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## Homework Statement

if the alpha is larger than 90 degree ( which means the resultant F is lean towards -x axis , then the angle between the Ax and the line from Ax to A will be less than 90 , am i right ?

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- Thread starter werson tan
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- #1

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if the alpha is larger than 90 degree ( which means the resultant F is lean towards -x axis , then the angle between the Ax and the line from Ax to A will be less than 90 , am i right ?

- #2

BvU

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No. It will be exactly 90##^\circ##

(the 90##^\circ## as indicated in figure 2-26 (a) )

(the 90##^\circ## as indicated in figure 2-26 (a) )

- #3

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why , i cant visualize it??No. It will be exactly 90##^\circ##

(the 90##^\circ## as indicated in figure 2-26 (a) )

- #4

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If the force is lean towards negative x axis , it means the angle between the force and positive x -axis is more than 90 degree, right ? then how is it possible for it to remain 90 degree?No. It will be exactly 90##^\circ##

(the 90##^\circ## as indicated in figure 2-26 (a) )

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ya , i mean the line from Ax to A, not about the line from the origin to A (i.e. A⃗ itself).

if the force is lean towards the negative x-axis then the , we would have three line , right , namely the line joining origin to Ax , A(which is lean towards negative x axis ) , and a line which join A to negative x-axis , right ? if it is so , then the i an undersatnd why the angle is 90 degree. if it so , the triangle formed is at opposite of the current one .

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BvU

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I think you've got it all right and correct, just that the wording in post #1 was rather unfortunate.

With nowadays visualization possibilities it's almost a stone-age approach to put things in words, but let's do it anyway:

let's look in the blue plane of figure 2-26 (a) for the case ##\alpha \in [\pi/2, \pi]## where all the angles in our story are

so the vertical line A_{yz} is in the yz plane and that plane is perpendicular to the x axis.

and as you can see the line from A_{x} to A (the green line) is parallel to that plane and therefore also perpendicular to the x axis. Hence my 90 ##^\circ##.

You of course in your post #1 meant to refer to the angle between ##\vec A## and ##\vec A_x## which as you can see is ##\pi-\alpha ## and thereby in the range 0 to 90##^\circ##.

With nowadays visualization possibilities it's almost a stone-age approach to put things in words, but let's do it anyway:

let's look in the blue plane of figure 2-26 (a) for the case ##\alpha \in [\pi/2, \pi]## where all the angles in our story are

[edit] I mean to say that all the angles are __in that plane__. Sorry for the ambiguity.

so the vertical line A

and as you can see the line from A

You of course in your post #1 meant to refer to the angle between ##\vec A## and ##\vec A_x## which as you can see is ##\pi-\alpha ## and thereby in the range 0 to 90##^\circ##.

Note that I forgot to draw the arrow above ##\vec A_x## and ##\vec A_{yz}##. Nobody is perfect, be we keep trying.

To be specific: in ##\vec A_x = A_x\, \hat\imath\ ##, ##\ A_x## is a number (negative for the alpha in the picture: ##A_x = |\vec A | \cos\alpha##

To be specific: in ##\vec A_x = A_x\, \hat\imath\ ##, ##\ A_x## is a number (negative for the alpha in the picture: ##A_x = |\vec A | \cos\alpha##

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