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The result for acceleration as obtained above is significant for this type of arrangement, where blocks are pulled down by a string passing over a pulley. The acceleration as derive is equivalent to :
$$\begin{array}{l}\Rightarrow a=\frac{F}{{m}_{1}+{m}_{2}}\end{array}$$
where $F={m}_{2}g$ . It suggests that we can treat the arrangement as composite mass of ( ${m}_{1}+{m}_{2}$ ) subjected to an external force “F” of magnitude “ ${m}_{2}g$ ”. This analysis helps in situations where more than one block is pulled by a hanging block. Take the example as shown in the figure :
In this case,
$$\begin{array}{l}\Rightarrow a=\frac{F}{{m}_{1}+{m}_{2}+{m}_{3}+{m}_{4}}=\frac{{m}_{4}g}{{m}_{1}+{m}_{2}+{m}_{3}+{m}_{4}}\end{array}$$
$$\begin{array}{l}\Rightarrow {T}_{1}={m}_{1}a;\phantom{\rule{2pt}{0ex}}{T}_{2}=({m}_{1}+{m}_{2})a;\phantom{\rule{2pt}{0ex}}{T}_{3}=({m}_{1}+{m}_{2}+{m}_{3})a;\end{array}$$
Important to note here is that the separate strings have different tensions.
Consider the two cases when a block is pulled down by a force in two different manners. In one case, string is pulled by another mass ${m}_{2}$ , hanging from it. In the second case, the string is pulled by a force equal to hanging weight ( ${m}_{2}g$ ). Now the question is : whether two situations are equivalent or different ?
As worked out earlier, the acceleration in the first case, ${a}_{1}$ , is :
$$\begin{array}{l}\Rightarrow {a}_{1}=\frac{{m}_{2}g}{{m}_{1}+{m}_{2}}\end{array}$$
In the second case, let the acceleration of the block be “ ${a}_{2}$ ”. Then force analysis of the block on the table is :
$$\begin{array}{l}\Rightarrow {m}_{1}{a}_{2}=T\end{array}$$
But,
$$\begin{array}{l}T={m}_{2}g\end{array}$$
Now, combining two equations, we have :
$$\begin{array}{l}\Rightarrow {a}_{2}=\frac{{m}_{2}g}{{m}_{1}}\end{array}$$
Evidently, ${a}_{2}>{a}_{1}$ . This result is expected because applying force as weight and as force are two different things. In the first case, mass of the body applying force is also involved as against the later case, when no "body" of any mass is involved as far as application of force is concerned. The bottom line is : “do not hang a weight simply to apply force”.
Further, the behavior of string is true to ideal string so long string is not intervened by other element, which is itself accelerating. Consider a pulley – string system, in which the pulley is accelerating as shown in the figure.
In such case, the accelerations of different parts of the string are not same. Note in the example shown above that one end of the string is fixed, but the other end of string is moving. As a matter of fact, the acceleration of block (hence that of the string at that end) is twice that of the pulley. The acceleration of the other end of the string, which is fixed, is zero. This aspect is explained in a separate module on movable pulley in the course.
The tension in the string having certain mass is not same everywhere. A part of the force is required to accelerate string as well. Usually,mass is distributed uniformly along the length of the string. We describe mass distribution in terms of "mass per unit length" and denotes the same as "λ" such that the mass of the string,"m" of length "L" is given as :
m = λL
If we take cross section at any point in the string, then the part of the string on each side forms an mass element. having mass proportional to the length of that part of the string. This aspect is brought out in the example given here.
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