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General Physics I:

Sunil Karna

Section 8.2 Examples A

Example 8.2.1.

A mass of 10 kg is supported by a steel wire \((Y = 20\times10^{10} N/m^{2})\) of iameter 0.5 mm and length 2 m.
  1. What is the stress in the wire?
  2. What is the strain of the wire?
  3. How much does the wire stretch?
  4. What is the effective force constant of the wire?
  5. How much energy is stored in the wire?
Solution.
Given: \(m = 10 \,kg, \quad Y = 20\times 10^{10} \,N/m^{2}, \quad r = 2.5\times 10^{-4} \,m,\quad L = 2 \,m.\)
  1. \begin{equation*} p = \frac{F}{A}, \end{equation*}
    \begin{equation*} \quad F = w = mg = 98 \,N, \end{equation*}
    \begin{equation*} \quad A =\pi r^{2} = 1.964\times 10^{-2} m^{2}, \end{equation*}
    \begin{equation*} \quad p = 4.99\times 10^{8} \,Pa \end{equation*}
  2. \begin{equation*} Y = \frac{p}{s}, \quad s = 2.50\times 10^{-3} \end{equation*}
  3. \begin{equation*} s = \Delta L/L,\quad \Delta L = 4.99\times 10^{-3} \,m \end{equation*}
  4. \begin{equation*} F = k x, \quad x = \Delta L, \quad k = 1.96\times 10^{4} \,N/m \end{equation*}
  5. \begin{equation*} E = 1/2 k x^{2}, \quad E = 0.245 \,J \end{equation*}

Example 8.2.2.

Water has a Bulk modulus of \(0.21\times 10^{10} \,Pa\text{.}\) What is the percent change in volume for a drop of water which circulates from the top of a lake to the bottom if the lake is 1000 m deep where the pressure has increased by 9.8 MPa?
Solution.
Given: \(B = 0.21\times 10^{10} \,Pa, \quad \Delta p = 9.8\times 10^{6} \,Pa.\)
\begin{equation*} B = \frac{\Delta p}{s_{v}}, \quad s_{v} = \frac{\Delta V}{V} = 4.67\times 10^{-3} = 0.467 \% \end{equation*}

Example 8.2.3.

A solid copper sphere of \(0.25 \,m^{3}\) volume is placed 30.5 m below the ocean surface where the pressure is \(3.00 \times 10^{5} \,N/m^{2}\text{.}\) What is the change in volume of the sphere? The bulk modulus for copper is \(14 \times 10^{10} \,N/m^{2}.\)
Solution.
Given: \(V=0.25 \,m^{3},\quad h=30.5 \,m,\quad p=3.00 \times 10^{5} \,N/m^{2},\) \(B=14 \times 10^{10} \,N/m^{2}.\)
\begin{equation*} Now,\quad B=-\frac{pV}{\Delta V} \end{equation*}
\begin{equation*} \Rightarrow\quad \Delta V = -\frac{pV}{B} = -\frac{(3.00 \times 10^{5})(0.25)}{(14 \times 10^{10} N/m^{2})} = - 5.36\times 10^{-7} m^{3} \end{equation*}
-ve sign shows that volume has decreased.

Example 8.2.4.

A copper and aluminum wires are welded together at their ends. The original length of each wire is 50.0 cm and each has a diameter of 0.750 mm. A mass of 10.0 kg is suspended from the combined wire. By how much will the combined wire stretch? Given: \((Y_{a} = 7.0\times 10^{10} \,N/m^{2}) \) and \((Y_{c} = 11.0\times 10^{10} \,N/m^{2}).\)
Solution.
Given: \(m=10 \,kg, \quad l_{c}=l_{a} = 0.5 \,m, \quad d_{c}=d_{a}=0.750\times10^{-3} \,m, g=9.8 \,m/s^{2}.\)
The combined stretch is gieven by
\begin{equation*} \Delta l = \Delta l_{c}+\Delta l_{a} = \frac{Fl_{c}}{AY_{c}}+\frac{Fl_{a}}{AY_{a}} \end{equation*}
\begin{equation*} = \frac{F}{A} \left[\frac{l_{c}}{Y_{c}}+\frac{l_{a}}{Y_{a}}\right] \end{equation*}
\begin{equation*} = \frac{10\times 9.8\times 4}{\pi\left(0.75\times10^{-3}\right)^{2}} \left[\frac{0.5}{11.0\times 10^{10}}+\frac{0.5}{7.0\times 10^{10}}\right] \end{equation*}
\begin{equation*} \therefore \quad \Delta l = 2.59\times10^{-3}m = 2.59 \,mm \end{equation*}

Example 8.2.5.

A brass cube, 5.00 cm on a side, is subjected to a tangential force. If the angle of shear is measured in radians to be 0.010 rad, what is the magnitude of the tangential force?
Solution.
Given: \(l=0.05 \,m, \theta = 0.010 \,rad, \sigma _{b}= 4\times 10^{10} \,Pa.\)
\begin{equation*} \sigma _{b} = \frac{F}{A\theta} \end{equation*}
\begin{equation*} \therefore\quad F = \sigma_{b}A\theta =(4\times 10^{10})(0.05)^{2} (0.010) =125 \,N \end{equation*}

Example 8.2.6.

A horizontal spring stretches 20.0 cm when a force of 10.0 N is applied to the spring. By how much will it stretch if a 30.0 N force is now applied to the spring? If the same spring is placed in the vertical and a weight of 10.0 N is hung from the spring, will the results change?
Solution.
Given \(\Delta l_{1} =20 \,cm=0.2 \,m, F_{1} =10.0 \,N, F_{2} =30.0 \,N,\text{.}\) The restoring force in a spring
\begin{equation*} F=-kx. \end{equation*}
\begin{equation*} \therefore\quad F_{1}=k\Delta l_{1} \end{equation*}
\begin{equation*} \text{or,}\quad k = \frac{F_{1}}{\Delta l_{1}} = \frac{10}{0.2} = 50 \,N/m \end{equation*}
Now,
\begin{equation*} \Delta l_{2} = \frac{F_{2}}{k} = \frac{30.0}{50} = 0.6 \,m = 60 \,cm \end{equation*}
If the applied force remains same then stretch made in the spring remains the same, orientation of spring does not have any effect on its elongation.

Example 8.2.7.

A copper rod, 0.400 cm in diameter, supports a load of 150 kg suspended from one end. Will the rod return to its initial length when the load is removed or has this load exceeded the elastic limit of the rod? Note: Elastic limit of copper = \(1.39\times 10^{8} \,N/m^{2}.\)
Solution.
Given: \(Stress_{max} = 1.39\times 10^{8} \,N/m^{2}, \quad d=0.400 \,cm= 0.004 \,m, \quad g=9.8 \,m/s^{2}\)
\begin{equation*} Stress = \frac{F}{A} \end{equation*}
\begin{equation*} \therefore \quad F_{max} = Stress_{max}A = (1.39\times 10^{8})(\pi d^{2}/4) = 1746.73 \,N \end{equation*}
Here the applied force, \(F =150\times 9.8=1470 \,N\text{,}\) is less then the maximum force the copper wire can hold, hence the rod regains its original length after the removal of applied force.
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