The formula is: $$T= \dfrac{\left(\sum (\text{coefficient of tension})_i\right) \cdot g_{effective}}{\sum \frac{(\text{coefficient of tension})_i^2}{m_i}}$$ I tried to prove this formula but not getting it. An example for this formula is below:

enter image description here

Explaining the Q:

1g + 2g means correction of tension is 1 for 4kg block and 2 for 2kg block. In the end, g is always the same. Then, in the denominator it is the addition of $\text{coefficient of }T^2 / \text{mass}$.

About $g$ effective on top, in case of inclined plane.its value would be $g \sin \theta$ if it is like this:

enter image description here

  • $\begingroup$ I think you need to provide the reference or a link to the source of this equation, otherwise, not many people recognize what is involved and your question. $\endgroup$ – r13 Apr 27 at 18:41

I'll be using subscript notation $_2$and $_4$ to denote the mass with 2 kg and 4 kg respectively. Also note that $a_2$ and $a_4$ both point upwards which I consider as the positive direction.

enter image description here

There are 2 equations of motion

  • for mass 2: $$m_2\cdot a_2 = -m_2 \cdot g +2T \tag{eoq:m2}$$
  • for mass 4kg $$m_4\cdot a_4 = -m_4 \cdot g +T \tag{eoq:m4}$$

Additionally you get the kinematic constraint (assuming the wire is not deformable as)

$$a_4 = -2a_2 \tag{KC}$$

you can rewrite $eoq:m2,\ eoq:m4$ as:

$$a_2 = -g +\color{red}{2} \frac{T}{m_2} $$ $$a_4 = -g +\color{red}{1}\frac{T}{m_4} $$ $$\color{red}{1} a_4 = -\color{red}{2}a_2 $$

With the $\color{red}{red}$ are the correction coefficients g, you are mentioning. Therefore: $$a_2 = -g +\color{red}{g_2} \frac{T}{m_2} $$ $$a_4 = -g +\color{red}{g_4}\frac{T}{m_4} $$ $$\color{red}{g_4} a_4 = -\color{red}{g_2}a_2 $$

Then you solve the system of the equations of motion with the kinematic constraint: $$ \begin{cases} a_2 = -g +g_2 \frac{T}{m_2} \\ \color{red}{a_4} = -g + g_4\frac{T}{m_4} \\ \color{red}{g_4 a_4 = -g_2 a_2} \end{cases} \Rightarrow \begin{cases} a_2 = -g +g_2 \frac{T}{m_2} \\ -\frac{g_2}{g_4}a_2 = -g + g_4\frac{T}{m_4} \end{cases} $$

$$ \begin{cases} a_2 = -g +g_2 \frac{T}{m_2} \\ a_2 = \frac{{g_4}}{{g_2}}\left(g - g_4\frac{T}{m_4}\right) \end{cases} \Rightarrow -g +g_2 \frac{T}{m_2} = \frac{{g_4}}{{g_2}}\left(g - g_4\frac{T}{m_4}\right) $$

$$ +g_2 \frac{T}{m_2} +\frac{g_4}{g_2} g_4\frac{T}{m_4} = +g +\frac{g_4}{g_2} g $$ $$ \left(\frac{g_2}{m_2} + \frac{1}{g_2} \frac{g_4^2}{m_4}\right)T = \left(1 +\frac{g_4}{g_2}\right) g $$ $$ T = \frac{ \left(1 +\frac{g_4}{g_2}\right) g }{\left(\frac{g_2}{m_2} + \frac{1}{g_2} \frac{g_4^2}{m_4}\right)}$$

To simplify this we multiply by $g_2 $ mnominator and denominator: $$ T = \frac{g_2}{g_2} \frac{ \left(1 +\frac{g_4}{g_2}\right) g }{\left(\frac{g_2}{m_2} + \frac{1}{g_2} \frac{g_4^2}{m_4}\right)}\rightarrow T = \frac{ \left(g_2 +g_4\right) g }{\left(\frac{g_2^2}{m_2} + \frac{g_4^2}{m_4}\right)}$$

So finally you get:

$$ T = \frac{ \left(\sum g_i\right) g }{\sum \frac{g_i^2}{m_i}} $$

POINT OF CAUTION Although this has been proven for this particular set of pulley, it is something I've never come across. And despite, solving it myself (and marveling at the elegance that someone actually contrived this), I would be very cautious in using this equation in another problem without cross-checking it.

  • $\begingroup$ So , what you did is amazing. Really , I’m super happy. I would like to know how did you think to approach this problem. What were your thoughts , did any mistake happened during the process and then how did you think to correct it. $\endgroup$ – Srijan M.T Apr 28 at 6:37
  • $\begingroup$ I only took the equations of motion, and solved for them. The greatest trouble I had was that, I didn't know how the correction coefficients were estimated/derived or what do they affect (as I said I've proven it, but its a completely new approach). So I had to go back to the kinematic constraint which initially I did not put the coefficient. After that the rest was just algebraic manipulations. $\endgroup$ – NMech Apr 28 at 6:45
  • $\begingroup$ From my end, I would like to know, who came up with this idea. Was it your lecturer's did you read it at a book? Mainly curious. $\endgroup$ – NMech Apr 28 at 6:46
  • $\begingroup$ My teacher told me.It is just used in exams to calculate T quickly. $\endgroup$ – Srijan M.T Apr 28 at 6:57
  • 1
    $\begingroup$ @SrijanM.T, Thanks for the update. Also, if you feel that this answer has been answered please considered accepting it. $\endgroup$ – NMech Apr 28 at 22:44

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