This is the question

This is the problem, basically. Now my question is about the approach. Mine, and my teacher's. I would like some help as to which one is correct...


As the whole chain is in equilibrium, the horizontal forces must be balanced. Therefore, tension in the thread must be equal to the total horizontal Normal Reaction by the fixed sphere.

Now, to calculate the $dN$ for a mass $dm$ taken at an angle of $\theta$ from the horizontal is $dm g \sin(\theta)$ and its component in horizontal direction is $dN \cos\theta$ after the integration you get (finally) $\ T = \frac{\lambda r g}{2}$

My Teacher's Approach:

The differential increase in the tension in the chain due to the mass $dm $ should be

$dmg \cos\theta$.

Now integrate this increase in tension to get the net increase in the tension from the point where the chain looses contact to the point where it is attached to the thread. So you will get option A.

According to me, the problem lies in the fact that you are integrating the $dT$ as a scalar, whereas it is a vector, but my teacher does not agree. He says that we are focusing on the magnitude, and the fact that its a vector is compensated because its magnitude is a function in $\theta$. But I think that the direction of the infinite $dT$s is ignored while integrating.

  • $\begingroup$ Good homework question. $\endgroup$ – BMS Feb 25 '14 at 17:14
  • $\begingroup$ The teacher's approach seems more straightforward to me, it leads directly to the differential equation $dT=\lambda gR\cos\theta d\theta$, which can be very easily integrated. In the limit the vector $dT$ points in precisely the tangential direction, so directionality is accounted for and preserved even though we are working with a scalar equation. The differential element must be in both tangential and normal equilibrium, but the tangential differential equation is by far the easiest to work with IMHO. $\endgroup$ – Bryson S. Sep 18 '14 at 1:20

$\def\d{{\,\prime}}$ $\def\nR{{\mathbb{R}}}$ $\def\l{\left}\def\r{\right}$ $\def\vf{{\vec{f}}}$ $\def\vc{{\vec{c}}}$ $\def\vr{{\vec{r}}}$ $\def\ve{{\vec{e}}}$ $\def\vF{{\vec{F}}}$ $\def\rmF{{\mathrm{F}}}$ $\def\rmC{{\mathrm{C}}}$ $\def\rmN{{\mathrm{N}}}$ $\def\rmT{{\mathrm{T}}}$ $\def\nN{{\mathbb{N}}}$

What your teacher tried to explain is most easily described with the help of the principle of virtual work.

The sphere poses a constraint to the chain. It defines the shape of the chain. The chain itself implicitly represents also a constraint (it does not lengthen under pull forces). The sphere does not move therefore it does not exert any work on the chain (we neglect friction here).

Let the curve $\vf:\nR\rightarrow\nR^3$ describe the shape of the constraint in dependence on the curve length. We have extended the domain of this function to full $\nR$ to avoid problems when the chain moves.

For our example we can use $$ \vf(s) = \begin{cases} (R,s) & @ s < 0\\ \l(R\cos\l(\frac sR\r),R\sin\l(\frac sR\r)\r) &@ s\in\l[0,\frac\pi2R\r]\\ \l(\frac{\pi}2R-s,R\r) &@ s > \frac\pi2R \end{cases} $$ Note, that we do not really need the actual representation. It is just added that we know about what we speak. The curve $\vf$ is represented with red color in the following picture together with its parameterization through $s$. The red curve is a graphical representation of $\vf$.

For simplicity we do not care about the bending of the chain if it slides off the sphere on the thread side just as if it were also supported by some shelf there.

If $s$ is the positioning parameter of the chain in space and $\xi\in [0,l]$ is the actual position of the chain element on the chain the chain can be described by $$ \vr(s,\xi) = \vf(s-\xi). $$ For $\xi=0$ the point $\vr(s,0)=\vf(s)$ is the beginning of the chain. For $\xi=l$ the point $\vr(s,l)=\vf(s-l)$ is the end of the chain. Values of $\xi$ in between of $0$ and $l$ address other points $\vr(s,\xi)$ on the chain. If we enlarge $s$ the chain glides up along the curve $\vf$ if we reduce $s$ the chain glides down along the curve $\vf$. To see this you can follow the chain beginning at $\vr(s,0)=\vf(s)$.

The placement of the chain from your picture results for $s=\frac\pi2 R$.

Excursion into mechanics with constraints (Principle of Work):

At first consider a single particle constraint to some curve. In reality the particle could be a very small and heavy locomotive on a curved rail track in the mountains (up and down).

The curve is given by a smooth function $\vf:\nR\rightarrow\nR^3$ in dependence of some parameter $s$ (e.g. the path length in one direction starting from some predefined point as for our above example).

The particle is affected by some free force $\vF_\rmF$ which would also exists if the constraint wouldn't be there. Furthermore, the particle is affected by the constraint force $\vF_\rmC$ which keeps the particle on the curve.

The constraint force counteracts the component $\vF_\rmN$ of the free force the normal direction of the curve.

enter image description here

The resulting force is the tangent component $\vF_\rmT$ of the free force.

The particle is in equilibrium if the resulting tangent force $\vF_\rmT=\vF_\rmF+\vF_\rmC$ is zero.

The nice thing of the principle of virtual work is that we do not need to calculate explicitly the normal force but we can restrict the equations to the tangent direction.

The tangent direction of the curve at some location $s$ is the derivative $\vf\,'(s)$ of $\vf(s)$. We can scale the tangent direction by some parameter $\delta s$. In this way we get a straight line parameterized by $\delta s$ which is tangent to the curve at the point $s$. Note, that close to $s$ the curve and the tangent line look very similar (inclusively the scale division for $s$ and $\delta s$).

enter image description here

For each given value $\delta s$ the vector $\delta\vr:=\vf\,'(s)\cdot \delta s$ is a tangent vector on the curve at $s$. The tangent quantities such as $\delta \vr$ and $\delta s$ are often denoted as virtual in physics. And if physicans talk about small deviations $\delta s$ they actually mean such tangent quantities.

The force balance in the equilibrium point is $$ \vec0 = \vF_\rmF + \vF_\rmC $$ If we multiply this equation by a tangent vector at the location $s$ of the particle we get $$ 0 = \delta\vr\cdot(\vF_\rmF + \vF_\rmC) = \delta\vr\cdot\vF_\rmF + \underbrace{\delta\vr\cdot\vF_\rmC}_{=0} $$ whereby the second term $\delta\vr\cdot\vF_\rmC$ is zero because the tangent vector $\delta\vr$ and the constraint force $\vF_\rmC$ are perpendicular to each other.

So our resulting equilibrium equation is $$ 0=\delta\vr\cdot \vF_\rmF = \delta s \vf\,'(s)\cdot \vF_\rmF. $$ We are only considering here the case of one degree of freedom $s$ where we can just set $\delta s=1$. (Note by the way, that if we had more parameters, e.g., $ s=( s_1, s_2)$, we would have to test for all possible tangent directions, e.g. $\delta s=(1,0)$ and $\delta s=(0,1)$.)

For $\delta s=1$ our equilibrium equation reads as $$ 0= \vf\,'(s)\cdot \vF_\rmF. $$ which is one scalar equation which can be solved for the scalar curve parameter $s$.

As I already mentioned above. The nice thing about our first application of the principle of virtual displacement is that we have eliminated the unknown constraint forces in the equation to be solved.

Now we consider a system of $n$ coupled particles with one degree of freedom $s$. All particles are constraint to $n$ curves $\vr_i = \vf_i(s)$ with $i=1,\ldots,n$.

Now, there must be additional internal constraint forces $\vF_{ij}$ between the particles else they would run on the curves $\vf_i$ independently and their positions could not be parameterized by the same parameter $s$.

Thereby, $\vF_{ij}$ is the force exerted by particle $j$ on particle $i$.

Because of actio=reactio we have $\vF_{ij} = -\vF_{ji}$. Therefore we only need the forces $\vF_{ij}$ with $i>j$ in our formulas.

Inclusively the internal constraint forces the equilibrium force balance for the $i$-th particle reads as $$ 0=\sum_{j=1}^{i-1} \vF_{ij} - \sum_{j=i+1}^n \vF_{ji} + \vF_{\rmC i} + \vF_{\rmF i}. $$ The $\vF_{\rmC i}$ are the constraint forces from the environment and $\vF_{\rmF i}$ are the free forces on the $i$-th particle as for our one-particle problem.

What are examples of internal constraint and how do the internal constraints work? A simple internal constraint would be a light staff with a heavy particle attached to each end. To show that this is not the only kind of internal constraint we can extend this example. We do not attach the particles directly but put them on two light wheels which are attached rotary to the ends of the staff such that they work together like a mechanism with one degree of freedom such that the motion of the particles is coupled and can be expressed by one parameter $s$.

enter image description here

In a difference coordinate system $\vr_j-\vr_i$ the point $\vr_j$ is constrained to the curve $\vf_j(s)-\vf_i(s)$ and we have the same situation as for the constrained single particle. The constraint force $\vF_{ji}$ is perpendicular to the tangent of the difference curve $$ (\delta\vr_j-\delta\vr_i)\perp \vF_{ji}\quad\Rightarrow\quad (\delta\vr_j-\delta\vr_i)\cdot \vF_{ji} = 0. $$ From the single-particle example we have some hope to eliminate the constraint forces by exploiting this orthogonality.

But, in the balance formula $0=\sum_{j=1}^{i-1} \vF_{ij} - \sum_{j=i+1}^n \vF_{ji} + \vF_{\rmC i} + \vF_{\rmF i}$ for the force on the $i$-th particle all internal constraint forces are mixed in. Furthermore, since we only have one parameter $s$ to determine the position of all particles we only need one formula.

For that reason let us try the sum over all the scalar products: $$ 0 = \sum_{i=1}^n \delta \vr_i \cdot \l(\sum_{j=1}^{i-1} \vF_{ij}- \sum_{j=i+1}^n \vF_{ji} + \vF_{\rmC i} + \vF_{\rmF i}\r) $$ $$ 0= \sum_{i=1}^n\sum_{j=1}^{i-1} \delta \vr_i \cdot\vF_{ij}- \sum_{i=1}^n\sum_{j=i+1}^n \delta \vr_i \cdot\vF_{ji} +\sum_{i=1}^n \delta \vr_i \cdot \vF_{\rmC i} + \sum_{i=1}^n \delta \vr_i \cdot\vF_{\rmF i} $$ From the single-particle problem we already know $\delta\vr_i\cdot \vF_{\rmC i}=0$ and there remains only the balance equation $$ 0= \sum_{i=1}^n\sum_{j=1}^{i-1} \delta \vr_i \cdot\vF_{ij}- \sum_{i=1}^n\sum_{j=i+1}^n \delta \vr_i \cdot\vF_{ji} + \sum_{i=1}^n \delta \vr_i \cdot\vF_{\rmF i} $$ Let us have a closer look onto the first two sums with the internal constraint forces $\vF_{ij}$ $$ \sum_{i=1}^n\sum_{j=1}^{i-1} \delta \vr_i \cdot\vF_{ij}- \sum_{i=1}^n\sum_{j=i+1}^n \delta \vr_i \cdot\vF_{ji} $$ For easier comparison of the terms in the two sums we exchange the names of the indexes in the second one: $$ \sum_{i=1}^n\sum_{j=1}^{i-1} \delta \vr_i \cdot\vF_{ij}- \sum_{j=1}^n \sum_{i=j+1}^n \delta \vr_j \cdot\vF_{ij} $$ That is just re-naming and does not change the value of the sum. Let us have a look onto the set of index pairs over which we have to sum in the first term and in the second one. We can do this graphically in the following diagram. There mark the locations $(i,j)$ of the index pairs which are included in the sum with points. I just draw some of these points and put a gray triangle onto the region where we would have to mark all grid points.

enter image description here

If we do the same for the second term we see that we obtain exactly the same triangle and therefore also the same set of index pairs.

If you prefer we can also check this in set-notation. The set of index pairs for the first term is: $$ \{(i,j)\in\nN^2 \mid 1\leq i\wedge i\leq n \wedge 1 \leq j\wedge j \leq i-1\} $$ Just transforming the last inequality: $$ =\{(i,j)\in\nN^2 \mid 1\leq i\wedge i\leq n \wedge 1 \leq j\wedge j+1 \leq i\} $$ Just reordering of the inequalities: $$ =\{(i,j)\in\nN^2 \mid 1 \leq j \wedge 1\leq i \wedge j+1 \leq i\wedge i\leq n\} $$ The inequalities $1\leq j$ and $j+1\leq i$ already imply $1\leq i$ therefore we can drop this inequality. Furthermore, $j+1\leq i$ implies the inequality $j\leq n$ which we can add without danger. This gives the set $$ =\{(i,j)\in\nN^2 \mid 1 \leq j \wedge j\leq n \wedge j+1 \leq i\wedge i\leq n\} $$ of index pairs for the second term.

Since the sets of index pairs for the sums are the same we can combine the summands under the same sum sign $$ \sum_{i=1}^n\sum_{j=1}^{i-1} \delta \vr_i \cdot\vF_{ij}- \sum_{j=1}^n \sum_{i=j+1}^n \delta \vr_j \cdot\vF_{ij} =\sum_{i=1}^n\sum_{j=1}^{i-1} \l( \delta \vr_i \cdot\vF_{ij}-\delta \vr_j \cdot\vF_{ij}\r) $$ and factor out the internal constraint forces $$ =\sum_{i=1}^n\sum_{j=1}^{i-1} \underbrace{\l( \delta \vr_i -\delta \vr_j \r)\cdot\vF_{ij}}_{=0} = 0 $$ Eventually the sum becomes zero with the orthogonality condition for the internal constraint forces $\l( \delta \vr_i -\delta \vr_j \r)\cdot\vF_{ij}=0$ which we discussed further above.

Exploiting this our balance equation $$ 0= \sum_{i=1}^n\sum_{j=1}^{i-1} \delta \vr_i \cdot\vF_{ij}- \sum_{i=1}^n\sum_{j=i+1}^n \delta \vr_i \cdot\vF_{ji} + \sum_{i=1}^n \delta \vr_i \cdot\vF_{\rmF i} $$ becomes just $$ 0 = \sum_{i=1}^n \delta \vr_i \cdot\vF_{\rmF i}. $$ The sum over the products of the virtual displacements $\delta\vr_i$ with the free forces $\vF_{\rmF i}$ is zero.

This formula is called the Principle of Virtual Work.

Just to get a formula usable for calculation we substitute $\delta\vr_i=\vf_i\,'(s)\cdot\delta s$ $$ 0 = \sum_{i=1}^n \delta s \vf_i\,'(s) \cdot\vF_{\rmF i} $$ and set $\delta s=1$ which gives $$ 0 = \sum_{i=1}^n \vf_i\,'(s) \cdot\vF_{\rmF i}. $$ This is one scalar formula for our one degree of freedom $s$ in the system.

If our system of particles is actually a continuum instead of discrete particles the index $i\in \{1,\ldots,n\}$ for $\vr_i(s)$ is replaced by a continuous parameter $\xi\in [0,l]$ with some length $l>0$ and we write $\vr(s,\xi)$ instead of $\vr_i(s)$.

Beside discrete forces $\vF_{\rmF i}$ which operate at certain places $\xi_{\rmF i}$ $i=1,\ldots,n_{\rmF}$ of the constrained continuum there can also act a distributed force per length $\vF_{\rmF}^\d(\xi,\vr)$ on the continuum.

To get a transition from the principle of virtual work for discrete particle systems to continuous particle systems one can start with a with a discretization of the continuum.

We dissect the continuum into $k$ sections $j=1,\ldots,k$ with length $\frac lk$. The parameter value for the $j$-th section runs from $\xi=\xi_{j-1}:=(j-1)\frac lk$ to $\xi_j:=j\frac{l}{k}$. Furthermore, we address each of the sections by some intermediate parameter $\xi^*_j$ with $\xi_{j-1}<\xi^*_j<\xi_j$. For an instance $\vr(s,\xi_j^*)$ can be the center of mass of the $j$-th section.

If the pieces are small enough, i.e., the number of pieces $k$ is high enough, a discrete particle system will be a good approximation of the system of pieces. We consider the points $\vr(s,\xi_j^*)$ at the intermediate parameter values $\xi_j^*$ as location of the $j$-th particle.

The free force caused by the distributed force per length on the $j$-th piece will be: $$ \int_{\xi_{j-1}}^{\xi_j} \vF_\rmF^\d(\vr(s,\xi))d\xi\approx \vF_\rmF^\d(\vr(s,\xi^*_j))\cdot(\xi_j-\xi_{j-1}) $$ Now we apply the principle of virtual work to this approximative particle system and obtain $$ 0 = \sum_{j=1}^{k}\delta \vr(s,\xi_j^*)\cdot \vF_\rmF^\d(\vr(s,\xi^*_j))\cdot(\xi_j-\xi_{j-1}) + \sum_{i=1}^{n_{\rmF}} \delta \vr(s,\xi_{\rmF i}) \cdot \vF_{\rmF i} $$ we expect the approximation to get better with higher with larger $k$.

For $k=1,2,\ldots$ we obtain a sequence of Riemann sums for the first sum which is converging to the integral $\int_0^l \delta\vr(s,\xi)\cdot\vF_{\rmF}^\d(s,\vr(s,\xi)) d\xi$ for $k\rightarrow\infty$.

This leads us to the principle of virtual work for a 1d-continuum with one degree of freedom $s$: $$ 0 = \int_0^l\delta\vr(s,\xi)\cdot\vF_{\rmF}^\d(s,\vr(s,\xi)) d\xi + \sum_{i=1}^{n_{\rmF}} \delta \vr(s,\xi_{\rmF i}) \cdot \vF_{\rmF i} $$ Thereby, the integral accounts for the distributed free force acting on the continuum and the sum accounts for the discrete free forces acting on it.

The application of the principle of virtual work delivers $$ F_s\delta s + \int_{\xi=0}^l \delta\vr(s,\xi)\cdot \vF_{\rm G}^\d d \xi = 0 $$ where $F_s$ is the supporting force of the chain, i.e., the thread force, $\vF_{\rm G}^\d = -\lambda g\ve_z$ is the weight force per length and $\delta\vr(\xi,s) = \frac{\partial \vr}{\partial s}(s,\xi)\delta s$ is the virtual vectorial displacement of the string.

$$ \l(F_s + \int_{\xi=0}^l \frac{\partial \vr}{\partial s}(s,\xi)\cdot \vF^\d_{\rm G} d \xi\r)\delta s = 0 $$ and if this should hold for virtual chain displacements $\delta s\neq 0$ we get the condition $$ F_s = -\int_{\xi=0}^l \frac{\partial\vr}{\partial s}(s,\xi)\cdot \vF^\d_{\rm G} d \xi $$ Note, that $\frac{\partial \vr}{\partial s}$ is just the unit vector in tangential direction of the chain. So, because of the constraint we are just integrating the tangential component $\frac{\partial \vr}{\partial s}(s,\xi)\cdot \vF^\d_{\rm G}$ of the weight force.

You could say that the constraint force of the sphere deviates the pull force in the chain.

The problem with your approach is that you are missing the deviation force of the chain in the calculation of the normal force $dN$. If you cut out a (finite) piece of chain then the ends of this piece will not have the same tangent direction because of the curvature of the sphere. If some force (the weight of the remainder of the chain) acts tangentially on the lower end of the piece it cannot be fully compensated by the tangent force at the higher end (the tangent vectors are linearly independent). The force difference must be compensated by a normal force on the shpere.

We can also just use another approach to check the result.

Assumption: no friction, no damping.

Potential energy of chain depending on the start angle $\theta$ on the sphere $$ E_{\rm pot}(\theta) = g\lambda\underbrace{(l-R\theta)}_{\small\mbox{length}}\underbrace{\left(-\frac{l-R\theta}2\right)}_{\small\mbox{height}}+\int_0^\theta g \underbrace{R\sin(\bar\theta)}_{\mbox{height}}\underbrace{\lambda R d\bar\theta}_{d m} = g\lambda\left(-\frac{(l-R\theta)^2}2 +R^2(1-\cos(\theta))\right) $$ The cut force in the thread on the chain-side is $$ F=\frac1{R}\frac{d E_{\rm pot}}{d\theta} = \frac{g\lambda}{R}\left(-R(R\theta-l) + R^2\sin(\theta)\right) $$ which gives at $\theta=\frac{\pi}2$ $$ F=g\lambda\left(l-R\frac{\pi}2+R\right) $$ Hopefully, without any mistakes. But, you should check. So, in the end this calculation delivers also solution A.

Note, a very good book on classical mechanics (with constraints) is Arnold's "Mathematical Methods of Classical Mechanics". This is a graduate text but the first chapters should already be readable for advanced students from high-school.

  • $\begingroup$ So basically, you mean to say that the differential mass cut out is a curve, hence, one must not say that dmg sin theta = dn $\endgroup$ – Saurabh Raje Feb 26 '14 at 6:39
  • $\begingroup$ But tell me, how can we integrate dT without vector calculus!!!??? It is not a scalar! $\endgroup$ – Saurabh Raje Feb 26 '14 at 6:46
  • $\begingroup$ I have added the reason for the scalar integration at the beginning of the answer. $\endgroup$ – Tobias Feb 26 '14 at 16:17
  • $\begingroup$ I am embarrassed to say that I honestly have no idea of what you have said in the beginning. I got a little bit, but the function in r,s; principle of virtual work does sound Greek. Can you please put it in simpler terms $\endgroup$ – Saurabh Raje Feb 26 '14 at 16:28
  • 1
    $\begingroup$ At first I added a picture and some text to make the geometry clear. Can you understand the geometry part now? $\endgroup$ – Tobias Feb 26 '14 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.