A 25-kg object is being lifted by two people pulling on the ends of a 1.15-mm-diameter nylon cord that goes over two 3.00-m-high poles 4.0 m apart, as shown in Fig. 9–86. How high above the floor will the object be when the cord breaks?

The stress value can be determined by dividing the ultimate strength's value by the factor of safety. Its value is directly related to the value of the ultimate strength.

Given data:

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

The diameter of the nylon cord, \(d = 1.15\;{\rm{mm}}\).

The height of the poles, \(H = 3\;{\rm{m}}\).

The distance between the poles, \(l = 4.0\;{\rm{m}}\).

The tension force in the cord can be calculated by using the following expression:

\({F_{\rm{T}}} = \left( {\frac{\pi }{4}{d^2}} \right){\sigma _{\rm{T}}}\)

Here, \({\sigma _{\rm{T}}}\) is the ultimate tensile strength of the nylon, and its value is \(500 \times {10^6}\;{\rm{Pa}}\).

Substitute the values in the above equation.

\(\begin{array}{l}{F_{\rm{T}}} = \left[ {\frac{\pi }{4}{{\left\{ {\left( {1.15\;{\rm{mm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{mm}}}}} \right)} \right\}}^2}} \right]\left( {500 \times {{10}^6}\;{\rm{Pa}}} \right)\\{F_{\rm{T}}} = 519.1\;N\end{array}\)

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

Here, \(h\) is the height of the object above the floor.

Apply the force equilibrium condition along the vertical direction.

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\2{F_{\rm{T}}}\sin \theta - mg = 0\\2{F_{\rm{T}}}\sin \theta = mg\\\theta = {\sin ^{ - 1}}\left( {\frac{{mg}}{{2{F_{\rm{T}}}}}} \right)\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}\theta = {\sin ^{ - 1}}\left[ {\frac{{\left( {25\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)}}{{2\left( {519.1\;N} \right)}}} \right]\\\theta = 13.65^\circ \end{array}\)

The distance of the object above the floor can be calculated as follows:

\(\begin{array}{c}\tan \theta = \frac{{\left( {3\;{\rm{m}}} \right) - h}}{{\frac{l}{2}}}\\2\left[ {\left( {3\;{\rm{m}}} \right) - h} \right] = \tan \theta \left( l \right)\\\left( {6\;{\rm{m}}} \right) - 2h = \tan \theta \left( l \right)\\h = \frac{{\left( {6\;{\rm{m}}} \right) - \tan \theta \left( l \right)}}{2}\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}h = \frac{{\left( {6\;{\rm{m}}} \right) - \tan \left( {13.65^\circ } \right)\left( {4\;{\rm{m}}} \right)}}{2}\\h = 2.51\;{\rm{m}}\end{array}\)

Thus, the distance of the object above the floor is \(2.51\;{\rm{m}}\).