HSC 2013 Test Paper - HELP with mechanics (1 Viewer)

[JoshuaPutra]

So I am having trouble with the understanding QUESTION 16 (b) parts (iv) and (v). I will link the test paper and the marking guidelines below but yet I am still having trouble with understanding the solutions. In general, when you are dealing with conical pendulums or even banked tracks, how do you resolve the forces into their vertical and horizontal components? ANY help is appreciated!

Mathematics Extension 2 HSC Examination Paper: http://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-hsc-maths-ext-2.pdf

Mathematics Extension 2 HSC Marking Guidelines: http://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-marking-guide-maths-ext-2.pdf

 

[Carrotsticks]

Have a read of this and let me know if any parts are ambiguous.

View attachment HSC 2013 Extension 2 Solutions.pdf

 

[JoshuaPutra]

It makes sense but when you are resolving the forces vertically, where do you get TsinB - Tsin (QPS) = mg? I understand why it is equal to mg but i am unsure of the LHS. Similarly for resolving the forces horizontally i don't understand the LHS in TcosB + Tcos (QPS) = mrw^2

But thank you for linking me that pdf, it offered a different method that explains it more in depth!

 

[Carrotsticks]

It makes sense but when you are resolving the forces vertically, where do you get TsinB - Tsin (QPS) = mg? I understand why it is equal to mg but i am unsure of the LHS. Similarly for resolving the forces horizontally i don't understand the LHS in TcosB + Tcos (QPS) = mrw^2

But thank you for linking me that pdf, it offered a different method that explains it more in depth!

Apologies for the stuffed up formatting! I think I sent you the draft by accident, but the content is still the same =)

Here is a better version View attachment HSC 2013 Extension 2 Solutions.pdf

When you draw the vertical vectors, you can see that the vertical component of T (from beta) is pointing upwards, so it is positive. However, the vertical component of T (from angle QPS) is pointing downwards, so we flip it upwards with the minus sign. In general, for resolving forces vertically, all forces going up = mg. So if you've got something that is pointing downwards, you make it point upwards by throwing in a minus sign.

For the horizontal part, it works just like any other conical pendulum problem. Have you done many problems with conical pendulums, where you have two strings?

You want to analyse what forces make up centripetal force (as it cannot exist by itself, it must be provided by something).

In this case, it is provided by the horizontal component of both T vectors. The horizontal components, added together, provide the centripetal force of the particle.

 

[JoshuaPutra]

I understand that! I'm just used to having the vertical forces including Cos. I guess it is different when there are two strings?

 

[JoshuaPutra]

Sorry to be pain but how do you go about drawing vertical and horizontal vectors? My teacher only touched on conical pendulums so it's not my strongest point!

 

[Carrotsticks]

I understand that! I'm just used to having the vertical forces including Cos. I guess it is different when there are two strings?

It depends from which angle you're looking at.

Best not to memorise that it 'should' be cos or sine etc. Just work it out case-by-case.

Students memorising that things 'should' be cos or sine etc were tripped up in the conical pendulums problem from the 2011 HSC, which had cosec in it!

Sorry to be pain but how do you go about drawing vertical and horizontal vectors? My teacher only touched on conical pendulums so it's not my strongest point!

See both vectors V? Pretend they are the hypotenuse(es?, hypoteni?) of two right angled triangles. Draw the other two sides with arrows as I've done below (excuse the shoddy diagram, I don't have my tablet with me atm).

So T is the dark bolded vector and the vertical/horizontal components are the smaller green ones. When you draw it out, you'll be able to see which one is pointing the 'wrong way'.

Mechanics is a topic best taught in person =/

 

[faisalabdul16]

It depends from which angle you're looking at.

Best not to memorise that it 'should' be cos or sine etc. Just work it out case-by-case.

Students memorising that things 'should' be cos or sine etc were tripped up in the conical pendulums problem from the 2011 HSC, which had cosec in it!



See both vectors V? Pretend they are the hypotenuse(es?, hypoteni?) of two right angled triangles. Draw the other two sides with arrows as I've done below (excuse the shoddy diagram, I don't have my tablet with me atm).

So T is the dark bolded vector and the vertical/horizontal components are the smaller green ones. When you draw it out, you'll be able to see which one is pointing the 'wrong way'.

Mechanics is a topic best taught in person =/

Bold part - agreed! It's hard learning mechanics online or even teaching it online

 

[kimsungho]

Oh that question. was just doing that paper. it's got to be one of the finest question BoS has made...