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Shouldn't it be -8sin105 instead of -8sin15? M1

Shouldn't it be -8sin105 instead of -8sin15?:s-smilie:
And if it's -8sin15 then why is it a sine then and not a cosine so shouldn't it be -8cos15? :redface:

(edited 7 years ago)
Untitledeed.png111.3KB
Reply 1
Avatar for B_9710
B_9710
18
The angle between the vertical and the 8N force is 15 degrees. It is some for resolving in the horizontal not cosine (in this case). You 'always 'use right angle triangles when resolving forces (pretty much, especially in cases like this) so you won't use obtuse angles, only acute angles.
See attached:

Untitledeed.png
Reply 3
Avatar for Adorable98
Adorable98
OP
15
Original post by B_9710
The angle between the vertical and the 8N force is 15 degrees. It is some for resolving in the horizontal not cosine (in this case). You 'always 'use right angle triangles when resolving forces (pretty much, especially in cases like this) so you won't use obtuse angles, only acute angles.


So say the diagram looked like this instead, so would the horizontal component be 10sin15 in this case?:s-smilie:
hw.png
No nothing about the two forces you've got left has changed so resolving horizantally (->) you would still get 10-8sin15.
Reply 5
Avatar for Adorable98
Adorable98
OP
15
Original post by plaguarist
No nothing about the two forces you've got left has changed so resolving horizantally (->:wink: you would still get 10-8sin15.


Original post by ghostwalker
See attached:

Untitledeed.png


Original post by B_9710
The angle between the vertical and the 8N force is 15 degrees. It is some for resolving in the horizontal not cosine (in this case). You 'always 'use right angle triangles when resolving forces (pretty much, especially in cases like this) so you won't use obtuse angles, only acute angles.

Thank you !! :smile::h:

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