9.1 Pulleys(application-i)

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Solving problems is an essential part of the understanding process.

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

Hints for solving problems

In some questions, we are required to find relation for the accelerations of different elements in the arrangement. We, therefore, need a framework or a plan to establish this relation. This relation is basically obtained by differentiating the relation of positions of the movable elements in the arrangement.

• Draw reference, which is fixed. This reference can be the level of fixed pulley or the ground.
• Identify all movable elements like blocks and pulleys (excluding static ones).
• Assign variables for the positions of movable elements from the chosen reference.
• Write down constraint relations for the positions of the elements. Usually, total length of the string is the “constraint” that we need to make use for writing relation for the positions.
• Differentiate the relation for positions once to get relation of velocities and twice to get relation of accelerations.

Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to the pulleys. The questions are categorized in terms of the characterizing features of the subject matter :

• Static or fixed pulley
• Moving pulley
• Combination or Multiple pulley system

Static or fixed pulley

Problem 1 : Pulley and strings are “mass-less” and there is no friction involved in the arrangement. Find acceleration of the block of mass 3 kg and tensions “ ${T}_{1}$ ” and “ ${T}_{2}$ ” as shown in the figure.

Solution 1 : We observe here that two strings are taught. Therefore, acceleration of the all constituents of the system is same. This situation can be used to treat the system as composed of three blocks only (we neglect string as it has no mass). The whole system is pulled down by a net force as given by :

$F=2g+3g-g=4g$

Total mass of the system is :

$m=1+2+3=6\phantom{\rule{1em}{0ex}}kg$

Applying law of motion, the acceleration of the system and hence acceleration of the block of mass 3 kg is :

$a=\frac{F}{m}=\frac{4g}{6}=\frac{40}{6}=\frac{20}{3}\phantom{\rule{1em}{0ex}}m/{s}^{2}$

In order to find tension “ ${T}_{1}$ ”, we consider FBD of mass of 1 kg. Here,

$\sum {F}_{y}={T}_{1}-1Xg=m{a}_{y}=1X\frac{20}{3}$

$⇒{T}_{1}=\frac{20}{3}+10=\frac{50}{3}\phantom{\rule{1em}{0ex}}N$

In order to find tension “ ${T}_{2}$ ”, we consider FBD of mass of 3 kg. Here,

$\sum {F}_{y}=3Xg-{T}_{2}=m{a}_{y}=3X\frac{20}{3}=20$

$⇒{T}_{2}=3X10-20=10\phantom{\rule{1em}{0ex}}N$

Problem 2 : Pulley and string are “mass-less” and there is no friction involved in the arrangement. The blocks are released from rest. After 1 second from the start, the block of mass 4 kg is stopped momentarily. Find the time after which string becomes tight again.

Solution 1 : We need to understand the implication of stopping the block. As soon as block of 4 kg (say block “B”) is stopped, its velocity becomes zero. Tension in the string disappears. For the other block of 2 kg (say block “A”) also, the tension disappears.

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