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M1 statics help

Please see the photos. It's question 4. Why does my 1st method not work but my second method does? You should be able to use either method to reach the answer. What am I doing wrong?
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In your first method you haven't resolved all of the forces, there should be a term for both tensions in each of those two equations :smile:
Original post by Jordan\
In your first method you haven't resolved all of the forces, there should be a term for both tensions in each of those two equations :smile:


I literally just figured that out! So I've been trying to work out the component of T1 acting in the T2 direction and vice versa.....is it just T1cos75? And then T2cos75 in the T1 direction? Or what?


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Original post by anoymous1111
I literally just figured that out! So I've been trying to work out the component of T1 acting in the T2 direction and vice versa.....is it just T1cos75? And then T2cos75 in the T1 direction? Or what?


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Actually would that not work because it's not a right angled triangle....so then it gets too complicated and it's easier to just used the 2nd method?


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Original post by anoymous1111
I literally just figured that out! So I've been trying to work out the component of T1 acting in the T2 direction and vice versa.....is it just T1cos75? And then T2cos75 in the T1 direction? Or what?


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Yes, see below.

Original post by anoymous1111
Actually would that not work because it's not a right angled triangle....so then it gets too complicated and it's easier to just used the 2nd method?


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If you have any two vectors a\mathbf{a} and b\mathbf{b} and the angle between them is θ \theta \ , then the component of a\mathbf{a} that acts in the direction of b\mathbf{b} is going to be acosθ\mathbf{a} \cos \theta and vice versa.

You can construct a right-angled triangle of the situation to convince yourself that this is the case. :smile:
Original post by Star-girl
Yes, see below.



If you have any two vectors a\mathbf{a} and b\mathbf{b} and the angle between them is θ \theta \ , then the component of a\mathbf{a} that acts in the direction of b\mathbf{b} is going to be acosθ\mathbf{a} \cos \theta and vice versa.

You can construct a right-angled triangle of the situation to convince yourself that this is the case. :smile:


Thank you! But I'm still getting the wrong answer. Could you write out the working for this question for me? I'd appreciate it soooooo much




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Original post by anoymous1111
Thank you! But I'm still getting the wrong answer. Could you write out the working for this question for me? I'd appreciate it soooooo much




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Here you go.

20150805_001308.jpg

Not quite sure why it's come out sideways, but oh well. :lol:
Original post by Star-girl
Here you go.

20150805_001308.jpg

Not quite sure why it's come out sideways, but oh well. :lol:


Thank you soooooo much that's so kind of you!


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Original post by anoymous1111
Thank you soooooo much that's so kind of you!


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No problem. :smile:
(edited 8 years ago)

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