Question

When a mass of 25 kg is hung from the middle of a fixed straight aluminum wire, the wire sags to make an angle of 12° with the horizontal as shown in Fig. 9–83. Determine the radius of the wire.

Short answer

The radius of the wire is \(3.5 \times {10^{ - 4}}\;{\rm{m}}\).

Step-by-step solution

Meaning of stress

Mathematically, the stress value may be calculated by dividing the resisting force acting on the body by the cross-sectional area of the object.

The radius of the wire can be calculated from the relationship between stress and strain.

Given information

Given data:

The mass,\(m = 25\;{\rm{kg}}\).

The angle made by the wire with the horizontal, \(\theta = 12^\circ \).

Evaluation of the tension force in the wire

The free-body diagram of the system can be drawn as follows:

Here, \(T\) is the tension in the aluminum wire; \(L\) is the original length of the wire; \(L + \Delta L\) is the length of the wire after the hanging of the mass.

According to Newton’s second law of motion, the expression for the net force on the mass can be written as follows:

\(\begin{array}{c}{F_{{\rm{net}}}} = mg\\2T\sin \theta = mg\\T = \frac{{mg}}{{2\sin \theta }}\end{array}\)

Substitute the values in the above expression.

\(\begin{array}{l}T = \frac{{\left( {25\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)}}{{2\sin \left( {12^\circ } \right)}}\\T = 589.2\;{\rm{N}}\end{array}\)

Evaluation of the fractional change in the length of the aluminum wire

The fractional change in the length of the aluminum wire can be determined using the following figure.

From the above figure, the cosine of the angle is as follows:

\(\begin{array}{c}{\rm{cos}}\theta = \frac{{\left( {\frac{L}{2}} \right)}}{{\left( {\frac{{L + \Delta L}}{2}} \right)}}\\{\rm{cos}}\theta = \frac{{\left( {\frac{L}{2}} \right)}}{{\left( {\frac{L}{2} + \frac{{\Delta L}}{2}} \right)}}\\{\rm{cos}}\theta = \frac{{\left( {\frac{L}{2}} \right)}}{{\left( {\frac{L}{2}} \right)\left[ {1 + \frac{{\Delta L}}{2}\left( {\frac{2}{L}} \right)} \right]}}\\\cos \theta = \frac{1}{{1 + \frac{{\Delta L}}{L}}}\end{array}\)

Solve further as follows:

\(\begin{array}{c}\frac{{\Delta L}}{L} = \frac{1}{{\cos \theta }} - 1\\\frac{{\Delta L}}{L} = \frac{1}{{\cos \left( {12^\circ } \right)}} - 1\\\frac{{\Delta L}}{L} = 2.23 \times {10^{ - 2}}\end{array}\)

Evaluation of the radius of the wire

The expression for the cross-sectional area of the wire is as follows:

\(A = \pi {r^2}\)

Here, \(r\) is the radius of the wire.

The expression for the young’s modulus is as follows:

\(\frac{T}{A} = e\left( {\frac{{\Delta L}}{L}} \right)\)

Here, \(A\) is the sectional area of the wire; \(e\) is the modulus of elasticity of aluminum, and its value is \(70 \times {10^9}\;{{\rm{N}} \mathord{\left/{\vphantom {{\rm{N}} {{{\rm{m}}^{\rm{2}}}}}} \right.} {{{\rm{m}}^{\rm{2}}}}}\).

The above expression can be written as follows:

\(\begin{array}{c}\frac{T}{{\pi {r^2}}} = e\left( {\frac{{\Delta L}}{L}} \right)\\{r^2} = \frac{T}{{\pi \left[ {e\left( {\frac{{\Delta L}}{L}} \right)} \right]}}\\r = \sqrt {\frac{T}{{\pi \left[ {e\left( {\frac{{\Delta L}}{L}} \right)} \right]}}} \end{array}\)

Substitute the values in the above expression to get the radius of the wire.

\(\begin{array}{l}r = \sqrt {\frac{{\left( {589.2\;{\rm{N}}} \right)}}{{\pi \left( {70 \times {{10}^9}\;{{\rm{N}} \mathord{\left/{\vphantom {{\rm{N}} {{{\rm{m}}^{\rm{2}}}}}} \right.} {{{\rm{m}}^{\rm{2}}}}}} \right)\left( {2.23 \times {{10}^{ - 2}}} \right)}}} \\r = 3.5 \times {10^{ - 4}}\;{\rm{m}}\end{array}\)

Thus, the radius of the wire is \(3.5 \times {10^{ - 4}}\;{\rm{m}}\).