tmorrall
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Can someone show me how to work out question 7 part v - a
http://papers.xtremepapers.com/OCR/M...tion_Paper.pdf
http://www.ocr.org.uk/Images/61150-m...me-january.pdf
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Stonebridge
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The (upwards) tension in the chain holding up the pulley must be in equilibrium with the downwards forces on it due to the tension in the string and its weight.

You are told in the previous part that the tension in the string is 5.88N and you know the pulley is of mass 5kg. So its weight is 4.9N

Total downwards force on pulley is therefore 5.88N + 5.88N + 4.9N
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tmorrall
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(Original post by Stonebridge)
The (upwards) tension in the chain holding up the pulley must be in equilibrium with the downwards forces on it due to the tension in the string and its weight.

You are told in the previous part that the tension in the string is 5.88N and you know the pulley is of mass 5kg. So its weight is 4.9N

Total downwards force on pulley is therefore 5.88N + 5.88N + 4.9N
Why is there two lots of tension? One lot either side of the pulley? But does the tension not act upward on the right and downward on the left?


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Stonebridge
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(Original post by tmorrall)
Why is there two lots of tension? One lot either side of the pulley? But does the tension not act upward on the right and downward on the left?


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No. Certainly not.
The tension in the string acts equally downwards on both sides.
This is always true in any question where you have a "light" string and a frictionless pulley.

Imagine I hung two balanced identical 1kg masses (9.8N weight) on a string either side of the pulley.
What is the total downwards force on the pulley?
If it isn't 19.6N where has the weight of one of the masses gone?
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tmorrall
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(Original post by Stonebridge)
No. Certainly not.
The tension in the string acts equally downwards on both sides.
This is always true in any question where you have a "light" string and a frictionless pulley.

Imagine I hung two balanced identical 1kg masses (9.8N weight) on a string either side of the pulley.
What is the total downwards force on the pulley?
If it isn't 19.6N where has the weight of one of the masses gone?
How does acceleration affect this? When you're solving for a and T, does tension not act against the weight?


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tmorrall
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Stonebridge
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(Original post by tmorrall)
How does acceleration affect this? When you're solving for a and T, does tension not act against the weight?


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When there is acceleration you apply F=ma to the individual masses on the string. F is the resultant force where T is the tension (upwards) and mg the weight downwards.
You solve for T and a. (or for whatever is the unknown)
The total downwards force on the pulley due to the tension in the string will be 2T
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