Further Mechanics Question

I'm having trouble with part a of this question.

An elastic string PA has natural length 0.5m and modulus of elasticity 9.8 N. The string PB is inextensible. The end A of the elastic string and the end B of the inextensible string are attached to two fixed points which are on the same horizontal level. The end P of each string is attached to a 2kg particle. The particle hangs in equilibrium below AB, with PA making an angle of 30° with AB and PA perpendicular to PB. Find:

a the length of PA (7 marks)
b the length of PB (2 marks)
¢ the tension in PB. (2 marks)

Looking at the mark scheme, surely, since cos uses CAH, the hypotenuse T1 should = 2g over cos60?

Am I being stupid or is this just an error in the textbook?

Mark Scheme: Question 10, Page 3 - https://www.activeteachonline.com/default/player/document/id/725771/external/0/uid/357726
(edited 9 months ago)
Original post by Amy.fallowfield
I'm having trouble with part a of this question.

An elastic string PA has natural length 0.5m and modulus of elasticity 9.8 N. The string PB is inextensible. The end A of the elastic string and the end B of the inextensible string are attached to two fixed points which are on the same horizontal level. The end P of each string is attached to a 2kg particle. The particle hangs in equilibrium below AB, with PA making an angle of 30° with AB and PA perpendicular to PB. Find:

a the length of PA (7 marks)
b the length of PB (2 marks)
¢ the tension in PB. (2 marks)

Looking at the mark scheme, surely, since cos uses CAH, the hypotenuse T1 should = 2g over cos60?

Am I being stupid or is this just an error in the textbook?

Mark Scheme: Question 10, Page 3 - https://www.activeteachonline.com/default/player/document/id/725771/external/0/uid/357726

Youre resolving forces so projecting the weight 2g (hypotenuse) onto PA, so it must be 2gcos(60). The right angle is on the extended line PA, not the vertical weight.
(edited 9 months ago)
Original post by mqb2766
Youre resolving forces so projecting the weight 2g (hypotenuse) onto PA, so it must be 2gcos(60). The right angle is on the extended line PA, not the vertical weight.

Thank you so much. That makes sense now.
Original post by Amy.fallowfield
Thank you so much. That makes sense now.

Its worth being clear about and "always" sketching the right triangle with the original force on the hypotenuse as the aim is almost always to produce two perpendicular projections (the two legs).

If you put the right angle on the vertical weight here, you can still project it onto the hypotenuse (extended PA) and subtract the horizontal to get the original weight, but theyre not perpendicular and both would be involved in equating to the tension (as youd project the horizontal onto PA as well) and youd end up doing more work/getting confused.
(edited 9 months ago)