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But the point is made. The accelerations of two ends of a string need not be same, when in contact with a moving body with a free end.
Problem : A block of mass, “m” is connected to a string, which passes over a smooth pulley as shown in the figure. If a force “F” acts in horizontal direction, find the accelerations of the pulley and block.
Solution : Let us consider that the acceleration of pulley is “a” in the direction of applied force. Now as analyzed before, acceleration of the block is “2a” and is in the same direction as that of pulley.
As motion is confined to x-direction, we draw free body diagram considering forces in x-direction only.
$\text{Free body diagram of pulley}$
The external forces are (i) force, F, (ii) tension, T, in the string and (ii) tension, T, in the string.
$$\begin{array}{l}\sum {F}_{x}=F-2T={m}_{p}{a}_{p}\end{array}$$
As mass of the pulley is zero,
$$\begin{array}{l}\Rightarrow F=2\mathrm{T}\\ \Rightarrow T=\frac{F}{2}\end{array}$$
This is an unexpected result. The pulley is actually accelerated, but the forces on it form a balanced force system. This apparent contradiction of force law is due to our approximation that pulley is “mass-less”. $\text{Free body diagram of block}$
The external force is (i) tension, T, in the string.
$$\begin{array}{l}\sum {F}_{x}=T=m{a}_{x}=2ma\end{array}$$
Combining two equations, we have :
$$\begin{array}{l}\Rightarrow a=\frac{F}{4m}\end{array}$$
Thus, acceleration of the pulley is F/4m and that of block is F/2m.
Multiple pulleys may involve combination of both static and moving pulleys. This may involve combining characterizing aspects of two systems.
Let us consider one such system as shown in the figure.
Let us consider that accelerations of the blocks are as shown in the figure. It is important to note that we have the freedom to designate direction of acceleration without referring to any other consideration. Here, we consider all accelerations in downward direction.
We observe that for given masses, there are five unknowns ${a}_{1},\phantom{\rule{2pt}{0ex}}{a}_{2},\phantom{\rule{2pt}{0ex}}{a}_{3},\phantom{\rule{2pt}{0ex}}{T}_{1}\phantom{\rule{2pt}{0ex}}\text{and}\phantom{\rule{2pt}{0ex}}{T}_{2}$ . Whereas the free body diagram corresponding to three blocks provides only three independent relations. Thus, all five unknowns can not be evaluated using three equations.
However, we can add two additional equations; one that relates three accelerations and the one that relates tensions in the two strings. Ultimately, we get five equations for five unknown quantities.
1: Accelerations
The pulley “A” is static. The accelerations of block 1 and pulley “B” are, therefore, same. The pulley “B”, however, is moving. Therefore, the accelerations of blocks 2 and 3 may not be same as discussed for the case of static pulley. The accelerations of blocks 2 and 3 with respect to moving pulley are different than their accelerations with respect to ground reference.
We need to know the relation among accelerations of block 1, 2 and 3 with respect ground. In order to obtain this relation, we first establish the relation among the positions of moving blocks and pulley with respect to some fixed reference. Then we can obtain relation among accelerations by taking second differentiation of position with respect to time. Important to understand here is that these positions are measured with respect to a fixed reference. As such, their differentiation will yield accelerations of the blocks with respect to fixed reference i.e. ground reference. Let the positions be determined from a horizontal datum drawn through the static pulley as shown in the figure.
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