# Homework Help: Questions about stress

1. Feb 16, 2017

### billy722

1. The problem statement, all variables and given/known data
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.
2. Relevant equations

3. The attempt at a solution

2. Feb 16, 2017

### Staff: Mentor

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.

3. Feb 16, 2017

### haruspex

4. Feb 16, 2017

### billy722

Yes,it do not have1/2

5. Feb 16, 2017

### 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
?

6. Feb 17, 2017

### haruspex

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?

7. Feb 17, 2017

### billy722

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

8. Feb 17, 2017

### haruspex

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?

9. Feb 17, 2017

### billy722

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?

10. Feb 17, 2017

### haruspex

No you don't,
Right, but for the N/tex expression of stress you need to take the density into account.
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.

11. Feb 17, 2017

### billy722

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

12. Feb 17, 2017

### haruspex

What units for that term?
That looks like 1mm. Why are you dividing the density by the cross-sectional area?
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.

13. Feb 17, 2017

### billy722

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

14. Feb 17, 2017

### haruspex

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.

15. Feb 17, 2017

### billy722

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

16. Feb 17, 2017

### haruspex

Right. So what will you use for the L2?

17. Feb 17, 2017

### billy722

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

18. Feb 17, 2017

### haruspex

Yes. So what is the mass per unit length?

19. Feb 17, 2017

### 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

20. Feb 17, 2017

### haruspex

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.

21. Feb 17, 2017

### billy722

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

22. Feb 17, 2017

### Staff: Mentor

Don't guess. Calculate it.

23. Feb 17, 2017

### billy722

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

24. Feb 17, 2017

### Staff: Mentor

That is correct.

25. Feb 17, 2017

### 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?