NotTim
I need help understanding diagrams.
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NotTim
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For Example
NotTim
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|dw:1359418634316:dw|
NotTim
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so where would the 45 degree angle go i nthe triangle?
NotTim
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i never did get it right
shubhamsrg
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|dw:1359448218371:dw|
shubhamsrg
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This should help ?
NotTim
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|dw:1359419770094:dw| @shubhamsrg ?
shubhamsrg
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45 will be special case where all angles be 45
Leme explain you a general case, with angle = x
NotTim
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say it wasn't 45...
shubhamsrg
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shubhamsrg
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Vincent-Lyon.Fr
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Hint: Except in geometrical solving of problems, NEVER EVER draw general diagrams with 45° angles.
Always draw 20° to 30° angles. It will save you a lot of time.
shubhamsrg
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shubhamsrg
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Vincent-Lyon.Fr
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NotTim
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|dw:1359420708200:dw| So I know what is going on, then this would be correct also?
Vincent-Lyon.Fr
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WHere does this direction point to?
If it is horizontal, then you are wrong again. "Your" angle is the other one.
NotTim
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@Vincent-Lyon.Fr how do you get that?
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just remember when parallel to the slope means: Force produced by the weight-> mgsinx
if its perpendicular ( normal ) to the slope, use mgcosx
NotTim
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the main slope, or the new, smaller slope?
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the original one, below the slope, angle use 45 degrees
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For x axis : is always sine
For y axis : is always cosine
NotTim
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wait. sorry, so hwo can i derive where the angle will be?
tcy
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just use this angle for both axis|dw:1359458196671:dw|
NotTim
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wait...so how do i know the angle for the hypothetical top triangle?
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NotTim
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Ok. To anyone else answering, I will be leaving within 5 minutes. Thamks for all the help guys! At the very least, I know where the angles go, even if I do not knw how to derive it.
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you have to know your trigonometry well and have to look at different angles to solve this problem
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For the force directed on x axis due to resultant force, F on x = Fcos(30)
For the force directed on y axis due to resultant force, F on y = Fsin(30)
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Then,
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