Calculating Stress: Understanding Q1(1) and N/tex | Homework Help

[billy722]

Homework Statement

in Q1(1), i can only calculate out stress=100N/(1/2*10^-3)^2 m^2=0.1 GPa is it right?
Also, can you teach me how to calculate other answer? i have no idea in N/tex.

Homework Equations

The Attempt at a Solution

 

[mfb]

in Q1(1), i can only calculate out stress=100N/(1/2*10^-3)^2 m^2=0.1 GPa is it right?

I guess the 1/2 is a typo? You should put brackets around the denominator.

Look up the definitions of the other units for the conversion factors.

 

[haruspex]

stress=100N/(1/2*10^-3)^2 m^2=0.1 GPa

I don't know where the 1/2 came from, but you successfully lost it again.

For tex and Den see http://www.swicofil.com/companyinfo/manualyarnnumbering.html

 

[billy722]

I guess the 1/2 is a typo? You should put brackets around the denominator.

Look up the definitions of the other units for the conversion factors.

Yes,it do not have1/2

 

[billy722]

Then mass of yarn=
(100*10^-3)(1.5*10^-2)
=0.0015
In tex=0.0015/1000;
100/(0.0015/1000) N/tex
In den=0.0015/9000;
100/(0.0015/9000) N/den
?

 

[haruspex]

Then mass of yarn=
(100*10^-3)(1.5*10^-2)
=0.0015

Please explain your calculation and state the units.
Why is the mass of the segment of yarn relevant? If the yarn were twice as long would the stress in N/tex be different?

 

[billy722]

Please explain your calculation and state the units.

The mass of yarn=
Long of yarn(100mm)*density(1.5 g/cm^3)?

 

[haruspex]

The mass of yarn=
Long of yarn(100mm)*density(1.5 g/cm^3)?

That calculation gives you a mass per unit area, not a mass. But don't bother correcting that, answer my other question: why is the total mass of the yarn interesting? What has it got to do with calculating the stress? If the yarn were twice as long, so twice as massive, would the strain in N/tex be any different?

 

[billy722]

why is the total mass of the yarn interesting? What has it got to do with calculating the stress? If the yarn were twice as long, so twice as massive, would the strain in N/tex be any different?

1. I need it to find the mass which in tex unit?
2.i just now stress=force/cross section area
3.no,it will same?

 

[haruspex]

I need it to find the mass which in tex unit?

No you don't,

stress=force/cross section area

Right, but for the N/tex expression of stress you need to take the density into account.

no,it will same?

Right, which is why the length of the yarn is irrelevant, so its total mass is irrelevant.
What you should care about is the mass per unit length. Calculate that.

 

[billy722]

Right, but for the N/tex expression of stress you need to take the density into account.

So,the stress=force/mass per length
Mass per length=(1.5*10^-2)/(1*10^-3)(1*10^-3)(100*10^-3)=150000 g/m
=150000/1000 g/tex=150 g/tex
=150000/9000 g/den=16.6667 g/ den
Stress=100/150
Stress=100/1.6667
I don't sure

 

[haruspex]

1.5*10^-2

What units for that term?

/(1*10^-3)

That looks like 1mm. Why are you dividing the density by the cross-sectional area?

150000 g/m

150kg per metre? What is this yarn made of, depleted uranium?

Start with a logical basis for the calculation. No numbers at this stage, just describe in words what is to be multiplied by what and divided by what.

 

[billy722]

Density per unit length=density per unit volume(1.5g/cm^3)/total length(100mm)

 

[haruspex]

Density per unit length=density per unit volume(1.5g/cm^3)/total length(100mm)

No. First, that makes no sense dimensionally. On the left you have mass/length (M/L) and on the right (M/L³)/L = M/L⁴.
Secondly, the length of the yarn can have no bearing on the relationship between mass per unit length and mass per unit volume. If the yarn were twice as long those two densities would not change.

 

[billy722]

M/L=(M/L^3)*L^2?

 

[haruspex]

M/L=(M/L^3)*L^2?

Right. So what will you use for the L²?

 

[billy722]

L^2=cross section area (1*1 mm^2)?

 

[haruspex]

L^2=cross section area (1*1 mm^2)?

Yes. So what is the mass per unit length?

 

[billy722]

1.5 g/cm^3 =1.5*(10^-6) g/m^3
1 mm^2=1*(10^-6) m^2
Mass per length=1.5*10^(-12) g/m

 

[haruspex]

1.5 g/cm^3 =1.5*(10^-6) g/m^3

It does help to think through what each statement is saying. You have written that a cubic cm has a mass of 1.5g but a cubic m, a vastly larger volume, will have a mass of only 1.5 micrograms.

 

[billy722]

Ok, mass per length=1.5*(10^12) g/m

 

[mfb]

Don't guess. Calculate it.

 

[billy722]

(L/(Cm)^3)*(mm)^2=
(L/(10^-2m)^3)*(10^-3m)^2=
(L/M^3)*(M^2)
=>1.5*1 g/m?

 

[mfb]

That is correct.

 

[billy722]

Then,
1.5g/m=1.5/1000 tex=1.5*10^-3 tex
1.5/9000 deg=1.6667*10^-4 den
Stress=100/(1.5*10^-3)=66666.67 N/tex
=100/(1.6667*10^-4)=599988 N/den?

 

[billy722]

Then,
1.5g/m=1.5/1000 tex=1.5*10^-3 tex
1.5/9000 deg=1.6667*10^-4 den
Stress=100/(1.5*10^-3)=66666.67 N/tex
=100/(1.6667*10^-4)=599988 N/den?

i think that's wrong, it should be right
Tex= g/(10^2m) ; Mass per length= g/m = (10^2g)/(10^2m) = (10^2g) / (m) Tex = (1.5*10^2g) / m = 150 Tex ; Stress= 100/150= 2/3 N/Tex
Den= g/ (9*10^3m) ; Mass per length= g/m = (9*10^3g)/(9*10^3m) = (9*10^3g)/ (m) den = (1.5*9*10^3g)/ m =13500 Den = 1/135 N/Den
Stress= N/m^2 ; Stress= [N/ (g/m)]*(g/m^3) = (100N/1.5)*(1.5*10^6)= 100MPa

 

[haruspex]

Tex= g/(10^2m)

tex = g/10³m. Other than that, your post #26 was along the right path.

 

[billy722]

tex = g/10³m. Other than that, your post #26 was along the right path.

Thank you,you help me learn it,not just ans

 

[haruspex]

Thank you,you help me learn it,not just ans

That's good to hear.