# A possible trick question involving d'Alemberts principle [closed]

A few days ago I encountered this question from an old dynamics exam and it really got me thinking. But so far I could not definitely answer it on my own. I am hoping that you all can give me some more insight into this.

## The Question

Given the following truss consisting of three massless pendulum rods $$S_1$$, $$S_2$$ and $$S_3$$, a beam of mass $$m_B$$ and a pointmass $$M$$. Gravity acts on the system.

• Find the tension in $$S_1$$ using the principle of d'Alembert!

## My thoughts

First of: note that this system is static, since it cannot move in any direction (assuming all beams - especially $$S_2$$ - are rigid). So even though it was in an dynamics exam, it is actually a problem involving statics.

But that is not my point. The question specificly asks to find $$S_1$$ using d'Alemberts principle. Which is kind of odd, considering the fact that the big leap forward that came with this principle is the fact, that you can safely ignore all constraint forces if you want to describe the dynamics of a system.

In the framework of d'Alemberts principle, all we know about $$S_1$$ is the fact that the work done by its possible virtual displacements is equal to zero. (Or is it $$\sum_{i=1}^{3} \vec S_i \cdot \delta \vec r = 0$$? I am not entirely sure.)

$$\vec S_1 \cdot \delta \vec r = 0$$

All we know about $$\vec S_1$$ is the fact, that if we would try to rotate the first pendulum by $$\varphi$$ it only has a $$y$$-component, since we actually cannot rotate the pendulum. (The cartesian coordinate system is placed in the top left support)

$$\vec S_1 = \begin{bmatrix} S_1 \sin(\varphi) \\ - S_1 \cos(\varphi) \end{bmatrix} = \begin{bmatrix} 0 \\ - S_1 \end{bmatrix}$$

Describing $$M+m_B$$ in their common center of mass using $$\varphi$$ as out generalised coordinate we get for $$\delta \vec r$$:

$$\vec r = \begin{bmatrix} l\sin(\varphi) + x_S \\ -l\cos(\varphi) \end{bmatrix}$$

$$\delta \vec r = \frac{\partial \vec r}{\partial \varphi}\delta \varphi = \begin{bmatrix} l\cos(\varphi) \\ l\sin(\varphi) \end{bmatrix} \delta \varphi$$

This leads to the following equation which unsurprisingly holds true since $$\varphi=0$$.

$$\vec S_1 \cdot \delta \vec r = -S_1 l \delta \varphi \sin(\varphi) = 0$$

But what now? Determining $$S_1$$ from it in terms of $$L$$, $$M$$, $$m_B$$ and $$g$$ is not possible. Or at least I don't see a way. And why should there be one? D'Alemberts principle eliminates the need to describe constraint forces. So why should it be possible to calculate them in the first place? Do I miss something, or is this indeed a "trick" question?

• Could you clarify on what you are doing? Because the d'Alembert principle states that the sum of virtual works (here, moments times corresponding angles and forces times corresponding displacements) = 0. So, say we assign M_1,M_2, M_3 moments to the pins 1,2, and 3. The virtual work from these moments all depend on the same $\phi$ angle that you defined. These will be equated such that $\sum M_i = 0$. This is the only virtual work that consists of the $\phi$ d.o.f. Other moments depend on the angle $\theta$ between $S_1$ and the main bar, this equation gives you $S_3$. Commented Mar 9, 2021 at 0:26
• [...] You can then solve for $S_1$ from $\delta y$ virtual works.This essentially brings you to Newton's formulation. Commented Mar 9, 2021 at 0:27

I think that the idea is to use D'Alembert to find $$S_2$$. Because only after it is known, $$S_1$$ and $$S_3$$ can be determined by balancing the forces and the distances in the beam.
• The question may not be perfectly framed, but one can make reasonable assumptions that are stated. For one, the idealization of rods implies that the system is in equilibrium (since we know the beam cannot move without snapping one of the rigid rods). The tension in $S_2$ has a horizontal component; this must be balanced by something that is not $S_1$ or $S_3$. Thus, we have constraints on our system that do not permit arbitrary tensions, under the assumption that the system is physically reasonable and in equilibrium. Commented Mar 8, 2021 at 19:03
• If your start by the asumption that the system is static and in equiblibrium (since it cannot move and is only under gravitational load) and you apply $\sum F_x=0$, $\sum F_y=0$ and $\sum M=0$ you can show that $S_2$ is indeed equal to $0$. Using these three conditions you can also arrive easily at a nice and determined answer for $S_1$ and $S_3$. But this was specificly not the question. Commented Mar 8, 2021 at 19:27