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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

i thought the rule was you use sin for vertical forces and cos for horizontal but apparently not. would appreciate some clarification

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(Original post by

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

**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

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#3

**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

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#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!

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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?

**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 T

_{1}cos35 - T

_{2}cos35 = 0 but the mark scheme used sin

(Original post by

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.

**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.

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when resolving forces horizontally i took it as T

**entertainmyfaith**)when resolving forces horizontally i took it as T

_{1}cos35 - T_{2}cos35 = 0 but the mark scheme used sin**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.

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#7

(Original post by

when resolving forces horizontally i took it as T

oh, i'll keep that in mind!! thanks

**entertainmyfaith**)when resolving forces horizontally i took it as T

_{1}cos35 - T_{2}cos35 = 0 but the mark scheme used sinoh, i'll keep that in mind!! thanks

- 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

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(Original post by

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 .

**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 .

Last edited by entertainmyfaith; 1 week ago

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**entertainmyfaith**)

when resolving forces horizontally i took it as T

_{1}cos35 - T

_{2}cos35 = 0 but the mark scheme used sin

oh, i'll keep that in mind!! thanks

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