HSC 2015 mechanics discussion (1 Viewer)

[720YoloSwagScope]

For the youngs module question I think it's a trick question as it asked for GPa and KN/mm^2 is MPa and I got 202MPa so I put in 0.202 GPa

Forgot to put :1 do you think we. Will lose a mark?

I made same mistake as you for :1 thing, who knows :c Hoping for the best, even though we won't ever get to know if we did or didn't get the mark.
And lol, I just converted all my units into m and N rather than mm or kN so I got like 202 GPA

 

[keepLooking]

Assuming this diagram is correct, and let the roller joint be A, and pin joint be B, then:

I'm not sure whether that 1200 is 1.5m away or 3m away. I remember getting 3450 for the roller though.

Just for you D94.

 

[720YoloSwagScope]

Paper is up :/ http://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-engineering-studies.pdf
You drew it all for nothing LOL

 

[keepLooking]

Paper is up :/ http://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-engineering-studies.pdf
You drew it all for nothing LOL

Fuck. Let's wait for D94 to post the solutions for everything

 

[D94]

I'm not sure whether that 1200 is 1.5m away or 3m away. I remember getting 3450 for the roller though.

yeah the diagram above was wrong, the 1.2kN force was 1.5m, so, 3450N is the correct answer.

 

[vitamin D]

how did everyone go for multiple choice ?

 

[D94]

For the winch question,

You should be able to draw the FBD, if not, then it's unlikely you got the next part correct....

So,

For the horizontal plane (parallel to the surface): T - F_f - mgsin25 = 0
For the vertical plane (perpend. to the surface): N - mgcos25 = 0 => N = mgcos25
Also, F_f = uN.

So, solving this, T - F_f - mgsin25 = 0

T - uN - mgsin25 = 0
T - u(mgcos25) - mgsin25 = 0

T = u(mgcos25) + mgsin25 = 1041.77N (where g = 10N)


For the 2nd winch question, taking moments about the centre = 0,
910(100) - F(350) = 0
So, F = 260N

But this is a frictionless pivot, so taking efficiency into account, F_min = F/n = 260/0.85 = 305.88N

Alternatively, you could have divided F by n to begin with.

 

[D94]

For the youngs module question I think it's a trick question as it asked for GPa and KN/mm^2 is MPa and I got 202MPa so I put in 0.202 GPa

Forgot to put :1 do you think we. Will lose a mark?

N/mm^2 is MPa.

Answer is 206.25GPa. You can just use 18kN and 0.5mm since it's all a linear relationship.

 

[keepLooking]

For the winch question,

You should be able to draw the FBD, if not, then it's unlikely you got the next part correct....

So,

For the horizontal plane (parallel to the surface): T - F_f - mgsin25 = 0
For the vertical plane (perpend. to the surface): N - mgcos25 = 0 => N = mgcos25
Also, F_f = uN.

So, solving this, T - F_f - mgsin25 = 0

T - uN - mgsin25 = 0
T - u(mgcos25) - mgsin25 = 0

T = u(mgcos25) + mgsin25 = 1041.77N (where g = 10N)


For the 2nd winch question, taking moments about the centre = 0,
910(100) - F(350) = 0
So, F = 260N

But this is a frictionless pivot, so taking efficiency into account, F_min = F/n = 260/0.85 = 305.88N

Alternatively, you could have divided F by n to begin with.

Okay, that is all five marks lost then. ;[