Chapter 9, Problem 79GP

To determine

To Find: The height of the object from the floor.

Answer to Problem 79GP

The height of the object from the floor when the string breaks, is $\text{2}\text{.34 m}$ .

Explanation of Solution

Given:

Height of the poles, $\text{h = 3 m}$

Distance between the poles $\text{= 4 m}$

Distance between a pole to object $\text{= 2 m}$

Tensile strength of Nylon $\text{= 500}\times {\text{10}}^{\text{6}}{\text{ Nm}}^{\text{-2}}$

Mass of object, $\text{m = 25 kg}$

Diameter of Nylon string, $\begin{array}{l}\text{d = 1 mm }\\ \text{= 0}\text{.001 m}\end{array}$

Area, $\begin{array}{l}\text{A = }\frac{\text{π}}{\text{4}}\times {\text{ d}}^{\text{2}}\text{ = }\frac{\text{π}}{\text{4}}\times \text{ (0}{\text{.001)}}^{\text{2}}\\ \text{= 7}\text{.85}\times {\text{10}}^{\text{-7}}{\text{ m}}^{\text{2}}\end{array}$

Formula used:

We have to find out the vertical angle $\theta$ , made by object with the poles. For Newton's second law of motion is used along with equilibrium conditions for the vertical components.

  $\begin{array}{l}{\text{ΣFy = 2F}}_{\text{T}}\text{y - m}\times \text{g = 0}\\ {\text{2F}}_{\text{T}}\times \text{sin(θ) = m }\times \text{ g}\\ {\text{F}}_{\text{T}}\text{ = Tensile strength }\times \text{ Area}\end{array}$

Calculation:

Substitute the given values;

  $\begin{array}{l}{\text{θ = sin}}^{\text{-1}}\left(\frac{\text{m}\times \text{g}}{\text{2}\times {\text{F}}_{\text{T}}}\right)\\ {\text{θ = sin}}^{\text{-1}}\left(\frac{\text{25}\times \text{9}\text{.8}}{\text{2}\times {\text{F}}_{\text{T}}}\right)\\ {\text{F}}_{\text{T}}\text{ = 500}\times {\text{10}}^{\text{6}}\times \text{7}\text{.85}\times {\text{10}}^{\text{-7}}\text{ = 393 N}\\ {\text{θ = sin}}^{\text{-1}}\left(\frac{\text{25}\times \text{9}\text{.8}}{\text{2}\times \text{393}}\right){\text{ = sin}}^{\text{-1}}\left(\text{0}\text{.312}\right)\text{ = 18}{\text{.16}}^{\text{0}}\end{array}$

This angle $\theta$ will help to find the height of the object from the floor.

  $\begin{array}{l}\text{tan(θ) = }\frac{\text{3-h}}{\text{2}}\\ \text{3-h = 2}\times \text{tan(θ)}\\ \text{3-h = 2}\times \text{tan(18}{\text{.16}}^{\text{0}}\text{)}\\ \text{3-h = 0}\text{.66}\\ \text{h = 3-0}\text{.66 = 2}\text{.34 m}\end{array}$

Conclusion:

Thus, the height of the object from the floor when the string breaks, is $\text{2}\text{.34 m}$ .