Post by: Alwin on November 13, 2013, 11:01:15 am
Here are my solns so far:
(will make some nice LaTeXed Worked solns later :) )
All done now, here: http://www.mediafire.com/download/d5a3w6a3z6nj85o/VCAA_Physics_2013_15.11.2013.pdf
#2 Rest still coming, then I'll look at all the detail studies
Motion
1. a) a = 1.73 m/s/s if you find the net force and divide by the mass mgsin(10)/m
a = 1.75 m/s/s if you use a constant acceleration formula x = ut + 1/2 a t^2
b) Using constant acceleration formulas, a= 0.194 so then Ffriction = 0.50(1.75-0.194)=0.78N
2. a) W = 20 N
b) T = a x m2 = 2.5 x 6 = 15 N
3. a) 6 x 2 = 6 v so speed is 2m/s
b) KE before = 36 J and KE after = 12 J so inelastic
c) I = Change momentum of p of m1 = 2 x 2 - 6 x 2 = -8 Ns. Same as 8Ns to the left
4. a) One arrow “N” or “R” perpendicular to the slope
One arrow straight down “W=mg”
One arrow parallel to horizontal pointing to the left
b) tan(x) = v^2/gr so tan(x) = 0.125 hence the angle is 7.13 degrees
You could do a “from scratch” approach by only 2 mark q. One for formula, one for ans
5. a) Net force always towards centre. At P, direction is D
b) Fc = T – W so mv^2/r = T - mg --> T = 2x49/1 + 2x10 = 118 N
Didn't read the question properly first time sorry if I scared anyone!
6. THE TRICK WITH THIS Q IS THAT KE IS NOT ON THE GRAPH
a) At Z KE = 0 and GPE = 0. So total energy = EPE = 20J
b) At any point KE + GPE + EPE = total energy = 20J
So at Y, 5 + 10 + KE = 20 so KE = 5 = 1/2 m v^2 so then v = 3.16 m/s
c) They At Q there already is spring potential energy since it is extended.
Use EPE = 1/2 k x^2 to prove this
7. a) Period is the length of one day since it stays in the same spot, so 24 hours
b) Use Kepler’s law or similar, converting units THEN minus the radius of the earth
R = 4.24 e7 so then Radius
Supposed to be just R = 4.24 E7 thanks again for picking that up!
c) Only apparent weightlessness applies
8. a) vertically and up is positive -15 = 10 t + 1/2 (-10) t^2 solving a quadratic of formula
Or find the time separately in sections. t = 3 sec
b) vertically: v^2 = u^2 + 2 a x --> v^2 = 100 + 20x15
thanks guys! the working was right... I put it in the calc wrong lolol... v = -20m/s
horizontally: v = 20 cos30 = 17.32
so then |v| = 26.46 m/s and angle = inverse tan = (-20/17.32) = -49.1 degrees
… although this seems like a spesh question
Electronics
9 Total resistance = 5, Vout = 15V. Total resistance = 15, Vout = 5V
10. Two 50 ohm in parallel then followed by two more resistors in series
11.a) R = 2.5 / 0.005 = 500 ohm
b) Rthermistor = 1.0/0.005= 200 ohm. From graph, 20 degrees
c) Rldr = 10/0.005 = 2000 ohm. From graph, 15 lux
d) Change 1: increase the temp, smaller Vout
Change 2: decrease the lighting, smaller Vout
12.a) Vled = 2.0V Vresistor = 0.010 x 450 = 4.5. EMF = 6.5 V
b) Resistor: electrical -> heat. LED: electrical -> light
13.a) 5/0.012 = 416.6
b) The signal is changed
Electric Power
14. Should look like two north poles facing each other
15.a) 1:6 ratio so 18.0 V
b) Peak = root(2) x 18 = 25.5 V
c) P=V^2/R = 18^2/1200 = 0.27 W
d) Explain general principles of how transformers work (one mark)
Relate it to this situation here with the DC input and the on and off position (one mark)
Explain what happens when it’s turned on using Lenz’s Law (one mark)
16.a) Current going from W to X. Use slap rule, Anticlockwise
b) F=nBIL so F = 20x0.500x0.50x0.050 = 0.25 N
c) Slips rings still make contact so there still is friction present (one mark maybe?)
Explain this is a DC motor and the direction of the current in the coil needs to be altered
every half turn which is the function of the slip ring (one mark)
Explain if a slip ring commutator was used in this motor the armature would oscillate
perpendicular to the field then stop turning. So the motor does not work (one mark)
17.a) EMF=change in flux/time = (0.6-0.2)/0.5 = 0.8V
I = 0.8/0.1=8A …note that I don’t think you’d get this formula if you did the sin(2pif) rule…
b) EMF zero when flux is max/min, think of derivatives if it helps. So t=0, 0,5, 1, 1.0, 1.5, 2.0, 2.5
c) Moving towards a North Pole.
To oppose this motion, a North Pole is induced on the side of the ring closest to the N Pole
So, induced field direction is down towards the magnet.
Using RH Grip rule, the current flows clockwise
Or, the magnetic field strength is increasing so magnetic flux increases as the ring moves down
To oppose this, the induced current will create a field in the opposite direction, etc etc
As it doesn’t specify if the speed is constant and VCE doesn’t require you to know about
strength of magnetic fields, I don’t think you would be penalised if you didn’t mention how
the current changes over time (decreases for those interested)
d) Flux is relative to the magnetic field strength/density. Technically strongest part is around the
poles, 1/6th the total length I think I read somewhere last year, but we assume it’s at the
middle of the magnet. When it’s far away, weaker field so less current induced (think of Lenz’s
Law if it helps).
So, first min of graph is A (given),the first max is
Aka A: t=0, 2, B: t=1.0 C: t=0.5, 1.5, 2.5
Sorry guys, didn't look at the diagram right... thought it was B in the middle goshThis is main reason I do crap on exams =P
18.a) R = Vloss/I = 24/6=4.0 ohm
b) Voutput = 1200/6.0 = 200V
c) Ploss = 4 x 6^2 =144. So 96/1200 = 12% -- cheers Jtc + nliu, my mistake sorry!
d) I = 10/2.0 = 5A. Voutput = 1200/5 = 240V
Light and Matter
19.a) E = 6.63 E-34 x 6.7E14 = 4.44 e-19
b) wavelength = 3e8/6.67e14 = 4.48e-7 m
20.a) An atom is at 3.19eV state? That’s interesting wording…
f = 3.19/4.14e-15 = 7.71e+14 Hz. Wavelength = 3e8 / 7.71e+14 = 3.89e-7 m
b) 588.63nm, so f = 3e8/588.63e-9 = 5.10e+14 Hz. So E = 5.10e+14 x 4.14e-15 = 2.11 eV
This exactly corresponds to electron falling (emission spectra) from 2.11eV level to ground state
So, there is a spectral line at 588.63 nm
21.a) Max KE = change of e x Stopping voltage = 1.6e-19 x 1.85 = 2.96e-19 J
b) Work function =
c) Less intensity, smaller max photocurrent but same stopping voltage
d) The frequency is less than the threshold frequency. Thus, no current produced
22.a) The intensity at the centre will be the greatest. This is because it is equidistant from both slits
Constructive interference occurs
b) wavelength (lambda) = fringe width x slit separation / distance from slits to screen
so, using a lower frequency means bigger wavelength so bigger fringe width. So D. further away
c) centre-dark-bright-dark-bright = 1.4e3 nm. centre-dark = 1.4e3 / 4 = 3.5e2 nm
d) Particle model only predicts two central light band (one mark)
Wave model predicts bands formed by interference patterns (one mark)
Young’s double slit experiment showed constructive and destructive interference so supports
the wave model (one mark)
Moto: never trust uncyclopedia (bonus ten marks)
23.a) E = 80e3 eV = hf so f= 80e3/4.14e-15 = 1.93e+19 and wavelength = 3e8/1.93e+19=1.55e-11m
p = 6.63e-34 / 1.55e-11 = 4.27e-23 kgm/s
b)
*****indicates I'm double checking my solns for these questions
Post by: Robert123 on November 13, 2013, 12:01:43 pm
Edit: for 7b I didn't think you have to take off the radius of earth because I believe it said from centre of earth.
Post by: eddybaha on November 13, 2013, 12:45:38 pm
Post by: Zoomzoom on November 13, 2013, 12:47:01 pm
For the verticale component of Q8b, I got 20 (sqrt 400) for vYea i got the same thing! which consequently makes the angle 49
Edit: for 7b I didn't think you have to take off the radius of earth because I believe it said from centre of earth.
Post by: Robert123 on November 13, 2013, 12:49:01 pm
Yea i got the same thing! which consequently makes the angle 49Yeah I got that angle as well.
Did anyone else remember their answer for the last light and matter question (same energy vs same momentum for same diffraction pattern). What was your reasoning?
Post by: vcestudent345 on November 13, 2013, 12:50:58 pm
wasn't the mass 2kg, so it'd just be sqrt5=2.24
..or did I read the question wrong :o
Post by: mmlp2 on November 13, 2013, 01:13:42 pm
b) T = Fc = m v^2/gr = 2 x 49/10x1 = 9.8 Newtons
Fcp + mg = 98 + 20 = 118 N
Can't be 9.8 newtons!
Post by: silverpixeli on November 13, 2013, 01:16:13 pm
3. b) KE before = 36 J and KE after = 24 J so inelastic
For KE after, I got 0.5mv^2=0.5(6)(2^2)=3*4=12J not 24J, can't check my working to see if I made a mistake though
5. b) T = Fc = m v^2/gr = 2 x 49/10x1 = 9.8 Newtons
Actually since it's a vertical circle and you're at the bottom of it, Fnet=T-mg so T=Fnet+mg=118N iirc (I hope that was right)
7. b) Use Kepler’s law or similar, converting units THEN minus the radius of the earth
R = 4.24 e7 so then Radius is 4.27 E7 – 6.4E6 = 3.60 E7
Did it specifically ask for altitude because I was almost certain it asked for distance from the centre meaning you wouldn't need to minus the radius (again, I hope)
8. b) vertically: v^2 = u^2 + 2 a x --> v^2 = 100 + 20x15 so v=-11.61 m/s
horizontally: v = 20 cos30 = 17.32
so then |v| = 26.45 m/s and angle = inverse tan = (-11.61/17.32) = -33.83
so angle 146.17 degrees I think… although this seems like a spesh question
You might have stuffed up the angle somewhere, I got 49.1 below the horizontal and so did some others so maybe when you used arctan it went into the second quadrant or something, might want to check with a diagram.
13.a) 5/0.012 = 416.6
It depended on which point you took, I assume other acceptible answers will be 400 (taking the point 20mV, -8V) and 375 (taking the point 8mV, -3V) so don't sweat if you didn't get 416.6 guys :)
Post by: Alwin on November 13, 2013, 01:25:37 pm
I'll go back and check my working for those questions :)
Post by: ~T on November 13, 2013, 02:19:31 pm
For KE after, I got 0.5mv^2=0.5(6)(2^2)=3*4=12J not 24J, can't check my working to see if I made a mistake thoughI think I agree with all of these points. The last one I'm not entirely sure, because I feel that because they stressed "linear region" there was only a set number of points to go from, but I can't remember what I got and I presume those points were all within that region and it was just rounding error (on the graph) that gives different results.
Actually since it's a vertical circle and you're at the bottom of it, Fnet=T-mg so T=Fnet+mg=118N iirc (I hope that was right)
Did it specifically ask for altitude because I was almost certain it asked for distance from the centre meaning you wouldn't need to minus the radius (again, I hope)
You might have stuffed up the angle somewhere, I got 49.1 below the horizontal and so did some others so maybe when you used arctan it went into the second quadrant or something, might want to check with a diagram.
It depended on which point you took, I assume other acceptible answers will be 400 (taking the point 20mV, -8V) and 375 (taking the point 8mV, -3V) so don't sweat if you didn't get 416.6 guys :)
Post by: hazmas on November 13, 2013, 02:28:17 pm
Post by: joey7 on November 13, 2013, 02:51:01 pm
Post by: lzxnl on November 13, 2013, 04:37:34 pm
Hey guys, I'm at school now looking at a spare copy of the exam (which I think my teacher wants back). I'll scan it when I get home and whatnot :)
Here are my solns so far:
(will make some nice LaTeXed Worked solns later :) )
Motion
7. a) Period is the length of one day since it stays in the same spot, so 24 hours
b) Use Kepler’s law or similar, converting units THEN minus the radius of the earth
R = 4.24 e7 so then Radius is 4.27 E7 – 6.4E6 = 3.60 E7
Electronics
12.a) Vled = 2.0V Vresistor = 0.010 x 450 = 4.5. EMF = 6.5 V
b) Resistor: electrical -> heat. LED: electrical -> light
13.a) 5/0.012 = 416.6
b) The signal is changed
Electric Power
*****indicates I'm double checking my solns for these questions
I have a few queries about these. For the motion gravity question, the exam says "from the centre of the earth"; no radius subtraction needed?
The question for the period of orbit has the answer box in seconds :p
Also, for the LED, isn't some energy dissipated as heat as well as light?
And then "signal is changed"; not a very enlightening response is it :P
Post by: Alwin on November 13, 2013, 04:51:42 pm
I have a few queries about these. For the motion gravity question, the exam says "from the centre of the earth"; no radius subtraction needed?
The question for the period of orbit has the answer box in seconds :p
Also, for the LED, isn't some energy dissipated as heat as well as light?
And then "signal is changed"; not a very enlightening response is it :P
Noted :) thanks + everyone else, I was in a bit of a rush when I did it and starting on the other core studies now I'm back home... hopefully with less mistakes :P
Post by: Scanlia on November 13, 2013, 04:58:27 pm
So I'll put up what I got so far, hopefully it'll help..
https://docs.google.com/file/d/0B_pcvLkZRgy2OHV3U0xFTlNkaWc/edit?usp=sharing
Cheers
Post by: Alwin on November 13, 2013, 05:04:57 pm
Alwin if you want a LaTeX template to get started with, I started making some solutions, until I realised I had to go out :(
So I'll put up what I got so far, hopefully it'll help..
https://docs.google.com/file/d/0B_pcvLkZRgy2OHV3U0xFTlNkaWc/edit?usp=sharing
Cheers
Haha thanks :) but I started mine already... up until I realised I hadn't read half the qs properly and was posting some wrong answers HAHA.
Thanks for the offer tho :D
Post by: BananaPi on November 13, 2013, 05:23:44 pm
edit: 17d: Was the order of A, B and C out (A-C-B-C-A)? I seem to recall that one of the questions had strange ordering.
edit: 18c: Just check this, a number of high-scoring students that I know got 12% for this one (144/1200).
Post by: Jtc on November 13, 2013, 05:33:04 pm
Post by: lzxnl on November 13, 2013, 05:45:50 pm
Hey guys, I'm at school now looking at a spare copy of the exam (which I think my teacher wants back). I'll scan it when I get home and whatnot :)But RMS voltage was 18 V using the transformer...so wouldn't the power be 18^2/1200=0.27?
Here are my solns so far:
(will make some nice LaTeXed Worked solns later :) )
Electric Power
14. Should look like two north poles facing each other
15.a) 1:6 ratio so 18.0 V
b) Peak = root(2) x 18 = 25.5 V
c) P=V^2/R = 3^2/1200 = 0.0075 W
d) Explain general principles of how transformers work (one mark)
Relate it to this situation here with the DC input and the on and off position (one mark)
Explain what happens when it’s turned on using Lenz’s Law (one mark)
Quote
17.a) EMF=change in flux/time = (0.6-0.2)/0.5 = 0.8VWasn't there a time domain given? Up to like 2.0s? And in those questions do we include the endpoints? The question didn't say.
I = 0.8/0.1=8A …note that I don’t think you’d get this formula if you did the sin(2pif) rule…
b) EMF zero when flux is max/min, think of derivatives if it helps. So t=0, 0,5, 1, 1.0, 1.5, 2.0, 2.5
Quote
c) Moving towards a North Pole.But C is the middle of the magnet...
To oppose this motion, a North Pole is induced on the side of the ring closest to the N Pole
So, induced field direction is down towards the magnet.
Using RH Grip rule, the current flows clockwise
Or, the magnetic field strength is increasing so magnetic flux increases as the ring moves down
To oppose this, the induced current will create a field in the opposite direction, etc etc
As it doesn’t specify if the speed is constant and VCE doesn’t require you to know about
strength of magnetic fields, I don’t think you would be penalised if you didn’t mention how
the current changes over time (decreases for those interested)
d) Flux is relative to the magnetic field strength/density. Technically strongest part is around the
poles, 1/6th the total length I think I read somewhere last year, but we assume it’s at the
middle of the magnet. When it’s far away, weaker field so less current induced (think of Lenz’s
Law if it helps).
So, first min of graph is A (given),the first max is B (middle of magnet), next min is C, then B etc
Aka A: t=0, 2 B: t=0.5, 1.5, 2.5, C: t=1.0
Quote
18.a) R = Vloss/I = 24/6=4.0 ohm
b) Voutput = 1200/6.0 = 200V
c) Ploss = 6 x 4^2 =96. So 96/1200 = 8%
Wait what? Ploss = I^2 = 6^2*4=144?
Post by: Alwin on November 13, 2013, 05:48:45 pm
But RMS voltage was 18 V using the transformer...so wouldn't the power be 18^2/1200=0.27?
Wasn't there a time domain given? Up to like 2.0s? And in those questions do we include the endpoints? The question didn't say.
But C is the middle of the magnet...
Wait what? Ploss = I^2 = 6^2*4=144?
frick.
I really should learn how to read questions :P . ty again (Y)
Post by: Uberjew on November 13, 2013, 05:53:44 pm
The graph tells you the magnetic flux through the ring, not induced in the ring.
Does that mean that when you're doing 17.d, A and B are both a minimum flux whereas C is max, thus shouldn't you have it go through:
A at t=0, 2
C at t=0.5,1.5,2.5
B at t=1
Post by: rany on November 13, 2013, 05:56:10 pm
Post by: ~T on November 13, 2013, 06:01:35 pm
frick.I've found that the most ridiculous subjects in terms of having to read the question carefully are definitely Methods and Physics. I'd blame it on the fact that you knocked both of them over last year :P
I really should learn how to read questions :P . ty again (Y)
Post by: randomposter on November 13, 2013, 06:02:06 pm
I said it would be twice as intense as the original laser beam due to constructive interference - it's an antinodal point.
Wasn't sure what they were looking for.
Post by: Robert123 on November 13, 2013, 06:02:58 pm
Post by: Chazef on November 13, 2013, 06:05:14 pm
Post by: ~T on November 13, 2013, 06:08:05 pm
What did you guys put for 'describe the intensity at the centre of the interference pattern'I'm not sure it works exactly like that, but maybe. I just said that it would be a bright spot due to constructive interference, and that it would be the brightest of the bright spots due to it being closer to the source(s) than the others.
I said it would be twice as intense as the original laser beam due to constructive interference - it's an antinodal point.
Wasn't sure what they were looking for.
For the gain question in electronics (second last), what were we meant to do with the curvy parts to the clipping, should that be included for calculations?Nope, just the linear region (the straight section). Pick two points in there that you can accurately take off the graph and then take the gradient.
Post by: Apink on November 13, 2013, 06:12:12 pm
Post by: BananaPi on November 13, 2013, 06:18:40 pm
Can someone explain 2b please?
I started with the entire system: mass=8kg, force causing acceleration=2*10N, Friction = 0N.
Next, calculate that the acceleration of the system a=F/m=20N/8kg=2.5ms^-1.
The tension is equal to the net force acting on mass 2, which is F=T=ma=6kg*2.5ms^-1, which is equal to 15N.
Hope that helps.
Post by: joey7 on November 13, 2013, 06:20:43 pm
Doesn't work function equal KE max - hf
b) Work function = KE max / hf = 2.96e-19/(6.63e-34 x 1e15) = 0.446
Post by: Apink on November 13, 2013, 06:24:45 pm
I started with the entire system: mass=8kg, force causing acceleration=2*10N, Friction = 0N.Thanks!
Next, calculate that the acceleration of the system a=F/m=20N/8kg=2.5ms^-1.
The tension is equal to the net force acting on mass 2, which is F=T=ma=6kg*2.5ms^-1, which is equal to 15N.
Hope that helps.
Post by: Robert123 on November 13, 2013, 06:28:35 pm
Doesn't work function equal KE max - hf
I thought it is hf- KE.
This is because the energy of a photon is equal to the sun of kinetic energy and the work function so work function equal the energy of a photon take away KE
Post by: sin0001 on November 13, 2013, 06:34:32 pm
Post by: randomposter on November 13, 2013, 06:36:13 pm
Anyone else get around ~2eV for the work function? o.O
yeah 2.29 I think.
Post by: lzxnl on November 13, 2013, 06:54:13 pm
What did you guys put for 'describe the intensity at the centre of the interference pattern'
I said it would be twice as intense as the original laser beam due to constructive interference - it's an antinodal point.
Wasn't sure what they were looking for.
It's actually four times as intense there; the amplitude doubles, but the intensity is proportional to the square of the amplitude (that's not part of this course). But you have the right idea.
Post by: ~T on November 13, 2013, 07:36:00 pm
20.a) An atom is at 3.19eV state? That’s interesting wording…For 20b, do you think it would matter if I went the other way? I started with the 2.11eV to ground state transition and then showed the wavelength of light emitted by this was 588.63nm. I'm presuming not, but I'm being pedantic about my solutions just in case.
f = 3.19/4.14e-15 = 7.71e+14 Hz. Wavelength = 3e8 / 7.71e+14 = 3.89e-7 m
b) 588.63nm, so f = 3e8/588.63e-9 = 5.10e+14 Hz. So E = 5.10e+14 x 4.14e-15 = 2.11 eV
This exactly corresponds to electron falling (emission spectra) from 2.11eV level to ground state
So, there is a spectral line at 588.63 nm
Post by: Professor Polonsky on November 14, 2013, 01:40:33 am
b) Work function = KE max / hf = 2.96e-19/(6.63e-34 x 1e15) = 0.446hmmm? Isn't it
Post by: AsianNerd on November 14, 2013, 09:29:14 am
Post by: silverpixeli on November 14, 2013, 09:31:36 am
U scared the shit outa me for first question... So both ways are acceptable?both methods are technically correct so they should both be accepted
hmmm? Isn't it?
Post by: Professor Polonsky on November 14, 2013, 10:41:12 am
Post by: ~T on November 15, 2013, 03:43:43 pm
22.d) Particle model only predicts one central light band (one mark)Hang on, doesn't the particle model predict two bright bands, with the light propagating linearly through the two slits?
Wave model predicts bands formed by interference patterns (one mark)
Young’s double slit experiment showed constructive and destructive interference so supports
the wave model (one mark)
Moto: never trust uncyclopedia (bonus ten marks)
Post by: Alwin on November 15, 2013, 05:53:53 pm
Hang on, doesn't the particle model predict two bright bands, with the light propagating linearly through the two slits?
yup, you're 100% correct. now you know why I didn't get any 50s last year... and probs never will :P
dem silly mistakes *shakes fist at the sky*
Post by: ~T on November 15, 2013, 11:06:14 pm
I'm feeling 10B for Synchrotron. I think. I'm fairly sure undulator light is spatially coherent.
Post by: Phy124 on November 16, 2013, 01:04:23 pm
Worked Solutions:
1. The young's modulus is given by the gradient of a stress-strain graph in its linear region
D
2. The force which the cable can support at its breaking point is given by
Therefore the mass is given by
C
3. The strain energy of the sample can be estimated by the area under the graph from the origin to the stretched strain value.
B
4. It is important for a cable in a crane not to undergo plastic deformation, this sample R would be chosen as it deforms elastically for the greatest range of strain values.
D
5. We know that
, 
and 
From this we can derive the equation
We have;
and from the linear region of the graph for cable R;
A
6. The torque about point P is given by the force provided by the mass multipled by the perpendicular distance.
C
7. To find the tension in SR we cneed to sum the torques about point P and equate them to zero, as the system is in rotational equilibrium.
C
8. To find the force extered by KL downwards on MN we can sum the torques about point K and equate them to zero, as once again this system is in rotational equilibrium.
The additional forces acting are that of the weight of the beam and the mass at point L, these both cause torques.
D
9. Personally I don't think any of the options would be appropriate placement of the beams, however, option A is the most appropriate of those given as for the mostpart the top of the beam will be in tension.
A
10. It can be noted that the mass at point G will elongate the beam FG, in order to counteract the tension in this beam both EF and FH must be in compression.
B
11. The arch is in compression and the weight of the truck on the road will cause the road to deflect downwards and as such the cables will elongate and will be in tension.
B
D
2. The force which the cable can support at its breaking point is given by
Therefore the mass is given by
C
3. The strain energy of the sample can be estimated by the area under the graph from the origin to the stretched strain value.
B
4. It is important for a cable in a crane not to undergo plastic deformation, this sample R would be chosen as it deforms elastically for the greatest range of strain values.
D
5. We know that
From this we can derive the equation
We have;
and from the linear region of the graph for cable R;
A
6. The torque about point P is given by the force provided by the mass multipled by the perpendicular distance.
C
7. To find the tension in SR we cneed to sum the torques about point P and equate them to zero, as the system is in rotational equilibrium.
C
8. To find the force extered by KL downwards on MN we can sum the torques about point K and equate them to zero, as once again this system is in rotational equilibrium.
The additional forces acting are that of the weight of the beam and the mass at point L, these both cause torques.
D
9. Personally I don't think any of the options would be appropriate placement of the beams, however, option A is the most appropriate of those given as for the mostpart the top of the beam will be in tension.
A
10. It can be noted that the mass at point G will elongate the beam FG, in order to counteract the tension in this beam both EF and FH must be in compression.
B
11. The arch is in compression and the weight of the truck on the road will cause the road to deflect downwards and as such the cables will elongate and will be in tension.
B