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ALL CIE PHYSICS DOUBTS HERE !!!

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ashish:

--- Quote from: Deadly_king on October 17, 2010, 01:14:11 pm ---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  ;)

--- End quote ---

yeah you are right but are my answers right?

Deadly_king:

--- Quote from: ashish on October 17, 2010, 01:18:32 pm ---yeah you are right but are my answers right?

--- End quote ---

Am not sure about it myself :-\

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

Since the system is in equilibrium,  TV = 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 TH =  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:

--- Quote from: Garfield on October 17, 2010, 02:15:26 pm ---I got the same answers as Deadly king for the first and second part.

3rd one I'm not sure of either  :-\

--- End quote ---

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:

--- Quote from: Deadly_king on October 17, 2010, 01:52:15 pm ---Am not sure about it myself :-\

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

Since the system is in equilibrium,  TV = 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 TH =  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 ???

--- End quote ---

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

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