# Pulleys  (Page 4/5)

 Page 4 / 5

Now, the lengths of the two strings are constant. Let they be ${L}_{1}\phantom{\rule{2pt}{0ex}}\text{and}\phantom{\rule{2pt}{0ex}}{L}_{2}$ .

$\begin{array}{l}{L}_{1}={x}_{0}+{x}_{1}\\ {L}_{2}=\left({x}_{2}-{x}_{0}\right)+\left({x}_{3}-{x}_{0}\right)={x}_{2}+{x}_{3}-2{x}_{0}\end{array}$

Eliminating ${x}_{0}$ , we have :

$\begin{array}{l}⇒{L}_{2}={x}_{2}+{x}_{3}-2{L}_{1}+2{x}_{1}\\ ⇒{x}_{2}+{x}_{3}+2{x}_{1}=2{L}_{1}+{L}_{2}=\text{constant}\end{array}$

This is the needed relation for the positions. We know that acceleration is second derivative of position with respect to time. Hence,

$\begin{array}{l}⇒{a}_{2}+{a}_{3}+2{a}_{1}=0\end{array}$

This gives the relation of accelerations involved in the pulley system.

2: Tensions in the strings

We can, now, find the relation between tensions in two strings by considering the free body diagram of pulley “B” as shown in the figure.

Here,

$\begin{array}{l}{T}_{1}=2{T}_{2}\end{array}$

This result appears to be simple and on expected line. But it is not so. Note that pulley "B" itself is accelerated. The result, on the other hand, is exactly same as for a balanced force system. In fact this equality of forces in opposite direction is possible, because we have considered that pulley has negligible mass. This aspect has been demonstrated in the force analysis of the example given earlier (you may go through the example again if you have missed the point).

3: Free body diagrams of the blocks

The free body diagrams of the blocks are as shown in the figure.

$\begin{array}{l}{m}_{1}g-{T}_{1}={m}_{1}{a}_{1}\\ {m}_{2}g-{T}_{2}={m}_{2}{a}_{2}\\ {m}_{3}g-{T}_{2}={m}_{3}{a}_{3}\end{array}$

Thus, we have altogether 5 equations for 5 unknowns.

There is one important aspect of the motions of blocks of mass " ${m}_{2}$ " and " ${m}_{3}$ " with respect to moving pulley "B". The motion of blocks take place with respect to an accelerating pulley. Thus, interpretation of the acceleration must be specific about the reference (ground or moving pulley). We should ensure that all measurements are in the same frame. In the methods, described above we have considered accelerations with respect to ground. Thus, if acceleration is given with respect to the moving pulley in an analysis, then we must first change value with respect to the ground.

Problem : In the arrangement shown in the figure, mass of A = 5 kg and mass of B = 10 kg. Consider string and pulley to be mass-less and friction absent everywhere in the arrangement. Find the accelerations of blocks.

Solution : Let the acceleration of block “A” be “a” in the downward direction. Let the tensions in the string be “ ${T}_{1}$ ” and “ ${T}_{2}$ ”. See figure below showing forces acting on the blocks and moving pulley. The force analysis of the block “A” yields :

$⇒5g-{T}_{1}=5a$ $⇒50-{T}_{1}=5a$

From constraint of string length, we see that :

${x}_{1}-{x}_{0}+{x}_{2}-{x}_{0}+{x}_{2}=L$ $⇒{x}_{1}+2{x}_{2}=L+2{x}_{0}=\text{Constant}$

Differentiating twice with respect to time, we get relation between accelerations of two blocks as :

$⇒{a}_{1}=-2{a}_{2}$ $⇒{a}_{2}=-\frac{{a}_{1}}{2}=-\frac{a}{2}$

Force analysis of the moving mass-less pulley yields :

$⇒{T}_{2}=2{T}_{1}$

Force analysis of the block B results (refer to the force diagram shown in the beginning) :

$⇒2{T}_{1}-10X10=10X\frac{a}{2}=5a$ $⇒2{T}_{1}-100=5a$

Simultaneous equations of forces on blocks A and B are :

$⇒50-{T}_{1}=5a$ $⇒2{T}_{1}-100=5a$

Solving for “a”, we have :

$⇒a=0$

Thus, accelerations of two blocks are zero. Note, however, that tensions in the strings are not zero.

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