# Why the total virtual work done by forces from constraints vanishes? (Perpendicularity of two or more particles)

My mechanics book claims that the total force on the $$i$$-th particle is $$F_i=K_i+Z_i \tag{2.5}$$where $$Z_i$$ is the force due to constraints and $$K_i$$ the real, dynamic force. Then, the book states without proof that $$\sum Z_i\cdot \delta \mathbf r_i=0.$$ The only reason why it is true that I can think of is that the constraining force must be "perpendicular" (or normal) to the surface on which the particles can move. However, the notion of perpendicularity only makes sense for one-particle systems restricted to a surface. It does not make sense, for example, for the system in which two particles are connected by a light, rigid rod.

This generalisation of perpendicularity to more than two particles does not seem clear to me. Is there any mathematically formal way to talk about this notion of "perpendicularity"? Or is there a better explanation for why the total work done by forces from constraints vanishes?

References:

1. Florian Scheck, Mechanics, p. 92.

Scheck is assuming that the principle of virtual work (PVW) holds, i.e. that the constraint forces do no virtual work. This in turn leads to d'Alembert's principle, cf. e.g. this & this Phys.SE posts.

One can prove the PVW for various classes of constraint forces. An important class is the constraint forces in a rigid body, cf. e.g. this & this Phys.SE posts.

I think you are referring to the D'Alembert principle, which excatly states that the constraining forces do not provide any contributions to the work, namely the virtual work vanishes

$$\sum Z_i \cdot \delta r_i.$$

And you are right that these forces are perpendicular to the surface on which the motion is constrained.

You can understand this in more detail when recognizing that a motion, in contrast to a freely moving particle in 3-dimensional space, can in principle be constraint. This constrained motion is induced out of constraining forces. You already made the nice example where the motion is constraint on a particular surface. When you now want to calculate the Newtonian equations of motion for your particle you will run into trouble since the constraining forces are often not known. You can either introduce generalized coordinates (this will lead to the Lagrange equations) or you can use D'Alemberts pricniple.

For holonomic constraints the constraining forces do not affect the motion on a surface $$g(r,t)= 0$$, thus these constraining forces are perpendicular to the surface defined by $$g(r,t)= 0$$.

• "a motion can constraint which result out of contraining forces" What does that mean? – Ma Joad Nov 25 '19 at 14:34
• Also, what if it is not a surface? e.g. a rod – Ma Joad Nov 25 '19 at 14:35
• I have edited this sentence in order to make it more clear. In summary: the constrained motion is induced by constraining forces. – Max Stammer Nov 25 '19 at 14:45
• Then it is still a surface, namely, the surface of the rod when the particles motion is constrained on the surface of this rod. Then you have to define your surface accordingly. – Max Stammer Nov 25 '19 at 14:46
• Okay. What about the case of two or more constraints? That cannot be seen as a surface... – Ma Joad Nov 25 '19 at 22:55

The equation of motion are:

$$m_1\,\ddot{x}_1=F_1+Z$$ $$m_2\,\ddot{x}_2=F_2-Z$$

or

$$\begin{bmatrix} m_1 & 0 \\ 0 & m_2 \\ \end{bmatrix}\,\underbrace{\begin{bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \\ \end{bmatrix}}_{\vec{\ddot{y}}}=\begin{bmatrix} F_1 \\ F_2 \\ \end{bmatrix}+\underbrace{\begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}}_{C_Z}\,Z\tag 1$$ the constraint equation

$$x_2-x_1=0\quad \Rightarrow \dot{x}_2=\dot{x}_1\tag 2$$

so $$x_1=q_1$$ is the generalized coordinate thus $$\vec{\dot{y}}=J\,\dot{q}_1$$

where

$$J=\begin{bmatrix} 1 \\ 1 \\ \end{bmatrix}$$

and eq. (1)


$$\begin{bmatrix} m_1 & 0 \\ 0 & m_2 \\ \end{bmatrix}\,J\,\ddot{q}=\begin{bmatrix} F_1 \\ F_2 \\ \end{bmatrix}+\underbrace{\begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}}_{C_Z}\,Z\tag 3$$

to eliminate the constraint force $$Z$$ we multiply eq(3) from the left with $$J^T$$

$$J^T\,C_Z=\begin{bmatrix} 1 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}=0$$

or

$$\delta x_1\,Z-\delta x_2\,Z=0$$

this is the Virtual work principle

we obtain the equations of motion

$$J^T\,M\,J\,\vec{\ddot{q}_1}=J^T\,\vec{F}$$

• Yes, you have derived that D'Alembert's principle is true in this case. But what about other systems? How can you say it is true for all systems with holonomic constraints? – Ma Joad Nov 25 '19 at 22:51
• You can use D’Alembert principle to eliminate the constraint forces for all systems with holonomic constraints ,by obtain the Jacobi Matrix J ,this is always possible out of the constraint equations – Eli Nov 26 '19 at 8:08
1. You are right. There is no guarantee that forces of constraint will do no work. For instance, if there is dissipation, there is an irreversible loss of mechanical energy of the system into heat. An example would be surface friction between solid surfaces, which in the general case cannot be incorporated into the Lagrangian formalism because there is no way to define a corresponding potential.

However, if the body is rolling then the point of contact always has zero velocity, and work done by forces of friction is zero $$-$$ note in this case, the frictional force is not zero, but the power generated due to it is. This is why one can use the usual Lagrangian formalism for cases wherein rolling is guaranteed.

$$\big[$$An example where you can incorporate dissipation is velocity dependent friction, and that's through the use of the Rayleigh Dissipation Function (cf. Goldstein, chapter 1); however, this is not the usual Lagrangian Mechanics anymore.$$\big]$$

1. So, in classical Lagrangian mechanics, the correct way to consider whether forces of constraint are doing any work or not is to decide if there is dissipation due to constraint forces. As you stated yourself $$-$$ the normal force doesn't do any work, only the parallel component does.

2. In the case of rigid body, one assumes that the internal energy of the rigid body (due to various short-ranged and long-ranged forces) is independent of the orientation of the rigid body. In classical mechanics, this assumption is often argued through internal forces that are assumed to be equal and opposite, and acting along the line joining any two parts. Due to the rigidity constraint $$-$$ which only allows perpendicular relative motion between any two parts of the rigid body $$-$$ the forces of constraint do no work. This explains the situation with the light rigid rod glued with two masses in your problem.

3. Imagine a non-holonomic constraint, for instance, a particle that is trapped in the interior of a fixed, undeformable, thin shell. The forces of constraint can do work when the particle collides with the walls of the shell. If the collisions cause loss of energy, the forces of constraint do work (during the course of a collision).

4. Imagine dropping a spinning ball vertically down on the floor $$-$$ when the ball rebounds, in a real scenario, its angular velocity changes. This is again an example of a parallel component doing work. On the other hand, if the ground/ball are deformable or non-rigid, then the constraint itself is not exact. Here you've all sorts of dissipation effects, as the oscillations induced due to collisions get damped and eventually get dissipated as heat & sound as the body relaxes, eg. a chain that is wrapped around near a hole in a table and slips vertically downwards.

5. There can be any number of interesting/mysterious phenomena if the bodies are charged (imagine insulated charged bodies for instance). The charges interact through the electromagnetic field that has its own dynamics. The usual assumptions of Newtonian mechanics break down in such cases, as the electromagnetic field can carry away energy and momentum.