# Sag in steel wire

## Homework Statement

:A steel wire of radius 0.4 × 10–3 m and length 1 m is tightly clamped between points A and B which are separated by 1 m and in the same horizontal plane. A mass is hung from the middle point of the wire such that the middle point sags by 2 cm from the original position.Compute the mass of body.young modulus of steel:20 × 10^10 N/m2 )
Relevant equations:=T= F =YAl/L
Attempt to a solution::
Elongation occurs as shown in figure. AD = BD = 0.50 m

In triangle ADC, Let AC = x
That's why x =√(50^2 + 2^2=√(2500 + 4)
= √(2504) = 50.03 cm
= 50.03 * 10^-2m
Elongation,l = x – 0.50
= 50.03 – 50 = 0.03 cm
Resolve T as shown in figure
2T cos (90 – theta) = mg
2T sin (theta) = mg
2T *2*10^-2 divided by x=mg
2T *2*10^-2 divided by 50.03 *10^-2=mg
2*Y*A* l divided by 50*10^-2*2
as we know T= F =YAl/L

2 *Y*pi r^2*0.03*10^-2 divided by 50*10^-2*2*divided by 50.03=m *9.8
That's why]m=2*20*10^10*3.14*(0.4*10^-3)^2*0.03*2 divided by 50*50.03*9.8
m=0.492 kg

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I want to understand the diagram.

Bystander
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I want to understand the diagram.
You've got the solution, which suggests you do understand the diagram. What questions have you got, or what bothers you?

NascentOxygen
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I want to understand the diagram.
The diagram shape is not drawn to an accurate scale, it is exaggerated, and this is okay if it makes the diagram easier to understand.

A and B are nails in the wall, and a steel wire is tightly stretched between A and B. When a weight is attached to that wire's midpoint, the wire stretches and sags into a V-shape. Steel wire is like a spring---it can be stretched by a force, and will return to its original length when that force is removed. The length that any column of steel will stretch can be calculated if you know its length, its cross-sectional area, and its Youngs Modulus. Every material has its own elasticity characteristic value known as its Young's Modulus. You can read up on it here: http://en.m.wikipedia.org/wiki/Young's_modulus

2*Y*A* l divided by 50*10^-2*2
as we know T= F =YAl/L
Here A=Area Which area should I take?According to the formula above we should take area on which restoring/deforming force acts.But I don't know on which area restoring/deforming force is acting.

Last edited:
Bystander
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"A" is the cross-sectional area of the wire. You can use the area of the unstretched wire since it's going to be reduced very little by the stretching, and going back to the stress-strain problem, it's not all that clear which area you actually should use without knowing a little more about how "Y" was determined (you'll recall that there were two ways that was shown, 1) for a starting cross-sectional area, and 2) for the area when it had been reduced by stretching).

According to the formula above we should take area on which restoring/deforming force acts
"A" is the cross-sectional area of the wire
So,we should take cross sectional area while calculating stress,not the area on which restoring/deforming force acts.

Bystander
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Both. They're not going to be enough different to be noticeable. Ignore me when I start throwing too much extra information. All you need on this problem is the cross-sectional area of the wire.

. Ignore me when I start throwing too much extra information.
OK.

They're not going to be enough different to be noticeable.
Means the cross sectional area and the area on which force acts both are one and the same thing.

Bystander
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Yes.

But how?I mean how tension force or restoring force acts on cross sectional area of wire?

Bystander
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The wire is the only thing being stretched, or allowed to relax to the horizontal under the influence of the restoring force if the weight is removed.

The wire is the only thing being stretched, or allowed to relax to the horizontal under the influence of the restoring force if the weight is removed.
So,that means restoring force acts on the wire ,and we denote this force in terms of per unit area.That's why we will take stress is equal; to force divided by cross sectional area of wire .Right,sir?

Bystander