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gracy
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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*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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