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Consider the above scenario: In the subsequent motion, we need to find the work done by tension on the (trolley + mass) system.

Solution: Suppose at an instant, the velocity of the trolley (and hence that of point $A$) is $\vec{V_{a}}$, and the velocity of the point mass is $\vec{V_{b}}$. Then the power delivered by tension to the system is $\vec{T}\cdot(\vec{V_{b}}-\vec{V_{a}}$). Now, tension is always directed along the string, and, the velocity of the mass relative to point $A$, (i.e. $\vec{V_{b}}-\vec{V_{a}}$) is always perpendicular to the string. So the dot product is zero, i.e., the power delivered to the string is zero at all times. Hence, the tension does no work on the system.

Consider another scenario: https://physics.stackexchange.com/a/571564/196626. Here, the work done by the normal reaction on the system is zero.

  • In situation 1, the work was zero, because the relative velocity was constrained to be perpendicular to the string. (string constraint)

  • In situation 2, the work was zero since the block was constrained to move along the wedge, (and hence perpendicular to the normal force). (contact constraint).

These two situations demonstrate what has often been the case in many classical mechanics questions: The work done by tension and normal forces on the system =0. In these two particular situations, the common thing seems to be that the tension (and normal) are constraining the elements of the system. It seems like the common link is that they are constraint forces. My question attempts at a generalization:

Can we claim (in general) that the work done by constraint forces on a system is always zero?

I somehow feel that the reason somehow lies in the meaning of the term "constraint" itself, but it's just a feeling.

  • $\begingroup$ Very relevant (essetially answering the question): D'Alembert's principle $\endgroup$
    – user258881
    Commented Aug 14, 2020 at 18:20
  • $\begingroup$ @FakeMod also, I dont see how that wikipedia page is relevant: The expression simply "omits" the constraint forces. It seems to me that the principle only involves non constraint forces, and in that light, i dont see how it answers the question... $\endgroup$
    – satan 29
    Commented Aug 14, 2020 at 18:46
  • 1
    $\begingroup$ See physics.stackexchange.com/q/8453 $\endgroup$
    – user258881
    Commented Aug 14, 2020 at 18:53

4 Answers 4


Can we claim (in general) that the work done by constraint forces on a system is always zero?

We can't, it's not true for a large class of constraints.

Counterexample 1. Non-holonomic constraints.

These are constraints that depend on more than just positions and time, such as $$ f(x_i, \dot{x_i}, t) ~=~ 0 $$ So here is a simple counterexample, suppose the constraint is $$ x - \dot{x}~=~0, $$ for a particle in one dimension with no other forces present. Then it will accelerate away from the origin, which means the constraint force is doing work.

Counterexample 2. Time-dependent constraints.

Suppose for example the constraint is $$ x_1 - x_2 = t^2 $$ for two particles in one dimension with no other forces present. Then the particles will accelerate away from each other, so the constraint force is doing work.

Holonomic time-independent constraints don't do work.

These are "normal" constraints, that only depend on positions, such as massless rigid rods enforcing distances, frictionless inclined planes, etc. We can see that they do zero work in several ways.

One way is to note that we can take the constraint to be enforced by a conservative force, with potential 0 for allowed positions and infinity for disallowed ones. Then if the system moves from one allowed configuration to another, the work done by that force is equal to minus the change in potential, which is zero.


According to The Feynman Lectures on Physics :

In motion with a fixed frictionless constraint, no work is done by the constraint because the forces of constraint are always at right angles to the motion. By the “forces of constraint” we mean those forces which are applied to the object directly by the constraint itself—the contact force with the track, or the tension in the string.

So I guess that eliminates all friction less constraint forces.(When considered from an appropriate frame of reference) You can still ask why is the constraint force perpendicular to the the body on which it acts and its because the surface is 'friction-less'. Because friction is essentially the tangential component of the constraint force acting on the body and when we say that friction is absent we are saying that the tangential component is zero.

Now in case there is friction(usual in real world), if it dissipates energy in the form of heat etc. then the work done by the constraint force is clearly non-zero. This is what usually happens in real-life scenarios(eg. Hinge forces which are a form of constraint). However even if the friction does not dissipate heat(improbable) the work done by it will turn out to be zero if we consider a the whole system on which the force is acting.This is just a consequence of Newtons Third Law. But usually this is not very clear.For example when we consider a block sliding on a surface we take the ground to be a body with infinite mass and disregard the work done on it.


Work done is defined by $$W = Fd$$

So as you can see, if a force does not cause any displacement, the work done by that force is 0. The definition of a constraint force is that it is a force that determines the path of an object when another force is applied, but doesn't cause displacement. However since it is determining the path this means that it can change the direction of the displacement, even if it can't cause it. Displacement involves direction and distance. So that means for a constraint force to do 0 work, it needs to act perpendicular to the direction of the displacement.

In terms of vectors, what is causing the displacement and therefore the work is the net force that is parallel to the direction of displacement. So what it comes down to then is this: does the constraint force always work perpendicular to the direction of the displacement? The answer is: not necessarily. It depends in which direction you consider the work to be done. If the constraint has a component parallel to the direction of the work done, then it is doing non 0 work, which can be both positive or negative. If it is perpendicular to the direction of the work done, then it is 0.

For example, in your first scenario, which I can't really understand from the diagram, but lets consider a ball hanging from a string. If you push the ball in a direction parallel to the ground, the constraint force will make it go in a circular path. Now the question is, in which direction will you consider the displacement? If you consider it to be the circular path, then the tension will always be perpendicular to the displacement and therefore work done by constraint force will be 0. However if you notice, the circular path is a constant change in direction, and therefore not displacement in a fixed direction. So then let's consider another way of looking at the same scenario. What if we considered displacement to be parallel to the primary force, which is perpendicular to the string and parallel to the ground. Then after you push the ball, when the ball is displaced along the circular path, its path is no longer parallel to the direction of the primary force, the tension of the string will make it move up slightly, and it will also have a slight component parallel but opposite to the direction of the primary force. The tension then will do some work, and the value will change depending on which direction you consider the displacement and how far you push the ball. If you consider the displacement to be ‘up’ the tension will have a positive work done, and if you consider it to be parallel to the primary force, it will be negative. Furthermore we are not even considering the elasticity of the string, which can also factored into the work done by constraint forces. So the answer is that the work done by a constraint force is usually 0, but not necessarily, depending on the direction of the displacement you choose. But calculating the work done by considering the direction of the displacement to be perpendicular to the constraint force is usually the most meaningful and useful value, and so that is what is usually done.


According to D'Alembert's principle the constraints forces doesn't affect the motion of the system, to see it , you have to generate the equation of motion with generalized coordinate.

your example

The position vector to the mass

$$\vec R_m= \left[ \begin {array}{c} x_{{A}} \left( t \right) +L\sin \left( \varphi \right) \\L\cos \left( \varphi \right) \end {array} \right] $$

the velocity is $$\vec v=\vec{\dot{R}}_m=\underbrace{\frac{\partial \vec R}{\partial \varphi}}_{J}\,\dot{\varphi}+\frac{\partial }{\partial t}\vec{R}$$

$$J= L\,\left[ \begin {array}{c} \cos \left( \varphi \right) \\ -\sin \left( \varphi \right) \end {array} \right] $$

and the tension force is:

$$\vec T=T\,\vec{\hat e}_L=T\, \begin{bmatrix} \sin(\varphi) \\ \cos(\varphi) \\ \end{bmatrix}$$

thus D'Alembert's principle.


the Newton EOM's

$$m\,J^T\,J\,\ddot{\varphi}=J^T\left(\vec F_u-m\,\frac{\partial ^2}{\partial t^2}\vec R\right)+\underbrace{J^T\,\vec T}_{=0}$$ where

$$\vec F_u= \begin{bmatrix} 0 \\ -m\,g \\ \end{bmatrix}$$

$${L}^{2}\ddot\varphi +g\,L\sin \left( \varphi \right) + \left( { \frac {d^{2}}{d{t}^{2}}}x_{{A}} \left(t \right) \right) L\cos \left( \varphi \right) =0$$

enter image description here

the free body diagram show you that the constraint force $F_{cx}=T\,\sin(\varphi)$ is a internal force, only external force (applied force ) can do work


we can claim that the work done by constraint forces on a system is always zero?


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