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first draw the free body diagram of ring showing all the forces acting on it
wat abt that 5N force? which component acts in the upward direction to oppose weight mg?
5cos30 =4.33N oH OK NOW THE FRICTIONAL FORCE =6.0-4.33=1.67N
yes but how? dont u want to know why?
add that 5cos@ component in ur free body diagram
yes so altogether? how does it look on the body?
its |dw:1332947814228:dw|clear on the body
thats it u did it! dont worry about horizontal component of 5N(5sin30) as the bar will balance this force by its normal force(anyway it is rigid) so for the next part wat is frictional force=? in terms of co-eff of friction
f=cefficient of friction xR
u can find it?
1.67=cofficientx4.33 coefficient=1.67/4.33 =0.385
yes u deserve a medal....
u have the anser for the 2nd question,i have a 2nd thought on it
but sadly the marking scheme says that R=5 SIN30 =2.5 And the constant of friction is 0.668
i thought so!
wat is normal reacftion exerted by a body?
normal recation=mg i think
no wat is it actually the concept /meaning behind it? normal=====perpendicular have u heard abt it?
normal reaction is at right angles with surfaces
right angle wit the surfaces in contact theree can be a normal force from a body in contact with another surface/body perpendicular to the plane in which both lie check this for any other place where u used normal reaction so here the only force normal to the plane of ring and touching tube is only 5sin30 get it?
wow awesome :)
thanks boss this was a clever question it needs a free and relaxed mind,otherwise if not so and if it was exam exam it was gonna give me a nice time
np anytime all u have to do is get wat u learn deep in your mind in a clear way:) thank u for helping me figure out the normal reaction part(it was tld to me long ago by my sis)
ok what are about if the surface was smooth would that change anything?
then will the upward force 5cos30 force balance the weight mg? will it be enough to keep it in equilibrium?
no it wouldnt and it wouldnt be enough to balance the equilibrum
so? it will fall down with a force mg-5co30
therefore force of friction will be that right
i have to go now post ur question here and send me the link bye!
ok thank u boss