You are Here: Home >< Maths

# Tension in a string watch

1. On the January 2006 OCR M1 Paper i don't quite understand question 6iii, please could someone explain?

http://blogs.thegrangeschool.net/mat...oklet-2010.pdf
The paper is on page 7 of the link above

https://211c25b87002c8b34761c926d793...20M1%20OCR.pdf
-Mark scheme
2. Your first link doesn't work for me however the URL says 2010 not 2006.
3. (Original post by Vikingninja)
Your first link doesn't work for me however the URL says 2010 not 2006.
It also says it's a "past paper booklet" so presumably means a past paper booklet till 2010?
4. Since P is zero, there is no force pulling the ring to the side, so it hangs below A, and because of the length of the string, it will be below B.
5. (Original post by Zacken)
It also says it's a "past paper booklet" so presumably means a past paper booklet till 2010?
Oh right didn't realise that. Never seen a booklet of several in one and content didn't work for me.
6. (Original post by ghostwalker)
Since P is zero, there is no force pulling the ring to the side, so it hangs below A, and because of the length of the string, it will be below B.
What I think is AR is only 0.4m so R can't be below B as AB is 0.5m so R will be 0.4m below A. Also I don't understand why the tension is 1N because the weight of R is 2N so since BR is slack, for it to hang in equilibrium Tension must be 2N instead of 1N to balance the weight of 2N?
7. (Original post by runny4)
What I think is AR is only 0.4m so R can't be below B as AB is 0.5m so R will be 0.4m below A. Also I don't understand why the tension is 1N because the weight of R is 2N so since BR is slack, for it to hang in equilibrium Tension must be 2N instead of 1N to balance the weight of 2N?
R is a smooth ring. It is not fixed to the string, but is free to slide. Does it make sense now?
8. (Original post by ghostwalker)
R is a smooth ring. It is not fixed to the string, but is free to slide. Does it make sense now?
I get the thing about it being able to slide know but I don't understand why the tension upwards isn't 2N to balance the weight of R which is 2N.
9. (Original post by runny4)
I get the thing about it being able to slide know but I don't understand why the tension upwards isn't 2N to balance the weight of R which is 2N.
Since it's free to slide and the length of the string is greater than the distance AB, then it hangs below B, and is held by two parts of the string - one either side of the ring.

Tension in each part is 1N, giving a total upward force of 2N.

10. (Original post by ghostwalker)
Since it's free to slide and the length of the string is greater than the distance AB, then it hangs below B, and is held by two parts of the string - one either side of the ring.

Tension in each part is 1N, giving a total upward force of 2N.

ok thank you

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 16, 2016
Today on TSR

### Three reasons you may feel demotivated right now

...and how to stay positive

### Can I get A*s if I start revising now?

Discussions on TSR

• Latest
Poll
Useful resources

## Make your revision easier

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE