<jupyter_start><jupyter_text>Free body diagram> <NAME>  > Laboratory of Biomechanics and Motor Control ([http://demotu.org/](http://demotu.org/))  > Federal University of ABC, Brazil Figures from  De motu animalium by <NAME> (1608-1679), the father of biomechanics, depicting a static analysis of forces acting on the human body.In the mechanical modeling of an inanimate or living system, composed by one or more bodies (bodies as units that are mechanically isolated according to the question one is trying to answer), it is convenient to isolate each body (be they originally interconnected or not) and identify each force and moment of force (torque) that act on this body in order to apply the laws of mechanics.**The free body diagram (FBD) of a mechanical system or model is the representation in a diagram of all forces and moments of force acting on each body, isolated from the rest of the system.**  The term free means that each body, which maybe was part of a connected system, is represented as isolated (free) and any existent contact force is represented in the diagram as forces (action and reaction) acting on the formerly connected bodies. Then, the laws of mechanics are applied on each body, and the unknown movement, force or moment of force can be found if the system of equations is determined (the number of unknown variables can not be greater than the number of equations for each body).How exactly a FBD is drawn for a mechanical model of something is dependent on what one is trying to find. For example, the air resistance might be neglected or not when modeling the movement of an object and the number of parts the system is divided is dependent on what is needed to know about the model.  The use of FBD is very common in biomechanics; a typical use is to use the FBD in order to determine the forces and torques on the ankle, knee, and hip joints of the lower limb (foot, leg, and thigh) during locomotion, and the FBD can be applied to any problem where the laws of mechanics are needed to solve a problem.  Equilibrium conditionsIn a static situation, the following equilibrium conditions, derived from Newton-Euler's equations for linear and rotational motions must be satisfied for each body under analysis using FBD:$$ \begin{array}{l l}\sum \mathbf{F} = 0 \\\\\sum \mathbf{M} = 0\end{array} $$That is, the **vectorial** sum of all forces and moments of force acting on the body must be zero hence the body is not moving or rotating.   And this must hold in all directions if we use the Cartesian coordinate system because the movement is independent in the orthogonal directions:$$ \sum \mathbf{F} = 0 \;\;\;\implies\;\;\; \begin{array}{l l}\sum F_x = 0 \\\sum F_y = 0 \\\sum F_z = 0\end{array} $$$$ \sum \mathbf{M} = 0 \;\;\implies\;\;\; \begin{array}{l l}\sum M_x = 0 \\\sum M_y = 0 \\\sum M_z = 0\end{array} $$Although many forces and moments of force can actuate on a rigid body, their effects in terms of motion, translation and rotation of the body, are the same of a single force acting on the body center of gravity and a couple of antiparallel forces (or simply, a force couple) that generates a moment of force. This moment of force has the particularity that is not around a fixed axis of rotation and because of that it is also called free moment or pure moment.  The next figure illustrates this principle:  Figure. Any system of forces applied to a rigid body can be reduced to a resultant force applied on the center of mass of the body plus a force couple.Based on the concept of force couple, we can make a distinction between torque and moment of force: torque is the moment of force caused by a force couple. We can also refer to torque as the moment of a couple. This distinction is more common in Engineering Mechanics.  The equilibrium conditions and the fact that a multiple force system can be reduced to a simpler system will be useful to guide the drawing of a FBD. Other important principles for a mechanical analysis are the principle of moments and the principle of transmissibility: Varignon's Theorem (Principle of moments)> *The moment of a force about a point is equal to the sum of moments of the components of the force about the same point.*   Note that the components of the force don't need to be orthogonal. Principle of transmissibility> *For rigid bodies with no deformation, an external force can be applied at any point on its line of action without changing the resultant effect of the force.***Example** (From Meriam 1997). For the figure below, calculate the magnitude of the moment about the base point *O* of the 600-N force in five different ways[.](http://ebm.ufabc.edu.br/wp-content/uploads/2013/02/torque.png)Figure. Can you calculate the torque of the force above by five different ways?One way:<jupyter_code>import numpy as np
M = np.cross([2,4,0], [600*np.cos(40*np.pi/180),
                       -600*np.sin(40*np.pi/180),0])
print("The magnitude of the moment of force is: %d Nm"
      %np.around(np.linalg.norm(M)))<jupyter_output>The magnitude of the moment of force is: 2610 Nm<jupyter_text>Let's start using the FBD to solve simple problems in mechanics. Example 1: Ball resting on the ground  Our interest is to draw the FBD for the ball:  Figure. Free-body diagram of a ball resting on the ground.W is the weight force due to the gravitational forces which act on all particles of the ball but the effects of the weight force on each particle of the ball are the same if we draw only one force (the ball weight) acting on the center of gravity of the ball.   N is the normal force, it is perpendicular to the surface of contact and prevents the ball from penetrating the surface.  The vector representing a force must be drawn with its origin at the point of application of the force. However, for sake of clarity one might draw the contact forces outside the body with the tip of the vector at the contact region (note that in the figure above, we had to draw small vectors to avoid the superposition of them).   It is not necessary to draw the body in the FBD, just the vectors representing the forces would be enough.The ball also generates forces on the ground (the ball attracts the Earth with the same force but opposite direction and pushes the ground downwards), but since the question was to draw the FBD for the ball only, we don't need to care about that.   When drawing a FBD, it's not necessary to portrait the body with details; one can draw a simple line to represent the body or nothing at all, only caring to draw the forces and moments of force in the right places.From the FBD, one can now derive the equilibrium equations:   In the vertical direction (there's nothing at the horizontal direction):$$ \mathbf{N}+\mathbf{W}=0 \;\;\; \implies \;\;\; \mathbf{N}=-\mathbf{W} $$ Where $\mathbf{W}=m\mathbf{g}\;\;(g=-10 m/s^2)$.This very simple case already illustrates a common problem when drawing a FBD and deriving the equations:   Is W negative or positive? Is $\mathbf{W}$ representing a vector quantity that we don't explicitly express knowledge about its direction or we will express its direction with a sign? Will we substitute the symbol $\mathbf{W}$ by $m\mathbf{g}$ or $-m\mathbf{g}$? And in any of these cases, is $g$ equals to $10$ or $-10 m/s^2$?This problem is known as double negative and it happens when we negate twice something we wanted to express as negative, for example, $\mathbf{W}=-m\mathbf{g}$, where $g=-10 m/s^2$. Be careful and consistent with the convention adopted. If in the reference system we chose, the upward direction is positive, the value of $\mathbf{W}$ should be negative.   So, if we write the equation as $\mathbf{N}+\mathbf{W}=0$, it's because we are representing the vectors and the value of $\mathbf{W}$ is either equal to $m\mathbf{g}$ (with $g=-10 m/s^2$) or $-m\mathbf{g}$ (with $g=10 m/s^2$).    It would be also correct to write the equation as $\mathbf{N}-\mathbf{W}=0$ where $\mathbf{W}$ is equal to $m\mathbf{g}$ (with $g=10 m/s^2$), but the problem with this convention is that we are constraining the direction of the vector to only one possibility, but this might not be always true.   The best option is to draw and write in the equations in vectorial form, without their signals and let the signals appear when the numerical values for these quantities are inputted.   But it is really a matter of convention, once a convention is adopted, you should grab it with all your forces!  Example 2: Person standing stillOur interest is to draw the FBD for a person standing still:  Figure. Free-body diagram of a person standing.The FBD for the standing person above is similar to the previous one; the difference is that now there are two surfaces of contact (the two feet). And under each foot, there is an area of contact, not just a single point.   In somewhat similar to the center of gravity, it is possible to compute a center of force, representing the point of application of the resultant vertical force of the ground on the foot (and it is even possible to compute a single center of force considering both feet).   $\mathbf{N_1}$ and $\mathbf{N_2}$ are the normal forces acting on the feet and they were arbitrarily drawn acting on the middle of the foot, but the subject could stand in such a way that the center of force could be in the toes, for example.    To exactly determine where the center of force is under each foot, we need an instrument to measure these forces and the point of application of the resultant vertical force. One instrument for that, used in biomechanics, is the force plate or force platform. In biomechanics, the center of force in this context is usually called center of pressure.  Let's derive the equilibrium equation only for the forces in the vertical direction:$$ \mathbf{N_1}+\mathbf{N_2}+\mathbf{W}=0 \;\;\; \implies \;\;\; \mathbf{N_1}+\mathbf{N_2}=-\mathbf{W} $$ Where $\mathbf{W}=m\mathbf{g}\;\;(g=-10 m/s^2).$This FBD, although simple, has already a complication to be solved: if the body is at rest, we know that the magnitude of the weight is equal to the magnitude of the sum of $\mathbf{N_1}$ and $\mathbf{N_2}$, but we are unable to determine $\mathbf{N_1}$ and  $\mathbf{N_2}$ individually (the person could stand with more weight on one leg than in the other). But the total center of force between $\mathbf{N_1}$ and $\mathbf{N_2}$ must be exactly in the line of action of the weight force otherwise a moment of force will appear and the body will rotate.     Example 3: Two boxes on the ground     Our interest is to draw the FBD for both boxes:  Figure. Free-body diagram of two boxes on the ground. Now we have to consider the forces between the two boxes.   Note that the forces $\mathbf{W_1}$ e $\mathbf{W_2}$ are the gravitational forces due to Earth. The boxes also attract each other but these forces are negligible and we don't need to draw them.    The FBD for box 1 is the same as we drew for the ball before. For the box 2, as it acts on box 1 with the contact force $\mathbf{N_1}$ (the normal force) that prevents the box 1 from penetrating the surface of box 2, box 1 reacts with the same magnitude of force but in opposite direction, this is the force $\mathbf{-N_1}$.   Remember from Newton's third law that the action and reaction force act on different bodies and these forces do not cancel each other.Let's derive the equilibrium equation only for the forces in the vertical direction:   For body 1:   $$ \mathbf{N_1}+\mathbf{W_1}=0 \;\;\; \implies \;\;\; \mathbf{N_1}=-\mathbf{W_1} $$ For body 2:$$ \mathbf{N_2}+\mathbf{W_2}-\mathbf{N_1}=0 \implies \;\;\; \mathbf{N_2}=-\mathbf{W_2}+\mathbf{N_1} \;\;\; \implies \;\;\; \mathbf{N_2}=-\mathbf{W_2}-\mathbf{W_1} $$ Where $\mathbf{W_1}=m_1\mathbf{g}\;\;\text{and}\;\;\mathbf{W_2}=m_2\mathbf{g}\;\;(g=-10 m/s^2).$Note that the magnitude of $\mathbf{N_1}$ is equal to the magnitude of $\mathbf{W_1}$ and the magnitude of $\mathbf{N_2}$ is equal to the sum of the magnitudes of $\mathbf{W_1}$ and $\mathbf{W_2}$.At the end of the first example it's written: "The best option is to draw and write in the equations, the vectors, without their signals and let the signals appear when the numerical values for these quantities are inputted."   If this is to be followed, why then the representation of $-\mathbf{N_1}$ acting on body 2 in the FBD?    The answer is that this is a different minus sign; it means that whatever is the value of this force, it should be the opposite of the normal force acting on body 1.  Example 4: One segment fixed to a base (no movement)Now we have a segment fixed to a base by a joint and we want to draw the FBD for the segment:  Figure. Free-body diagram of one segment and one joint.  On any joint one can have a force and a moment of force (but a joint with a free axis of rotation, offering no resistance to rotation, generates no moment of force, only joint forces).    For this example, we guessed the most general case, a joint with a force and a moment of force, but we may find later that one of these quantities is zero, this is ok.   And we arbitrarily chose the directions for joint force and moment of force; the actual directions don't matter now, but if we can choose any one, let's be positive!Forces:   $$ \mathbf{F} + \mathbf{W} = 0 \;\;\;\implies\;\;\;\begin{array}{l l}\mathbf{F_{x}} + 0 = 0 \;\;\;\implies\;\;\; \mathbf{F_{x}} = 0 \\\mathbf{F_{y}} + \mathbf{W} = 0 \;\;\;\implies\;\;\; \mathbf{F_{y}} = -\mathbf{W} \\\end{array} $$Moments of force around the center of gravity of the segment:   $$ \mathbf{M}+\mathbf{r_{cgp}}\times\mathbf{F}=0 \;\;\; \implies \;\;\; \mathbf{M}+\frac{\ell}{2}W=0 \;\;\; \implies \;\;\; \mathbf{M}=-\frac{\ell}{2}m\mathbf{g} $$ Remember that the direction of $\mathbf{M}$ is perpendicular to the plane of the FBD, it is in the $\mathbf{z}$ direction, because of the cross product.Where $\mathbf{r_{cgp}}$ is the position vector from the center of gravity to the proximal joint and $\mathbf{W}=m\mathbf{g}\;(g=-10 m/s^2)$.We can calculate the moment of force around any point that the condition of equilibrium must hold. In this example, we could have calculated the moment of force direct around the joint; in this case the weight causes a moment of force and the joint force does not.To find the direction of the moment of force might be difficult sometimes. If we had numbers for all these variables, we could simply do the mathematical operations and let the results indicate the direction.   For example, suppose the center of gravity is at the origin and that the bar has mass 1 kg and length 1 m.    Writing the equilibrium equation for the moments of force around the center of gravity, the moment of force at the joint is:<jupyter_code>m, l, g = 1.0, 1.0, -10.0 # m = 1kg, l = 1m, g = -10 m/s2
r = [-l/2-0, 0, 0] # note how the position vector from origin to F is created: tip minus tail
F = [0, -m*g, 0]   # this is the joint force not the gravitational force
M = -np.cross( r, F )
print("The moment of force at the joint is (in Nm):", M)<jupyter_output>The moment of force at the joint is (in Nm): [-0. -0.  5.]<jupyter_text>The moment of force $\mathbf{M}$ is positive (counterclockwise direction) to balance the negative moment of force due to the gravitational force.   Note that $\mathbf{M}$ is a vector with three components, where only the third component, in the $\mathbf{z}$ direction, is nonzero (because $\mathbf{r}$ and $\mathbf{F}$ where both in the plane $\mathbf{xy}$).   And writing the equilibrium equation for the moments of force around the joint, the moment of force at the joint is:<jupyter_code>m, l, g = 1.0, 1.0, -10.0 # m = 1kg, l = 1m, g = -10 m/s2
r = [l/2-0, 0, 0] # note how the position vector from origin to W is created: tip minus tail
F = [0, m*g, 0]   # this is the gravitational force
M = -np.cross( r, F )
print("The moment of force at the joint is (in Nm):", M)<jupyter_output>The moment of force at the joint is (in Nm): [-0. -0.  5.]
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