# resolving forcesWatch

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#1
i've been reading through my revision guide and completing questions on resolving forces but one thing i seem to be mixing up is when to use sin or cos when resolving horizontally/vertically.
i thought the rule was you use sin for vertical forces and cos for horizontal but apparently not. would appreciate some clarification 0
1 week ago
#2
(Original post by entertainmyfaith)
i've been reading through my revision guide and completing questions on resolving forces but one thing i seem to be mixing up is when to use sin or cos when resolving horizontally/vertically.
i thought the rule was you use sin for vertical forces and cos for horizontal but apparently not. would appreciate some clarification You can't just follow a rule like that - you need to understand the trigonometry. Do you have an example of a question where you get mixed up?
0
1 week ago
#3
(Original post by entertainmyfaith)
i've been reading through my revision guide and completing questions on resolving forces but one thing i seem to be mixing up is when to use sin or cos when resolving horizontally/vertically.
i thought the rule was you use sin for vertical forces and cos for horizontal but apparently not. would appreciate some clarification This is correct - so long as you are using the angle between the vector and the horizontal. If you are using the angle between the vector and the vertical, it's the other way around.
0
1 week ago
#4 Hope this helps. You just use trigonometry and rearrange for the horizontal and vertical components. I always draw a triangle and work it out as if it’s a trigonometry question - much easier than having to remember it!
1
#5
(Original post by Sir Cumference)
You can't just follow a rule like that - you need to understand the trigonometry. Do you have an example of a question where you get mixed up? when resolving forces horizontally i took it as T1cos35 - T2cos35 = 0 but the mark scheme used sin (Original post by Pangol)
This is correct - so long as you are using the angle between the vector and the horizontal. If you are using the angle between the vector and the vertical, it's the other way around.
oh, i'll keep that in mind!! thanks 0
1 week ago
#6
(Original post by entertainmyfaith)
when resolving forces horizontally i took it as T1cos35 - T2cos35 = 0 but the mark scheme used sin The horizontal component of the force is the opposite side from the angle of 35 degrees (when you construct a triangle) hence why you use sine.

As said before, you need to understand trigonometry rather than follow a rule which doesn't always hold.
2
1 week ago
#7
(Original post by entertainmyfaith) when resolving forces horizontally i took it as T1cos35 - T2cos35 = 0 but the mark scheme used sin oh, i'll keep that in mind!! thanks A force with magnitude F can always be split into two perpendicular components:

- The component adjacent to the angle is - The component opposite to the angle is If you think of it in terms of GCSE trig, the original force is the hypotenuse of a right angled triangle and the two components are the other two sides. Here I have split the force on the right into two perpendicular components. The red component is adjacent to the angle 35 so it is equal to and the blue component is opposite the angle so it is . If instead you consider the angle 55 (90 - 35) then the blue component is since it is now adjacent to the angle and the red component is - you get the same value if you use 35.

You may prefer a triangle instead of the above so you could visualise it like this: The same applies here : the blue horizontal component is opposite the angle 35 so must be and the vertical component is .
Last edited by Sir Cumference; 1 week ago
2
#8
(Original post by Sir Cumference)
A force with magnitude F can always be split into two perpendicular components:

- The component adjacent to the angle is - The component opposite to the angle is If you think of it in terms of GCSE trig, the original force is the hypotenuse of a right angled triangle and the two components are the other two sides. Here I have split the force on the right into two perpendicular components. The red component is adjacent to the angle 35 so it is equal to and the blue component is opposite the angle so it is . If instead you consider the angle 65 (90 - 35) then the blue component is since it is now adjacent to the angle and the red component is - you get the same value if you use 35.

You may prefer a triangle instead of the above so you could visualise it like this: The same applies here : the blue horizontal component is opposite the angle 35 so must be and the vertical component is .
oh this makes it a lot easier to understand with the lines!! thank you Last edited by entertainmyfaith; 1 week ago
0
1 week ago
#9
(Original post by entertainmyfaith) when resolving forces horizontally i took it as T1cos35 - T2cos35 = 0 but the mark scheme used sin oh, i'll keep that in mind!! thanks It's the cosine of the angle you 'go across' to the direction you are resolving in . so here you are going across 90- 35 so it is cos 55 [or sin 35]
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