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Hardest Question Ever in M1
Part c). No chance today if the question is like that! - attached.
Thanks for any help
Thanks for any help
That's not so bad :/
As long as you realise that the reaction is zero, and can think of something to put for one of those annoying 'wordy' questions, it's ok.
The worst one I found was from a solomon paper (L maybe?) where there were two triangles with different angles, and a ball rolling down each.
EDIT: Are you asking for an answer or just saying that it's hard? I'm confused.
As long as you realise that the reaction is zero, and can think of something to put for one of those annoying 'wordy' questions, it's ok.
The worst one I found was from a solomon paper (L maybe?) where there were two triangles with different angles, and a ball rolling down each.
EDIT: Are you asking for an answer or just saying that it's hard? I'm confused.
Speleo
That's not so bad :/
As long as you realise that the reaction is zero, and can think of something to put for one of those annoying 'wordy' questions, it's ok.
The worst one I found was from a solomon paper (L maybe?) where there were two triangles with different angles, and a ball rolling down each.
EDIT: Are you asking for an answer or just saying that it's hard? I'm confused.
As long as you realise that the reaction is zero, and can think of something to put for one of those annoying 'wordy' questions, it's ok.
The worst one I found was from a solomon paper (L maybe?) where there were two triangles with different angles, and a ball rolling down each.
EDIT: Are you asking for an answer or just saying that it's hard? I'm confused.
It all seems like a blur to me! A solution for it would be good
Thanks xHow did you get the paper? Did you blag a copy from school or something lol.
a) 0 N, as it is about to tilt over the other hinge.
b) Taking moments about D,
5 x 1500 = 2W (W here is the weight)
=> W = 3750 N
c) Draw the new diagram.
With new info, taking moments about C, we have:
5 x 1000 = Wx => x = 5000/W [1]
Adapting old info, taking moments about D, we have:
7500 = W(4-x) (instead of 2W) [2]
Subsitituting [1] in [2], we have:
7500 = 4W - 5000
=> W = 3125 N
d) Using [1], we have x = 5000/3125 = 1.6 N
e) Modelling as a rod assumes all mass of the log is concentrated on a line, assuming it has length only and ignoring its width and breadth.
Hope this helps.
b) Taking moments about D,
5 x 1500 = 2W (W here is the weight)
=> W = 3750 N
c) Draw the new diagram.
With new info, taking moments about C, we have:
5 x 1000 = Wx => x = 5000/W [1]
Adapting old info, taking moments about D, we have:
7500 = W(4-x) (instead of 2W) [2]
Subsitituting [1] in [2], we have:
7500 = 4W - 5000
=> W = 3125 N
d) Using [1], we have x = 5000/3125 = 1.6 N
e) Modelling as a rod assumes all mass of the log is concentrated on a line, assuming it has length only and ignoring its width and breadth.
Hope this helps.
/\Shaz\/
a) 0 N, as it is about to tilt over the other hinge.
b) Taking moments about D,
5 x 1500 = 2W (W here is the weight)
=> W = 3750 N
c) Draw the new diagram.
With new info, taking moments about C, we have:
5 x 1000 = Wx => x = 5000/W [1]
Adapting old info, taking moments about D, we have:
7500 = W(4-x) (instead of 2W) [2]
Subsitituting [1] in [2], we have:
7500 = 4W - 5000
=> W = 3125 N
d) Using [1], we have x = 5000/3125 = 1.6 N
e) Modelling as a rod assumes all mass of the log is concentrated on a line, assuming it has length only and ignoring its width and breadth.
Hope this helps.
b) Taking moments about D,
5 x 1500 = 2W (W here is the weight)
=> W = 3750 N
c) Draw the new diagram.
With new info, taking moments about C, we have:
5 x 1000 = Wx => x = 5000/W [1]
Adapting old info, taking moments about D, we have:
7500 = W(4-x) (instead of 2W) [2]
Subsitituting [1] in [2], we have:
7500 = 4W - 5000
=> W = 3125 N
d) Using [1], we have x = 5000/3125 = 1.6 N
e) Modelling as a rod assumes all mass of the log is concentrated on a line, assuming it has length only and ignoring its width and breadth.
Hope this helps.
The bit highlighted in red. Are you allowed to do that? Because the 2 models are different?
Srathmore
The bit highlighted in red. Are you allowed to do that? Because the 2 models are different?
His test was under initial assumptions that were wrong, so yes, of course you can in light of this knowledge. It is the assumptions that have changed, and not the results. Besides, I can actually remember this exact question; I did it 2 or 3 years ago.
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