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## 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

2 *Y*pi r^2*0.03*

[/B]

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 kgThat'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