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Question: The breaking stress of aluminum is \(7.5 \times {10^8}dyne.c{m^{ - 2}}\). Find the maximum length of...

The breaking stress of aluminum is 7.5×108dyne.cm27.5 \times {10^8}dyne.c{m^{ - 2}}. Find the maximum length of aluminum wire that can hang vertically without getting broken. The density of aluminum is 2.7g.cm32.7g.c{m^{ - 3}}. (Given g=980cm.s2g = 980cm.{s^{ - 2}}).

Explanation

Solution

The breaking stress means when the maximum amount of force can be applied to the material, before it's breaking. The material cannot withstand any additional stress and so breaking occurs. There may be changes in length while the stress is vertically on the material. We are going to find the maximum length of the aluminum. Formula used:
stress(σ)=Load(W)Area(A)stress\left( {{\sigma _{}}} \right) = \dfrac{{Load\left( W \right)}}{{Area\left( A \right)}}
Where WW - the weight of the aluminum
AA - Cross-sectional area of aluminum wire

Complete answer:
We have the known values of breaking stress, the density of the aluminum. The load means the weight of the aluminum to be calculated.
The weight of the aluminum, W=m×gW = m \times g
Where m- the mass of the aluminum wire
gg - acceleration due to gravity
The mass of the aluminum, m=A×l×ρm = A \times l \times \rho
Where A- cross-sectional area of aluminum wire
ll - length of the aluminum wire
ρ\rho - the density of the aluminum wire.(ρ=2.7g.cm3)\left( {\rho = 2.7g.c{m^{ - 3}}} \right)
Therefore the weight of the aluminum becomes,
W=A×l×ρ×gW = A \times l \times \rho \times g
The breaking stress of aluminum,
α=WA\alpha = \dfrac{W}{A}
Substituting the known values, the stress equation becomes
α=A×l×ρ×gA\alpha = \dfrac{{A \times l \times \rho \times g}}{A}
α=l×ρ×g\Rightarrow \alpha = l \times \rho \times g [The breaking stress of aluminum is 7.5×108dyne.cm27.5 \times {10^8}dyne.c{m^{ - 2}}and its density is 2.7g.cm32.7g.c{m^{ - 3}}.]

7.5×108=l×2.7×980 \Rightarrow 7.5 \times {10^8} = l \times 2.7 \times 980
l=7.5×1082.7×980\Rightarrow l = \dfrac{{7.5 \times {{10}^8}}}{{2.7 \times 980}}
l=283446.712\Rightarrow l = 283446.712
l=2.834×105cm\Rightarrow l = 2.834 \times {10^5}cm
From the above calculations, the maximum length of the aluminum wire can be found as l=2.834×105cml = 2.834 \times {10^5}cm.

Note: The aluminum material is a good conductor of heat and electricity. so that it can be used in electrical transmission systems. Due to its properties of malleable, corrosion-resistant, ductile, non-magnetic, and lightweight, it can be used for aerospace applications. The property of aluminum, malleable means the ability of the material to be pressed into the desired shape. The property ductility means the length of the material can be increased by reducing its thickness, here the maximum tensile stress can be applied.