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Young modulus and moments question

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image.jpgLooking for some help on question 5, not sure what to do exactly
Original post by Purple K
Attachment not found
image.jpgLooking for some help on question 5, not sure what to do exactly


maybe use F=kΔLF=k \Delta L?? not too sure about the first one

second one use the formula for strain which = ΔLL\dfrac{\Delta L}{L}

3rd obviously use formula for young modulus which is E=FLAΔL E=\dfrac{FL}{A \Delta L}
use values from question and the tension you worked out in part A
then use formula for strain which is strain=ΔLLstrain= \dfrac{\Delta L}{L}
Do you know the answers?
I don't know if I'm correct but if I had to answer it in an exam I would get

Tension in both = 49.05N

Extension = 1.7mm

Diameter = 8.57x10^-4 m

But like I said, these are just guesses.
Reply 3
Hmm I haven't been able to look through it yet with my teacher but I got some different answers so I'm not entirely sure
Original post by MHD30
Do you know the answers?
I don't know if I'm correct but if I had to answer it in an exam I would get

Tension in both = 49.05N

Extension = 1.7mm

Diameter = 8.57x10^-4 m

But like I said, these are just guesses.



I like your answers. One should be able to obtain an exact answer for the diameter if algebraic manipulation is used - 2×0.60 \sqrt{2} \times 0.60 mm.

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