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```
A man sits on a chair supported by a rope passing over a friction less fixed pulley .The man who weighs 1000N exerts a force of 450N on the chair downwards whule pulling the rope on the other side . If the chair weighs 250N tgen acceleration of chair is?
A man sits on a chair supported by a rope passing over a friction less fixed pulley .The man who weighs 1000N exerts a force of 450N on the chair downwards whule pulling the rope on the other side . If the chair weighs 250N tgen acceleration of chair is?

```
3 years ago

Aman Sharma
40 Points
```							Suppose acceleration is in direction of man going downwards so by drawing the diagram and using fbd consider g to be 10 so the mass of the person is 100kg and 125kg of the chair , we get the following 2 equations 1 T- 450N (force applied by the man) = 45a (450÷g i.e  mg÷mg = m) and 2-   [mg (of man) + mg of chair = (mass of man + mass of chair) so putting values the equation are as follow 1)T - 450N =  45a  , 2) 1250 - T = 125 a.  Solving equation via elimination method we get a= 4.70
```
3 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions