Yatayyat
Badges: 14
Rep:
?
#1
Report Thread starter 1 year ago
#1
Here is the Q itself: https://imgur.com/DCZdKXD

I don't know how to find what the new length and radius of 'X' has to be given that it has to a produce an extension of e/4 which is a quarter of the original extension that we can get with with 'W'

What equation do I have to relate to, to answer this Q?

I was thinking that if F = kx

where f = force, k = spring constant and x = extension

Then using the wire X which is of the same material as wire W means that wire X and W should have the same k (spring constant)

x = F / k

So with W it is: e = mg/k

And with X it is: e/4 = mg/k (but surely the mass of this wire is different to W)

Do we have to use the fact that mass is known to be mass = density * vol

And vol can be rewritten as vol = cross sectional area * length

If so how would I implement that into the Q to find what the length and radius of X should be?

Any help would be really grateful! Thanks!
0
reply
BobbJo
Badges: 12
Rep:
?
#2
Report 1 year ago
#2
Young's Modulus, E = Stress / Strain = FL/(Ax)

E = FL/(Ax)
FL = EAx
x = FL/EA
Same load, so F is the same
Same material, so E is the same
hence x is proportional to L/A
The ratio L/A for X needs to be 1/4 of that of W
This gives D

The wires being of the same material does not mean their spring constants are the same
m in mg refers to the mass of the load, not of the wire
Last edited by BobbJo; 1 year ago
0
reply
Yatayyat
Badges: 14
Rep:
?
#3
Report Thread starter 1 year ago
#3
(Original post by BobbJo)
Young's Modulus, E = Stress / Strain = FL/(Ax)

E = FL/(Ax)
FL = EAx
x = FL/EA
Same load, so F is the same
Same material, so E is the same
hence x is proportional to L/A
The ratio L/A for X needs to be 1/4 of that of W
This gives D

The wires being of the same material does not mean their spring constants are the same
m in mg refers to the mass of the load, not of the wire
Thanks this makes a lot more sense to me now!
Didn't think I needed to involve the use of stress, strain and young modulus into this but solely just by using Hooke's law.
Thanks again!
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (22)
17.32%
I'm not sure (3)
2.36%
No, I'm going to stick it out for now (41)
32.28%
I have already dropped out (3)
2.36%
I'm not a current university student (58)
45.67%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise