In constructing a large mobile, an artist hangs an aluminum sphere of mass 5.1 k

Question

In constructing a large mobile, an artist hangs an aluminum sphere of mass 5.1 kg from a vertical steel wire 0.51 m long and 2.4×103 cm2 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 11.3 kg .

A-

Compute the tensile strain for the top wire.

Express your answer using two significant figures.

B-

Compute the tensile strain for the bottom wire.

Express your answer using two significant figures.

C-

Compute the elongation strain for the top wire.

Express your answer using two significant figures.

D-Compute the elongation strain for the bottom wire.

Explanation / Answer

I am assuming that steel wire is massless.
A)
Taking both sphere and cube including bottom wire as system, top wire is balancing both the weights.
So, tension in top wire= (m1+m2)*g
= (5.1 + 11.3) * 9.8
=160.72 N
Area = pi*r^2 = pi*(2.4*10^-5 )^2 m^2 = 1.81*10^-9 m^2
I think you need to calculate tensile strength
tensile strength = tension / area
= 160.72 / 1.81*10^-9
=8.88*10^10 N/m^2
Answer: 8.88*10^10 N/m^2
B)
So, tension in bottom wire= (m2)*g
= ( 11.3) * 9.8
=110.74 N
I think you need to calculate tensile strength
tensile strength = tension / area
= 110.74 / 1.81*10^-9
=6.12*10^10 N/m^2
Answer: 6.12*10^10 N/m^2

C)
young modulus of aluminium = 69*10^9 N/m^2
use:
tensile strain = tensile stress / young mudolus
= 8.88*10^10 / 69*10^9
=1.29
Answer: 1.29
D)
young modulus of steel = 200*10^9 N/m^2
use:
tensile strain = tensile stress / young mudolus
= 6.12*10^10 / 200*10^9
=0.306
Answer: 0.306

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