# 9.2 Pulleys (application - ii)

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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.

## Representative problems and their solutions

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

• Change of force direction
• External force on pulley system
• Horizontal pulley system
• Vertical pulley system
• Incline and pulley system

## Change of force direction

Problem 1 : A person is pulling a block of 50 kg at a uniform speed with the help of a pulley and rope arrangement as shown in the figure. If the person weighs 60 kg, find the normal force that the person exerts on the surface underneath.

Solution : In order to find normal force applied by the person on the surface underneath, we need to carry out force analysis of the external forces on the person. The surface and the person apply normal force on each other, which is equal in magnitude but opposite in direction. The forces on the person are (i) weight, “mg”, acting downward (ii) normal force, “N”, acting upward and (iii) tension, “T”, acting upward.

The free body diagram of the person is shown on the right side of the figure shown below.

$\sum {F}_{y}⇒T+N=mg$

$⇒N=60\phantom{\rule{2pt}{0ex}}X\phantom{\rule{2pt}{0ex}}10-T=600-T$

It is clear that we would need the value of tension, “T”, to solve the equation for the normal force. We can find the value of “T” by considering the FBD of the block being raised. The FBD of the block is shown on the left side of the figure above. Note that we consider that person is raising the block without acceleration (at uniform velocity as given in the question).

$\sum {F}_{y}=T-50g=0$

$⇒T=50g=50\phantom{\rule{2pt}{0ex}}X\phantom{\rule{2pt}{0ex}}10=500\phantom{\rule{1em}{0ex}}N$

Putting this value in the equation for normal force, we have :

$⇒N=600-T=600-500=100\phantom{\rule{1em}{0ex}}N$

## External force on pulley system

Problem 2 : In the arrangement shown in the figure, the strings are taught initially. An external force of 600 N is applied in the vertically upward direction. Considering strings and pulleys to be “mass – less” and all contacts without friction, determine the accelerations of the blocks.

Solution : The forces on each of the blocks are (i) weight of the block (ii) tension in the string and (iii) normal force.

When a block is lifted, the normal force becomes zero as the contact between surfaces is broken. The external forces on the block are, then, only two, comprising of its weight and tension in the string. It is, therefore, clear that we need to know the tension in the string to know the accelerations of the blocks.

#### Questions & Answers

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