Resolving forces question

https://e4c57ffe72e64565dd23ce74ff8c96ff1cc0bd77.googledrive.com/host/0B1ZiqBksUHNYc2g4c29zSXplTmc/Set-D/Resolving.pdf

I don't understand part b (ii) of the first question. How would you draw the force diagram and which is the resultant force when F is vertical? Every time I try to draw it, I end up getting the wrong answer because I use Tan(theta) = Opp/Aj but according to the mark scheme, you should use Cos(theta) = Adj/Hyp.... but I don't understand why.

Reply 1

Zacken

Original post by jessyjellytot14

You're talking about b(ii), that's just resolving the tension vertically. There's a

30^{\circ}

degree angle between the tension and the vertical, so resolving the tension in the vertical direction gives you

T \cos 30^{\circ}

Reply 2

KaylaB

Original post by jessyjellytot14

I just basically drew up what Zacken said, along with a little explanation to how we find horizontal and vertical components when the force acts at an angle

Reply 3

Zacken

Original post by KaylaB

I just basically drew up what Zacken said, along with a little explanation to how we find horizontal and vertical components when the force acts at an angle

That's very neat! :love:

Reply 4

KaylaB

Original post by Zacken

That's very neat! :love:

Thank you

Spoiler

Reply 5

Zacken

Original post by KaylaB

Thank you

Spoiler

Mine never has and I've done a lot of maths. :rofl:

Reply 6

jessyjellytot14

Original post by KaylaB

I just basically drew up what Zacken said, along with a little explanation to how we find horizontal and vertical components when the force acts at an angle

Oh wow thank you for taking the time to draw it out :smile:

! In the triangle you drew would y represent the weight force?

Reply 7

Student403

Original post by KaylaB

I just basically drew up what Zacken said, along with a little explanation to how we find horizontal and vertical components when the force acts at an angle

X

Whoa that drawing :redface:

Reply 8

jessyjellytot14

Original post by Zacken

You're talking about b(ii), that's just resolving the tension vertically. There's a

30^{\circ}

degree angle between the tension and the vertical, so resolving the tension in the vertical direction gives you

T \cos 30^{\circ}

Thanks, I would rep you but it won't let me

Reply 9

Zacken

Original post by jessyjellytot14

Thanks, I would rep you but it won't let me

Don't worry about it.

y

would represent the horizontal component of the tension, nothing more.

Reply 10

KaylaB

Original post by jessyjellytot14

Oh wow thank you for taking the time to draw it out :smile:

! In the triangle you drew would y represent the weight force?

Original post by Zacken

Don't worry about it.

y

would represent the horizontal component of the tension, nothing more.

Yeah exactly, y is the horizontal component of the tension. I just labelled it y so that you could work out what it could be using the triangle, which is

T \sin 30^{\circ}

Original post by Student403

Whoa that drawing :redface:

It's one of the only things I'm good for in mechanics :innocent:

Reply 11

Student403

Original post by KaylaB

Yeah exactly, y is the horizontal component of the tension. I just labelled it y so that you could work out what it could be using the triangle, which is