In constructing a large mobile, an artist hangs an aluminum sphere of mass 5.9 k

Question

In constructing a large mobile, an artist hangs an aluminum sphere of mass 5.9 kg from a vertical steel wire 0.53 m long and 2.9×103 cm^2 in cross-sectional area. On the bottom of the sphere he attaches a similar steel wire, from which he hangs a brass cube of mass 13.3 kg .

Part A

Compute the tensile strain for the top wire.

Express your answer using two significant figures.

Part B

Compute the tensile strain for the bottom wire.

Express your answer using two significant figures.

Part C

Compute the elongation strain for the top wire.

Express your answer using two significant figures.

Part D

Compute the elongation strain for the bottom wire.

Express your answer using two significant figures.

Explanation / Answer

Given that

mass of the aluminimum sphere m1=5.9 kg

mass of the brass cube m2=13.3 kg

area of cross section of wire A=2.9*10^-7 m^2

length of the steel wire L=0.53 m

now we find the tensile strain for top of the wire

the total tension T=(m1+m2)g=(5.9+13.3)*9.8=188.16 N

Young's modulus of steel wire, E = 20*10^10 Pa
Tensile strain = T/(A*E)
Tensile strain = 188.16 /(2.9×10^7 * 20 * 10^10 Pa)
Tensile strain = 188.16/(2.9×10^7 m^2 * 20 * 10^10 Pa)
Tensile strain = 3.24*10^-3 N

now we find the tensile stain for bottom of wire

tension in the bottom wire T2=m2g=13.3*9.8=130.34 N

the tensile strain for bottom of wire =130.34/2.9*10^-7*20*10^10

=2.25*10^-3 N

now we find the elongation strain for the top wire

the elongation strain for the top of wire =3.24*10^-3*0.53=1.7*10^-3 m

now we find the elongation strain bottom of wire

the elongation strain bottom of wire =2.25*10^-3*0.53=1.2*10^-3 m

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