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Man sits on a chair supported by a rope passing over a frictionless fixed pulley. The man who weighs 1 , 000 N exerts a force of 450 N on the chair downwards while pulling the rope on the other side. If the chair weighs 250 N then the acceleration of the chair is .?
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Man sits on a chair supported by a rope passing over a frictionless fi...
Understanding the Forces Involved
Let's break down the forces acting on the system.
- Weight of the Man: 1,000 N
- Force Exerted by the Man: 450 N (downward)
- Weight of the Chair: 250 N
The total downward force acting on the chair and the man combined is:
- Total Downward Force = Weight of the Man + Weight of the Chair + Force Exerted by the Man
- Total Downward Force = 1,000 N + 250 N + 450 N = 1,700 N
Calculating the Tension in the Rope
The man pulls the rope with a force of 450 N. Due to the pulley system, this force creates a tension in the rope, which is equal to 450 N.
Net Force Acting on the System
Now, to find the net force acting on the chair:
- Net Force on Chair = Tension in the Rope - Weight of the Chair
- Net Force on Chair = 450 N - 250 N = 200 N (acting upwards)
Applying Newton's Second Law
Using Newton's second law (F = ma), we can find the acceleration of the chair:
- Total Mass of the System:
- Weight of the Chair = 250 N
- Weight of the Man = 1,000 N
- Total Weight = 1,000 N + 250 N = 1,250 N
- Total Mass = Total Weight / g (where g = 9.81 m/s²)
- Acceleration Calculation:
- Total Mass = 1,250 N / 9.81 m/s² ≈ 127.4 kg
- Net Force = 200 N
- Acceleration = Net Force / Total Mass = 200 N / 127.4 kg ≈ 1.57 m/s²
Conclusion
The acceleration of the chair is approximately 1.57 m/s² upwards. This indicates that the chair is accelerating upward due to the net force created by the man's pull on the rope.
Let's break down the forces acting on the system.
- Weight of the Man: 1,000 N
- Force Exerted by the Man: 450 N (downward)
- Weight of the Chair: 250 N
The total downward force acting on the chair and the man combined is:
- Total Downward Force = Weight of the Man + Weight of the Chair + Force Exerted by the Man
- Total Downward Force = 1,000 N + 250 N + 450 N = 1,700 N
Calculating the Tension in the Rope
The man pulls the rope with a force of 450 N. Due to the pulley system, this force creates a tension in the rope, which is equal to 450 N.
Net Force Acting on the System
Now, to find the net force acting on the chair:
- Net Force on Chair = Tension in the Rope - Weight of the Chair
- Net Force on Chair = 450 N - 250 N = 200 N (acting upwards)
Applying Newton's Second Law
Using Newton's second law (F = ma), we can find the acceleration of the chair:
- Total Mass of the System:
- Weight of the Chair = 250 N
- Weight of the Man = 1,000 N
- Total Weight = 1,000 N + 250 N = 1,250 N
- Total Mass = Total Weight / g (where g = 9.81 m/s²)
- Acceleration Calculation:
- Total Mass = 1,250 N / 9.81 m/s² ≈ 127.4 kg
- Net Force = 200 N
- Acceleration = Net Force / Total Mass = 200 N / 127.4 kg ≈ 1.57 m/s²
Conclusion
The acceleration of the chair is approximately 1.57 m/s² upwards. This indicates that the chair is accelerating upward due to the net force created by the man's pull on the rope.
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Man sits on a chair supported by a rope passing over a frictionless fixed pulley. The man who weighs 1 , 000 N exerts a force of 450 N on the chair downwards while pulling the rope on the other side. If the chair weighs 250 N then the acceleration of the chair is .?
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Man sits on a chair supported by a rope passing over a frictionless fixed pulley. The man who weighs 1 , 000 N exerts a force of 450 N on the chair downwards while pulling the rope on the other side. If the chair weighs 250 N then the acceleration of the chair is .? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Man sits on a chair supported by a rope passing over a frictionless fixed pulley. The man who weighs 1 , 000 N exerts a force of 450 N on the chair downwards while pulling the rope on the other side. If the chair weighs 250 N then the acceleration of the chair is .? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Man sits on a chair supported by a rope passing over a frictionless fixed pulley. The man who weighs 1 , 000 N exerts a force of 450 N on the chair downwards while pulling the rope on the other side. If the chair weighs 250 N then the acceleration of the chair is .?.
Man sits on a chair supported by a rope passing over a frictionless fixed pulley. The man who weighs 1 , 000 N exerts a force of 450 N on the chair downwards while pulling the rope on the other side. If the chair weighs 250 N then the acceleration of the chair is .? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Man sits on a chair supported by a rope passing over a frictionless fixed pulley. The man who weighs 1 , 000 N exerts a force of 450 N on the chair downwards while pulling the rope on the other side. If the chair weighs 250 N then the acceleration of the chair is .? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Man sits on a chair supported by a rope passing over a frictionless fixed pulley. The man who weighs 1 , 000 N exerts a force of 450 N on the chair downwards while pulling the rope on the other side. If the chair weighs 250 N then the acceleration of the chair is .?.
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