# Work done by tension

## The problem statement, all variables and given/known data

Consider the following arrangement: Calculate the work done by tension on $2\,\rm kg$ block during its motion on circular track from point $A$ to point $B$.

## The attempt at a solution

We know that work done by a force is product of force and displacement. We know the displacement of point of application as $4\,\rm m$. How to find the work done by the tension as it is not constant it is variable!

2nd attempt (calculus approach)

The $2\,\rm kg$ block moves along the circle, so its speed is $\dfrac{R\mathrm d\theta}{\mathrm dt}$.

It pulls the string, the length of the string between point $O$ and the block can we obtained with simple geometry at any position $\theta$ (ignoring the size of the pulley). The total length of the string is unchanged, so the speed of the $1\,\rm kg$ block is $\frac{\mathrm dL}{\mathrm dt}$. I'm looking for help after this.

• you need not take tension, otherwise you won't be able to do any work :P – Nix Jul 10 '13 at 7:58
• i have to calculate the work done by tension – maths lover Jul 10 '13 at 7:59
• probably you are from physicsforums – Faiz Iqbal Feb 24 '16 at 11:03

Work done by tension on BOTH the blocks can be regarded as 0. This can be said by the Virtual Work Method.

The virtual work method:

Consider that block 1(mass $2kg$) displaces by a certain $d\vec{s_1}$. Infinitesimal work done on the block 1 by tension will be given by $$dW_1 = \vec T.d\vec s_1=Tds_1\cos\theta_1$$ Similarly, for block 2 we can say that $$dW_2=\vec T.d\vec s_2=Tds_2\cos\theta_2$$

Using string constraint, we can say that displacement of each block along the string is zero(because the string is inextensible). So we get $$ds_1\cos\theta_1+ds_2\cos\theta_2=0$$ Notice that I have used the same $\theta$ for each block as in tension because the direction along the string is the direction along tension vector.
Net work done by tension thus becomes $$dW_T=T(ds_1\cos\theta_1+ds_2\cos\theta_2)$$ $$\therefore dW_T=0$$ $$W_T=0$$

The solution to the actual problem:

If we apply $W=\Delta K$ on the system of the two blocks from initial position to the final position where block 1 is at the bottom of the cicrular arc,we get $$m_1g\Delta h_1+m_2g\Delta h_2=\frac 12 m_1 v_1^2+\frac 12 m_2 v_2^2$$ I do not include work done by tension on the system because i proved it to be 0.

We now need to find a relation between $v_1$ and $v_2$. We can do this by applying string constraint. $$v_1\cos\theta=v_2$$ where $\cos\theta=\frac 35$(by geometry). $$\therefore \frac 35 v_1 = v_2$$ Substituting $v_2$ in terms of $v_1$ in the above equation, we can find $v_1$.
Then applying $W=\Delta K$ on block 1 only we get $$W_T + W_{gravity} = \frac 12 m_1v_1^2$$ Substitute $W_{gravity}$ and $v_1$ in this equation and find $W_T$.

• you have taken v1 for which body? – maths lover Jul 10 '13 at 14:57
• how to apply string constraint directly i had to use calculus. – maths lover Jul 10 '13 at 15:00
• I have taken $v_1$ for 2kg mass. – udiboy1209 Jul 10 '13 at 15:13
• ya its correct. – maths lover Jul 10 '13 at 15:15
• yes they are similar – maths lover Jul 10 '13 at 16:44

OK, we got off on the wrong foot here. As I mentioned, I was kind of short on time, and therefore I made a lot of shortcuts that, upon closer reading, are not allowed.

So, let's start over. Given the good answer by udiboy above (which was what I was after initially), I'll complement his answer by following the more complicated approach.

The tension will do positive work because of the drop of the 2 kg mass ($W_2$), and negative work because of the rise of the 1 kg mass ($W_1$). Given that the tensile force is always the same on both ends of the string, the total work done by the tension will be zero, because both will simply cancel each other out.

You can get to this result by evaluating the more general expression for work done by a force:

$$W = \int_{C_1} \mathbf{F} \cdot d\mathbf{s}_1 + \int_{C_2} \mathbf{F} \cdot d\mathbf{s}_2$$

with $\mathbf{F}$ the tension force vector and $d\mathbf{s}$ the displacement vector, and $C_2$ the circular contour that $m_2 = 2\,\text{kg}$ follows, and $C_1$ the linear contour that $m_1 = 1\,\text{kg}$ follows.

These integrals are not so easy to evaluate, since $\mathbf{F}$ changes in both magnitude and direction. Naturally, the circular tract is the most complex piece of this puzzle, so, some definitions first: It is fairly straightforward to see that

$$d\mathbf{s}_2 = R \cdot d\theta \cdot \mathbf{e}_\theta$$

In general, $|\mathbf{F}|$ depends on $\theta$, so remember that $\mathbf{F} = \mathbf{F}(\theta)$. Since we have a dot product in the integrand, and the second vector only has a component in the $\mathbf{e}_\theta$ direction, the contribution of $\mathbf{F}_R$ to the work $W$ will be zero.

A straightforward exercise in trigonometry gives

\begin{align} \mathbf{F}_\theta(\theta) &= |\mathbf{F}(\theta)| \cdot \cos\left( \theta - \arctan\left(\frac{R(1 - \cos\theta)}{h_0 + R\sin\theta}\right) \right) \\ &=|\mathbf{F}(\theta)| \cdot \Psi \end{align}

where $\Psi$ is the cosine term, substituted for brevity. A straightforward exercise in dynamics gives

$$|\mathbf{F}(\theta)| =m_1\left(\frac{ \Xi - m_1g}{m_1+m_2}\right) + m_1g$$

where

$$\Xi = \frac{m_2g(h_0 + R\sin\theta)}{\sqrt{(h_0+R\sin\theta)^2+(R-R\cos\theta)^2}}$$

The second integral thus takes the form

$$-\int_{\theta=0}^{\theta = \pi/2} R|\mathbf{F}(\theta)|\Psi d\theta$$

For the first integral, there is a complication due to the varying magnitude of $\mathbf{F}$ and the relation between $\theta$ and $d\mathbf{s}_1$. Since we compute $|\mathbf{F}(\theta)|$ inside the second integrand, it is actually easiest to combine both integrals, and compute the work from a single integration over $\theta$.

For this, we need a relation between $d\mathbf{s}_1$ and $d\theta$. Knowing that the total vertical rise of mass $m_1$ equals

$$L = \sqrt{(h_0 + R\sin\theta)^2+(R-R\cos\theta)^2} - h_0$$

which makes the infinitesmal rise equal to

\begin{align} d\mathbf{s}_1 &= dL \\ &= \frac{dL}{d\theta}d\theta\\ &= \frac{R\cos\theta(h_0 + R\sin\theta) + R\sin\theta(R-R\cos\theta)}{L} \end{align}

Then, the total work done by the tension can be computed by

$$W = \int_{\theta=0}^{\theta=\pi/2} |\mathbf{F}(\theta)| \left(\frac{dL}{d\theta} - R\cdot\Psi\right)d\theta$$

Now, I'm sure there's a bunch of simplifications possible, but I just substituted $h_0 = 1\,\text{m}$, $R=3\,\text{m}$ and evaluated the integral numerically. What I get is:

$$W = 0$$

Further inspection shows that $R\cdot\Psi = dL/d\theta$ for all $\theta$, which provides a nice check on the correctness of all of the above.

If you carry out the integration on only one of these terms, you get:

\begin{align} W &= -\int_{\theta=0}^{\theta=\pi/2} |\mathbf{F}(\theta)| \cdot R\cdot\Psi\,d\theta \\ &= -\int_{\theta=0}^{\theta=\pi/2} |\mathbf{F}(\theta)| \cdot \frac{dL}{d\theta}\,d\theta \\ &\approx -50.9787\,\text{J} \end{align}

which roughly agrees with the value you mentioned.