ALL CIE PHYSICS DOUBTS HERE !!!

[ashish]

Do you mind to explain how you got your answers

b)(iii) It's not necessary that resultant is zero. The frictional force between the bed and the body of the skier may be greater or equal to the horizontal force.

@ ashish : Weight cannot be the answer since the latter acts vertically downwards ans will have no horizontal component 

yeah you are right but are my answers right?

[Deadly_king]

yeah you are right but are my answers right?

Am not sure about it myself

Let horizontal component of the tension be T_H and the vertical component to be T_V while T is the tension in the string.

Since the system is in equilibrium,  T_V = 8(9.81) -----> T x sin 40 = 8(9.81)
Therefore the tension in the string is 8(9.81)/sin 40 = 122 N

Horizontal component will be T_H =  T x cos 40 = 122 x cos 40 = 93.5 N

I have a feeling I made a mistake somewhere but can't really find out where
Can anyone confirm the answer

[nid404]

I got the same answers as Deadly king for the first and second part.

3rd one I'm not sure of either 

[Deadly_king]

I got the same answers as Deadly king for the first and second part.

3rd one I'm not sure of either  

Alright then.

For the third one, am pretty sure both answers(mine and Asif) will be accepted.

But I'll prefer mine since the question has not stated that the system is in limiting equilibrium. If that was the case, then asif's answer would have been best.

[ashish]

Am not sure about it myself

Let horizontal component of the tension be T_H and the vertical component to be T_V while T is the tension in the string.

Since the system is in equilibrium,  T_V = 8(9.81) -----> T x sin 40 = 8(9.81)
Therefore the tension in the string is 8(9.81)/sin 40 = 122 N

Horizontal component will be T_H =  T x cos 40 = 122 x cos 40 = 93.5 N

I have a feeling I made a mistake somewhere but can't really find out where
Can anyone confirm the answer

i have a doubt, the 122 N will be the tension in the inclined string  which will cause the upward lift?

[Deadly_king]

i have a doubt, the 122 N will be the tension in the inclined string  which will cause the upward lift?

I don't think the upward lift is applied here.

The tension along the whole piece of string will be the same. (122 N)

[lana]

hey guys !! =]
do you have any advice on how to prepare for the practical paper 3 ?? and could someone please explain to me an easy way to find uncertainties ??
thank you...

[ashish]

I don't think the upward lift is applied here.

The tension along the whole piece of string will be the same. (122 N)

hey deadly i kept on thinking on it ,the tension in the string cannot be 122N throughout it has to be 9.81*8 !

[S.M.A.T]

Do you mind to explain how you got your answers

b)(iii) It's not necessary that resultant is zero. The frictional force between the bed and the body of the skier may be greater or equal to the horizontal force.

@ ashish : Weight cannot be the answer since the latter acts vertically downwards ans will have no horizontal component  

DK   frictional force cannot be greater than the horizontal force,if it was greater than the body will start to move to the left .But u can still say that this is not the maximum frictional force that the bed can apply on the body because the question did not said that  the body is in limiting equilibrium so if horizontal force increase than the frictional force will increase.
Since the body did not move(i.e. in equilibrium) that means frictional force=horizontal force
I hope u understood what i said

[nid404]

Limiting friction is greater than the horizontal force. Friction can take any value from 0 to uR.

[S.M.A.T]

Limiting friction is greater than the horizontal force. Friction can take any value from 0 to uR.

yes this is what i meant to say 

So when an object is in equilibrium,the frictional force will take a value equal to the horizontal force

[nid404]

yes this is what i meant to say 

So when an object is in equilibrium,the frictional force will take a value equal to the horizontal force

Bingo

[S.M.A.T]

OK, I think this is slightly harder than the previous ones that I am struggling with.  I know that the tension is the same all round the system, but I don't know which directions each one are (could someone draw the arrows to show them, thanks!)

My (wrong) answers:

a) For equlibrium to be achieved, forces are balanced, zero resultant.

b) i) 141N

    ii) 131N

    iii) No idea!  Difference is not 0N, I'm stuck!

Many thanks in advance!

DK here is the working

b)i and ii

[Deadly_king]

yes this is what i meant to say  

So when an object is in equilibrium,the frictional force will take a value equal to the horizontal force

Understood you perfectly br0

I had the same thinking in mind but i think i misplaced my words. When I said frictional force is greater than the tension, I was referring to the maximum constant value that the friction may reach

By the way good job

[ashish]

DK here is the working

b)i and ii

nice work thought i posted the same working check my post , lol the answer was simple it was just that i didn't concentrate .... EXAMMMMMMMMMMMMMMMMMMs ss