angle $\theta$ with gravity $\vec{g}$ present as shown. The coefficient of friction between the object and incline is $\mu$. Draw the free-body diagram on the object, and write down the corresponding Newton's 2nd Law for the object.

We have four forces: gravity, the normal force due to the incline, and friction. We will use a dot to represent the object as it is a point mass. Recall we also have to define a coordinate system, so we will choose a convenient set of basis vectors. The tangential/normal basis vectors $\hat{e}_t, \hat{e}_n$ work best for this problem. Therefore, our free-body diagram is the following:

Now, we can proceed with Newton's 2nd Law: \[\begin{aligned} \sum \vec{F}_{ext} &= m\vec{a} \\ m\vec{g} + \vec{F}_s + \vec{N} &= m\vec{a} \end{aligned}\]

Decomposing the vector equation into scalar equations in $\hat{e}_t, \hat{e}_n$ : \[\begin{aligned} \hat{e}_t: \qquad F_s + mg\sin\theta = ma_t \\ \hat{e}_n: \qquad N - mg\cos\theta = ma_n \end{aligned}\]

Example Problem: Uniform circular motion of a particle

A particle of mass $m$ is undergoing uniform circular motion due to gravity from point $O$, which is fixed. Another force $\vec{F}$ is present. Draw the free-body diagram on the particle and write down the corresponding Newton's 2nd Law for the particle.

We will choose the polar basis to model this

There are two forces present: Gravity, which is the force that allows for circular motion to be possible, and another force present. Think of this force as something that represents some kind of thrust on a rocket modeled as a point mass. Circular motion is still possible without it, however, the addition of it helps understand and model complex dynamical systems better. All in all, the free-body diagram is as follows:

Now, we can proceed with Newton's 2nd Law: \[\begin{aligned} \sum \vec{F}_{ext} &= m\vec{a} \\ m\vec{g} + \vec{F} &= m\vec{a} \end{aligned}\]

Decomposing the vector equation into scalar equations in $\hat{e}_t, \hat{e}_n$ : \[\begin{aligned} \hat{e}_r&: \qquad F_r - mg = ma_r \\ \hat{e}_\theta&: \qquad F_\theta = ma_\theta \end{aligned}\]

We can now substitute in the acceleration in polar found here: \[\begin{aligned} \vec{a} &= (\ddot{r} - r\dot\theta^2)\hat{e}_r + (r\ddot\theta + 2\dot{r}\dot\theta)\hat{e}_\theta \\ &= a_r \hat{e}_r + a_\theta \hat{e}_\theta \end{aligned}\]

Substituting: \[\begin{aligned} F_r - mg &= m\ddot{r} - mr\dot\theta^2 \\ F_\theta &= mr\ddot\theta + 2m\dot{r}\dot\theta \end{aligned}\]

Since $r$ is constant, $\dot r = 0$ and $\ddot r = 0$. Therefore: \[\begin{aligned} F_r - mg &= - mr\dot\theta^2 \\ F_\theta &= mr\ddot\theta \end{aligned}\]

There's a reason choosing a coordinate system is important when solving kinetics problems. We can verify that $F_\theta = mr\ddot\theta$ is true using Newton's 2nd Law in rotation. However, we have to choose an appropriate axis to apply it about, since the choice of an axis matters. We will choose point $O$ to be our reference point. Using that: \[\begin{aligned} \sum \vec{M}_O = I_O\vec{\alpha} \end{aligned}\]

where $I_O = mr^2$ (see here), and $\alpha = \ddot\theta$. Now the external moment about point $O$ is as follows: \[\begin{aligned} \vec{M}_O &= \vec{r} \times \vec{F} + \vec{r} \times m\vec{g} \\ &= r\hat{e}_r \times (F_r \hat{e}_r + F_\theta \hat{e}_\theta) + r\hat{e}_r \times (-mg \hat{e}_r) \\ &= F_r r (\hat{e}_r \times \hat{e}_r) + F_\theta r (\hat{e}_r \times \hat{e}_\theta) - mgr (\hat{e}_r \times \hat{e}_r) \\ &= F_\theta r \, \hat{k} \end{aligned}\]

Substituting everything: \[\begin{aligned} F_\theta r &= mr^2 \ddot\theta \\ F_\theta &= mr \ddot\theta \end{aligned}\]

Precisely as we obtained from linear Newton's 2nd Law.

Rigid bodies

We will extend our understanding of drawing free-body diagrams on a single mass point to a body of continuously distributed mass, or rigid bodies. To solve the kinetics problem of rigid bodies, we will utilize Euler's equations .We will continue to call the equations Newton's equations, as they are simply an extension of Newton's equations from point masses to rigid bodies. Since rigid bodies can rotate about their own axis (namely, center of mass), Euler's 2nd Law is often used to compute angular quantities alongside the rigid body equations.

Example Problem: Rolling without slipping kinetics

A circular rigid body of mass $m$ and radius $r$ is rolling without slipping on a flat 2D surface under the influence of gravity $\vec{g}$. Another force $\vec{F}$ is applied at the top of the disk at point $P$ as shown. Draw the free-body diagram on the body and write down the corresponding Newton's 2nd Law for the body.

We have four forces: gravity at point $C$, normal force from the ground at point $Q$, force $\vec{F}$ at point $P$, and friction at point $Q$. Therefore, the free-body diagram is the following:

Now, we can proceed with Newton's 2nd Law. Recall Newton's 2nd Law for rigid bodies is about the center of mass. Therefore: \[\begin{aligned} \sum \vec{F}_{ext} &= m\vec{a}_C \\ m\vec{g} + \vec{F}_s + \vec{N} + \vec{F} &= m\vec{a}_C \end{aligned}\]

Applying Newton's 2nd Law in rotation (Euler's 2nd Law). We will choose the center of mass as the axis of rotation: \[\begin{aligned} \sum \vec{M}_{ext} &= I_C \, \vec \alpha \\ \vec{r}_{CP} \times \vec{F} + \vec{r}_{CQ} \times \vec{F}_s &= I_C \, \vec \alpha \end{aligned}\]


Trusses are structures made up of long, slim members connected at their ends. In this section, we will analyze a simplified version of planar trusses called simple trusses, which consist of two-force members connected by frictionless joints/pins.

In complicated trusses, there are often what is known as zero-force members, which are members that carry no load or force. The rules for identifying zero-force members are the following:

Rules for identifying zero-force members.

  • If two non-collinear members meet at a joint with no loading, then both members are zero-force members.
  • If three forces meet at a joint, and two of them are collinear, then the third is a zero-force member.

These rules will help us simplify trusses and solve for the internal forces in the desired members.

Drawing free-body diagrams on trusses is simple. As mentioned in the beginning, we will choose the entire truss as our system and draw the free-body diagram to determine the reaction forces on pinned/roller joints. Note we will need more equations and free-body diagrams on individual truss members to fully determine the reaction forces. See statics reference pages on how that is done. The following is a diagram showcasing eliminating zero-force members as well as the free-body diagram on the truss.


Deformable bodies and internal forces

Neither point masses nor rigid bodies can physically exist, as no body can really be a single point with no extent, and no extended body can be exactly rigid. Despite this, these are very useful models for mechanics and dynamics.

In the instance of rigid bodies, there must be internal forces keeping those bodies together. For example, internal forces in truss members.

The following truss showcases the internal forces between truss members.

These internal forces between truss members (action-reaction pairs) are the forces present in two-force members that were discussed earlier. To solve for those forces, see statics for method of joints and method of sections.

However now, imagine if we the member at an arbitrary point along its length. To maintain equlibrium, there must be forces at the cut, that are equal and opposite to the external forces (note those external forces are actually the action-reaction pairs, external to the member, but internal to the entire truss).

Click and drag the point to move it along the length of the rod

Force: tension compression

If we now examine an L-shaped two-force member, and cut at an arbitrary point along its length, we will find that the internal force shown earlier in the horizontal rod is not enough to maintain equilibrium, as the two rigid-bodies resulting from the cut can also now rotate. Therefore, there must be an internal moment that prevents the rotation. This also makes sense as two-dimensional rigid bodies have 3 degrees of freedom. Two in translation, and one in rotation.

Click and drag the point to move it along the length of the rod

Force: tension compression