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Mechanics 1 OCR friction help

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 :smile:). 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
Reply 1
Avatar for steve2005
steve2005
17
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'

Snipping Tool is probably already on your computer, if not http://windows.microsoft.com/en-US/windows-vista/Use-Snipping-Tool-to-capture-screen-shots



Uploaded with ImageShack.us
(edited 13 years ago)
Reply 2
Avatar for 3.5Nando
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
Reply 3
Avatar for Kasc
Kasc
13
Your original answer does not work because it doesn't take in to account the added downwards force from the string when calculating friction.
Reply 4
Avatar for Eloades11
Eloades11
OP
17
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 :smile:

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
Original post by Eloades11
Thats not the right answer, thanks for trying though :smile:



I see, I got it now, thanks


what was the answer?
Reply 6
Avatar for Eloades11
Eloades11
OP
17
Original post by Notsocleverstudent
what was the answer?


39.8
Original post by Eloades11
39.8


how did you get that?
Argh I wasn't getting an answer all this time cos I took g to be 10 :mad:. We take g=10 for m1. That is all :getmecoat:.
Original post by Notsocleverstudent
how did you get that?


R=9.8X5+Tsin30...
Original post by Cool story bro
R=9.8X5+Tsin30...


Ye i did

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


But how do you get your answer from that? :confused:
Take T common
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

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