x Turn on thread page Beta
 You are Here: Home >< Maths

# Mechanics 1 OCR friction help watch

1. A heavy ring of mass 5kg is threaded on a fixed rough horizontal rod. The coefficient of friction between the rod and the ring is 0.5. A light string is attached to the ring and pulled downwards with a force acting at a constant angle of 30 degrees to the horizontal(can't get the diagram up, don't know how to ). The magnitude of the force is T newtons, and is gradually increased from 0. Find the value of T that is just sufficient to make the equilibrium limiting.

Here's what I done, weight = 5 x 9.8 = 49,
Fr = 0.5 x 49 = 24.5
If equilibrium is limiting then the resloved horizontal force = 24.5
I resolved the force horizontally and got T cos30 = 24.5
so this gives T = 24.5/cos30 ?
This gives the wrong answer, I think I missed out a force somewhere, can someone explain which force, how to get it and check if I done anything else wrong?

Thanks
2. One way of posting a drawing is to use Excel and then take a screen shot using Snipping Tool and then upload to http://www.imageshack.us/

If using Excel choose the preference 'no grid lines'

3. check this:

resolve vertically:
5g + T sin 30 = R
5g + T/2 = R

resolve horizontally:
Fr= T cos 30 =( sqrt (3) x T )/2

where Fr= uR

sub in:

T cos 30 = 0.5 ( 5g + T sin 30 )

rearrange....

T ( sqrt(3) +1 )= 49

T = 49 / (sqrt(3) +1) = 17.94 N
4. Your original answer does not work because it doesn't take in to account the added downwards force from the string when calculating friction.
5. (Original post by 3.5Nando)
check this:

resolve vertically:
5g + T sin 30 = R
5g + T/2 = R

resolve horizontally:
Fr= T cos 30 =( sqrt (3) x T )/2

where Fr= uR

sub in:

T cos 30 = 0.5 ( 5g + T sin 30 )

rearrange....

T ( sqrt(3) +1 )= 49

T = 49 / (sqrt(3) +1) = 17.94 N
Thats not the right answer, thanks for trying though

(Original post by Kasc)
Your original answer does not work because it doesn't take in to account the added downwards force from the string when calculating friction.
I see, I got it now, thanks
Thats not the right answer, thanks for trying though

I see, I got it now, thanks
7. (Original post by Notsocleverstudent)
39.8
39.8
how did you get that?
9. Argh I wasn't getting an answer all this time cos I took g to be 10 . We take g=10 for m1. That is all .
10. (Original post by Notsocleverstudent)
how did you get that?
R=9.8X5+Tsin30...
11. (Original post by Cool story bro)
R=9.8X5+Tsin30...
Ye i did

Tcos(30) = 0.5 x (49 + Tsin30)

12. Take T common
13. (Original post by Cool story bro)
Take T common
I just realised I was doing Sin(30)/cos(30) and was gettign tan(30)

:/ im not worthy of mechanics

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: March 20, 2011
Today on TSR

### Happy St Patrick's day!

How are you celebrating?

### Stay at sixth form or go to college?

Discussions on TSR

• Latest
Poll
Useful resources

### 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

## 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