# Maths help! Exam in 2 days:(

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Thread starter 3 years ago
#1
Can someone please help with 5 iii - I dunno what they mean in the ms when it says "Resolve horiz and establish tensions equal E1
Resolve vert to show inconsistency " -how are tensions equal horizontally?
http://mei.org.uk/files/papers/ouikyuj45y.pdf
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3 years ago
#2
(Original post by Kalabamboo)
Can someone please help with 5 iii - I dunno what they mean in the ms when it says "Resolve horiz and establish tensions equal E1
Resolve vert to show inconsistency " -how are tensions equal horizontally?
http://mei.org.uk/files/papers/ouikyuj45y.pdf
Resolving horizontally for the box we have: Sines cancel and so tensions are equal.
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Thread starter 3 years ago
#3
(Original post by ghostwalker)
Resolving horizontally for the box we have: Sines cancel and so tensions are equal.
Thanks but I thought that the tensions AB and BC are the same anyway cause only the angle has changed - so the vertical or horizontal components must've changed to keep the tensions the same
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3 years ago
#4
(Original post by Kalabamboo)
Can someone please help with 5 iii - I dunno what they mean in the ms when it says "Resolve horiz and establish tensions equal E1
Resolve vert to show inconsistency " -how are tensions equal horizontally?
http://mei.org.uk/files/papers/ouikyuj45y.pdf
In response to your first (general) question, it means that we show the horizontal components of the tensions in AB and BC to be equal, so there is no net horizontal force on the box, but there is a net vertical force, due to the weight as well as the vertical components of the two tensions.

For your second question, we're given that alpha = 60, which matches the 60 degree angle between the weight and BC ("String BC is at 60 degrees to the vertical"). Thus if we resolve, we get tension in AB * cos(90-60) = tension in BC * sin(60), so since cos(90-x) = sin(x) for all x, the trig terms cancel and we get tension in AB = tension in BC. Thus their vertical components will be equal and cancel out, just like their horizontal components, but therefore there will be a net force downwards from the weight, meaning that the box cannot be in equilibrium, as required.
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3 years ago
#5
(Original post by Kalabamboo)
Thanks but I thought that the tensions AB and BC are the same anyway cause only the angle has changed - so the vertical or horizontal components must've changed to keep the tensions the same
No, the whole point is that there's no reason why the tensions should be the same, and indeed in part (ii) (the case where alpha = 30 degrees), you should have found that the tensions were different. The tensions change based on the angles, not the other way round.
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Thread starter 3 years ago
#6
(Original post by Prasiortle)
In response to your first (general) question, it means that we show the horizontal components of the tensions in AB and BC to be equal, so there is no net horizontal force on the box, but there is a net vertical force, due to the weight as well as the vertical components of the two tensions.

For your second question, we're given that alpha = 60, which matches the 60 degree angle between the weight and BC ("String BC is at 60 degrees to the vertical" . Thus if we resolve, we get tension in AB * cos(90-60) = tension in BC * sin(60), so since cos(90-x) = sin(x) for all x, the trig terms cancel and we get tension in AB = tension in BC. Thus their vertical components will be equal and cancel out, just like their horizontal components, but therefore there will be a net force downwards from the weight, meaning that the box cannot be in equilibrium, as required.
Thanks! I think I am getting this but one thing that it is bugging me is why do you equate them like Ghostwalker did:

This bit's confusing cause we are told that they are not in equilibrium so we can't really equate them.
Here is what I've jot down on my paper -sorry if it's not a lot of progress . Can you please help me oh it's not letting me post it for some reasonn 0
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3 years ago
#7
(Original post by Kalabamboo)
Thanks! I think I am getting this but one thing that it is bugging me is why do you equate them like Ghostwalker did:

This bit's confusing cause we are told that they are not in equilibrium so we can't really equate them.
Here is what I've jot down on my paper -sorry if it's not a lot of progress . Can you please help me oh it's not letting me post it for some reasonn It's a proof by contradiction.

If alpha = 60, then resolving horizontally the tensions must be equal.

Resolving vertically, the tensions cannot be equal - due to the weight of the box.

This is a contradiction, the tensions cannot be equal and not equal at the same time, hence alpha cannot be 60.
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3 years ago
#8
(Original post by Kalabamboo)
Thanks! I think I am getting this but one thing that it is bugging me is why do you equate them like Ghostwalker did:

This bit's confusing cause we are told that they are not in equilibrium so we can't really equate them.
Here is what I've jot down on my paper -sorry if it's not a lot of progress . Can you please help me oh it's not letting me post it for some reasonn We assume that the box is in equilibrium, and thus the net horizontal force should be 0. But then, from this assumption, we show that the net vertical force is non-zero, so the box is not in equilibrium, contradicting our initial assumption. Thus our initial assumption that the box is in equilibrium must have been wrong. It's a proof by contradiction.
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Thread starter 3 years ago
#9
(Original post by ghostwalker)
It's a proof by contradiction.

If alpha = 60, then resolving horizontally the tensions must be equal.

Resolving vertically, the tensions cannot be equal - due to the weight of the box.

This is a contradiction, the tensions cannot be equal and not equal at the same time, hence alpha cannot be 60.
Ah proof So we just say Tba and Tbc is the same despite the fact that an angle has changeed?
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Thread starter 3 years ago
#10
(Original post by Prasiortle)
We assume that the box is in equilibrium, and thus the net horizontal force should be 0. But then, from this assumption, we show that the net vertical force is non-zero, so the box is not in equilibrium, contradicting our initial assumption. Thus our initial assumption that the box is in equilibrium must have been wrong. It's a proof by contradiction.
Ah proof So we just say Tba and Tbc is the same despite the fact that an angle has changeed?
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3 years ago
#11
(Original post by Kalabamboo)
Ah proof So we just say Tba and Tbc is the same despite the fact that an angle has changeed?
You have to understand how a proof by contradiction works. We're not asserting that the tensions are the same; we assume that they're the same, and then add what the consequences of the assumption are.
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