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# M1 help

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1. https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

Very last part of the last question

Tension is 6.53N

I've seen 2Tcos45 used in some maths videos from examsolutions but it wasn't correct.

Anyone have any idea on how to answer it and why it's answered like that?

2. (Original post by Kay Fearn)
https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

Very last part of the last question

Tension is 6.53N

I've seen 2Tcos45 used in some maths videos from examsolutions but it wasn't correct.

Anyone have any idea on how to answer it and why it's answered like that?

Have you checked that the tension you've calculated in part a is correct?Well basically each part of the string has tension and if you resolve perpendicular to the pulley and find the perpendicular components of the tension you should get 2Tcos (alpha). If tan alpha = 4/3, then cos alpha = 3/5. Substitute this into 2Tcos(90-alpha/2) and solve. Its not always going to be cos 45. I believe the video you watched by exam solutions was a 90 degree angle with a pulley. this has a slope with a pulley

The reason why its 90-alpha/2 is because the slope forms a triangle. The angle that the pulley lies on is thus equal to 180 - 90 - alpha. which is 90 - alpha. Then we divide by two because we need to resolve for each tension
3. (Original post by Kay Fearn)
https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

Very last part of the last question

Tension is 6.53N

I've seen 2Tcos45 used in some maths videos from examsolutions but it wasn't correct.

Anyone have any idea on how to answer it and why it's answered like that?

2Tcos45 is for when the tension from P is perpendicular to the tension from Q. It doesn't work when there's a slope.

This video is great for explaining the three types of force on a pulley questions that often come up.

I do recommend understanding this topic instead of relying on a formula.

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