Solve Elasticity Problem: Young Modulus, Diameter 0.6mm

[Dumbledore211]

Homework Statement

In the fig AB is a wire of diameter 0.6mm and Young modulus of the material 2x10^11Nm^-2. C is the mid point of the wire. How much will the point B go down?

Homework Equations

Y=FL/Al where L= original length of the wire and l=extension in the wire

The Attempt at a Solution

Here the entire system is in equilibrium which means that there is no net force and there is no extra acceleration caused by the system. ∑Fy=0. The distance by which the point B goes down depends on the overall extension of the wire caused by the two masses. What I tried to do is find out l1 at point C where I use L=1.5 l1= FL/AY=1.0398x10^-5m
l2= 1.0398x10^-5
Total distance= l1 +l2= 2x1.0398x10^-5m= 2.0796x10^-5m. But the answer is 3.12x10^-3m. Where did I go wrong?

 

[nasu]

What is the meaning of this?
" L=1.5 l1= FL/AY=1.0398x10^-5m"
How is L equal to 1.5 l1, to start with.

 

[Dumbledore211]

Sorry what I meant is L=1.5 and small L sub 1= FL/AY since the first mass is attached at the mid point of L which is actually 3cm

 

[nasu]

And what did you take as F? And what did you get for area?
You should show all steps. Otherwise how can someone figure out where the error is?

 

[lightgrav]

How can the bottom half stretch the same amount as the top half?
Isn't the top half the same diameter wire, but holding up twice as much mass?

 

[Dumbledore211]

The top half is holding up the same amount of mass as the bottom half according to the figure. I took F to be equal 39.2N as it is 4kg-wt. The area in each case is π(0.3x10^-3)^2. This begs the question does the cross- sectional area change for each half as there is extension in each case..

 

[nasu]

The tensions in the strings are not the same.
Write the balance of the forces for the body attached in the middle of the string and you will see it.

 

[Dumbledore211]

@Nasu, okay I get what you are saying. The body attached in the middle is apparently in equilibrium which means that it's weight and the tension acting downward nullify the tension acting upward. So the tension in the upper part is more. But, we are not allowed to use the equation of tension which is T=λl/L because there is no modulus of elasticity. So, what equation should we use to find the different tensions in each halve of the string

 

[Tanya Sharma]

Do F=ma on the two masses to find the tensions in the two wires .Let T1 be the tension in the upper wire and T2 be the tension in the lower wire .

For the upper wire T1 = T2 +mg
For the lower wire T2 = mg

 

[Dumbledore211]

@Tanya Sharma, I do get the answer which is 3.119419727X10^-5 which is 3.12x10^-5m≈ but the actual answer is 3.12x10^-3m

 

[Tanya Sharma]

3.12x10^-5 looks alright to me .

 

[nasu]

Are you sure the length is 3 cm and not 3 m?

 

[Dumbledore211]

I think it should be 3.12X10^5m because there are often minor printing mistakes in our books. Thank you.Your help is appreciated

 

[Tanya Sharma]

Yes...Either the answer is 3.12x10^-5m or as nasu has pointed the length is 3m in which case the book answer is correct.