Force and Extension

I'm having a hard time understanding question 19. May someone please explain it to me? http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s13_qp_13.pdf

Reply 1

lerjj

Original post by asadmoosvi

The young modulus is going to be the same for both the scale model and the real thing. Set up an equation and follow it through from that.

Reply 2

asadmoosvi

Original post by lerjj

The young modulus is going to be the same for both the scale model and the real thing. Set up an equation and follow it through from that.

I tried that, but I'm not getting the right answer. Can you please solve it and show me how you did it? I'd really appreciate it!

Reply 3

lerjj

Original post by asadmoosvi

I tried that, but I'm not getting the right answer. Can you please solve it and show me how you did it? I'd really appreciate it!

What is the right answer? I may have gotten it wrong myself...

EDIT: I get 10^2, if this is right then I'll write up a full answer for you, otherwise I'll check my working as it took a little while and I may have made a few mistakes.

(edited 9 years ago)

Reply 4

asadmoosvi

Original post by lerjj

Yeah, that's the right answer. I'd love a full answer with explanation, please. :smile:

(edited 9 years ago)

Reply 5

lerjj

Original post by asadmoosvi

Yeah, that's the right answer. I'd love a full explanation, please. :smile:

Ok, so we know that the young's modulus is the same, setting up our equation we get:

general equation

Right, that much is quite easy to do (but really ugly). The bit which you could make mistakes is this:

conversions for the ten times linear scale factor

cancelling and solving

(edited 9 years ago)

Reply 6

asadmoosvi

Original post by lerjj

Ok, so we know that the young's modulus is the same, setting up our equation we get:

[br]\begin{array}{rcll}[br]E_1 &=& E_2[br]\\ \dfrac{\text{Stress}_1}{\text{Strain}_1} &=& \dfrac{\text{Stress}_2}{\text{Strain}_2}[br]& \implies \dfrac{F_1L_0,1}{A_1 \Delta L_1} = \dfrac{F_2L_{0,2}}{A_2 \Delta L_2}[br]\end{array}[br]

Right, that much is quite easy to do (but really ugly). The bit which you could make mistakes is this:

[br]\begin{array}{lr}[br]F_2=1000F_1 & \text{(Because the mass is ten times larger in each direction)}[br]\\ A_2=100A_2 & \text{(Because the cable has ten times the diameter and so 100X the cross sectional area)}[br]\\ L_{0,2}=10L_{0,1} & \text{(Because the starting cable is ten times longer)}[br]\end{array}[br]

[br][br]\therefore \dfrac{1000F \times 10 L_0}{100A \times \Delta L_2} = \dfrac{F \times L_0}{A \Delta L_1}

Unparseable latex formula:

\text{This cancels to:} [br]\dfrac{100FL_0}{A \Delta L_2}=\dfrac{FL_0}{A \Delta L_1}[br]\text{dividing everything by} \dfrac{FL_0}{A} [br]\text{you get:}[br]\dfrac{100}{\Delta L_2}= \dfrac{1}{\Delta L_1} [br]\imples \Delta L_2=100 \Delta L_1}[br]\therefore frac{\Delta L_2}{\Delta L_1}=100=10^2[br]

Thank you. I was taking the mass for the model as ten times smaller. I wasn't taking it as ten times smaller in each direction. So if the mass of the actual crane is 1000, the mass of the model will be 10. Got it.

(edited 9 years ago)

Reply 7

lerjj

Original post by asadmoosvi

Yeah, I took a while to realise i had to do that... going to spend a little bit of time trying to pretty up my LaTeX answer as it's kinda all over the place atm. A whole line of code isn't showing up, it should end with a [TeX \therefore sign. Oh well...