<< Chapter < Page | Chapter >> Page > |
There is a very useful technique to simplify the solution involving "mass-less" string and pulley. As string has no mass, the motion of the block-string system can be considered to be the motion of a system comprising of two blocks, which are pulled down by a net force in the direction of acceleration.
Let us consider two blocks of mass " ${m}_{1}$ " and " ${m}_{2}$ " connected by a string as in the previous example. Let us also consider that ${m}_{2}>{m}_{1}$ so that block of mass " ${m}_{2}$ " is pulled down and block of mass " ${m}_{1}$ " is pulled up. Let "a" be the acceleration of the two block system.
Now the force pulling the system in the direction of acceleration is :
$$F={m}_{2}g-{m}_{1}g=\left({m}_{2}-{m}_{1}\right)g$$
The total mass of the system is :
$$m={m}_{1}+{m}_{2}$$
Applying law of motion, the acceleration of the system is :
Clearly, this method to find acceleration is valid when the block - string system can be combined i.e. accelerations of the constituents of the system are same. We can check the efficacy of this technique, using the data of previous example. Here,
$${m}_{1}=10\phantom{\rule{1em}{0ex}}kg;\phantom{\rule{1em}{0ex}}{m}_{2}=20\phantom{\rule{1em}{0ex}}kg$$
The acceleration of the block is :
$$a=\frac{\left({m}_{2}-{m}_{1}\right)g}{\left({m}_{1}+{m}_{2}\right)}=\frac{\left(20-10\right)X10}{\left(10+20\right)}$$
$$a=\frac{10}{3}=3.33\phantom{\rule{1em}{0ex}}m/{s}^{2}$$
Moving pulley differs to static pulley in one important respect. The displacements of pulley and block, which is attached to the string passing over it, may not be same. As such, we need to verify this aspect while applying force law. The point is brought out with clarity in the illustration explained here. Here, we consider a block attached to a string, which passes over a mass-less pulley. The string is fixed at one end and the Pulley is pulled by a force in horizontal direction as shown in the figure.
In order to understand the relation of displacements, we visualize that pulley has moved a distance “x” to its right. The new positions of pulley and block are as shown in the figure. To analyze the situation, we use the fact that the length of string remains same in two situations. Now,
Length of the string, L, in two situations are given as :
$$L=\mathrm{AB}+\mathrm{BC}=\mathrm{AB}+\mathrm{BD}+\mathrm{ED}$$
$$\Rightarrow \mathrm{BC}=\mathrm{BD}+\mathrm{ED}$$
$$\Rightarrow \mathrm{ED}=\mathrm{BC}-\mathrm{BD}$$
Now, displacement of the block is :
$$\Rightarrow \mathrm{CE}=\mathrm{CD}-\mathrm{ED}=\mathrm{CD}-\mathrm{BC}+\mathrm{BD}=\mathrm{BD}+\mathrm{BD}=2\mathrm{BD}=2x$$
Note that for every displacement “x” of pulley, the displacement of block is 2x. We can appreciate this fact pictorially as shown in the figure below :
Further, as acceleration is second derivative of displacement with respect to time, the relation between acceleration of the block ( ${a}_{B}$ ) and pulley ( ${a}_{P}$ ) is :
This is an important result that needs some explanation. It had always been emphasized that the acceleration of a taut string is always same through out its body. Each point of a string is expected to have same velocity and acceleration! What happened here ? One end is fixed, while other end is moving with acceleration. There is, in fact, no anomaly. Simply, the acceleration of the pulley is also reflected in the motion of the loose end of the string as they are in contact and that the motion of the string is affected by the motion of the pulley.
Notification Switch
Would you like to follow the 'Physics for k-12' conversation and receive update notifications?