- #1
billy722
- 19
- 0
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.
I guess the 1/2 is a typo? You should put brackets around the denominator.billy722 said:in Q1(1), i can only calculate out stress=100N/(1/2*10^-3)^2 m^2=0.1 GPa is it right?
I don't know where the 1/2 came from, but you successfully lost it again.billy722 said:stress=100N/(1/2*10^-3)^2 m^2=0.1 GPa
Yes,it do not have1/2mfb said: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.
Please explain your calculation and state the units.billy722 said:Then mass of yarn=
(100*10^-3)(1.5*10^-2)
=0.0015
The mass of yarn=haruspex said:Please explain your calculation and state the units.
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 said:The mass of yarn=
Long of yarn(100mm)*density(1.5 g/cm^3)?
1. I need it to find the mass which in tex unit?haruspex said: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?
No you don't,billy722 said:I need it to find the mass which in tex unit?
Right, but for the N/tex expression of stress you need to take the density into account.billy722 said:stress=force/cross section area
Right, which is why the length of the yarn is irrelevant, so its total mass is irrelevant.billy722 said:no,it will same?
So,the stress=force/mass per lengthharuspex said:Right, but for the N/tex expression of stress you need to take the density into account.
What units for that term?billy722 said:1.5*10^-2
That looks like 1mm. Why are you dividing the density by the cross-sectional area?billy722 said:/(1*10^-3)
150kg per metre? What is this yarn made of, depleted uranium?billy722 said:150000 g/m
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.billy722 said:Density per unit length=density per unit volume(1.5g/cm^3)/total length(100mm)
Right. So what will you use for the L2?billy722 said:M/L=(M/L^3)*L^2?
Yes. So what is the mass per unit length?billy722 said:L^2=cross section area (1*1 mm^2)?
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 said:1.5 g/cm^3 =1.5*(10^-6) g/m^3
i think that's wrong, it should be rightbilly722 said: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?
tex = g/103m. Other than that, your post #26 was along the right path.billy722 said:Tex= g/(10^2m)
Thank you,you help me learn it,not just ansharuspex said:tex = g/103m. Other than that, your post #26 was along the right path.
That's good to hear.billy722 said:Thank you,you help me learn it,not just ans
Q1(1) refers to the first quartile of a stress-strain curve. It is the point at which 25% of the maximum stress is reached. This value is used to calculate a material's yield strength.
N/tex is a unit used to measure the number of filaments per tex, where tex is the mass in grams of 1000 meters of fiber. It is commonly used to measure the linear density of materials such as yarn or thread.
Stress is calculated by dividing the force applied to a material by the cross-sectional area of the material. This value is typically expressed in units of force per unit area, such as N/m^2 or Pa.
Stress and strain are directly proportional to each other within the elastic region of a material. This means that as stress increases, so does strain, and vice versa. However, once the material reaches its yield point, the relationship between stress and strain becomes nonlinear.
Calculating stress is crucial in understanding the mechanical properties of materials. It allows scientists to determine a material's yield strength, which is important for designing and testing the strength and durability of various products. Additionally, stress calculations can help identify potential weaknesses or failure points in a material, leading to improvements in its design and performance.