[Smiley Face :)]

Hi, working on this q for a while... needhelp please...

A heaving ring of mass 5 kg is threaded on a fixed rough rod. Coefficient of friction between rod and ring is 1/2... A light string is attatched to the string and pulled down at 30 degrees to the horizontal. The magnitude of force T from the light rope is increases from 0. Find the value of T that is just suffiecient to make the equilibrium limiting..

I worked out friction was 24.5 (5x9.8x0.5)
Then I worked out the the horizontal force of the string would be Tcos(30)
So i did
Tcos(30)- 24.5 = 0
Thus, T =28.3 N
The answer is actually 39.8... Where am i going wrong please??

[TheMagicMan]

(Original post by Smiley Face :))
Hi, working on this q for a while... needhelp please...

A heaving ring of mass 5 kg is threaded on a fixed rough rod. Coefficient of friction between rod and ring is 1/2... A light string is attatched to the string and pulled down at 30 degrees to the horizontal. The magnitude of force T from the light rope is increases from 0. Find the value of T that is just suffiecient to make the equilibrium limiting..

I worked out friction was 24.5 (5x9.8x0.5)
Then I worked out the the horizontal force of the string would be Tcos(30)
So i did
Tcos(30)- 24.5 = 0
Thus, T =28.3 N
The answer is actually 39.8... Where am i going wrong please??

Remember that when you resolve vertically you need to take into account the vertical component of the tension

[Smiley Face :)]

(Original post by TheMagicMan)
Remember that when you resolve vertically you need to take into account the vertical component of the tension

So would I have to go 5g-Tsin30=0?

[TheMagicMan]

(Original post by Smiley Face :))

So would I have to go 5g-Tsin30=0?

I would draw all the forces acting on the ring in a diagram: there is a reaction force acting upwards, gravity acting downwards, a frictional force horizontally, and the tension acting obliquely (is that a word?). Then resolve for some simlutaneous equations.

[Smiley Face :)]

(Original post by TheMagicMan)
I would draw all the forces acting on the ring in a diagram: there is a reaction force acting upwards, gravity acting downwards, a frictional force horizontally, and the tension acting obliquely (is that a word?). Then resolve for some simlutaneous equations.

I still can't do it.... Would it possible for u to add the solutions in please??

[TheMagicMan]

(Original post by Smiley Face :))
I still can't do it.... Would it possible for u to add the solutions in please??

We aren't allowed to do full solutions:

Resolving horizontally you should get that

You can then use this to substitute into your vertical equation

[Smiley Face :)]

(Original post by TheMagicMan)
We aren't allowed to do full solutions:

Resolving horizontally you should get that

You can then use this to substitute into your vertical equation

Thankkkks... now I need to figure out how to get that verticle eqn .... hmm