Home / Engineering & CS /The mine skip is being hauled to the surface over the curved track by the cable wound around the 30 -in. drum, which turns at the constant clockwise speed of \( 115 \mathrm{rev} / \mathrm{min} \). The shape of the track is designed so that \( y=x^{2} / 28 \), where \( x \) and \( y \) are in feet. Calculate the magnitude of the total acceleration of the skip as it reaches a level of \( 2.9 \mathrm{ft} \) below the top. Neglect the dimensions of the skip compared with those of the path. Recall that the radius of curvature is given by Answer: \( a= \) \( \mathrm{ft} / \mathrm{sec}^{2} \)

Question The mine skip is being hauled to the surface over the curved track by the cable wound around the 30 -in. drum, which turns at the constant clockwise speed of \( 115 \mathrm{rev} / \mathrm{min} \). The shape of the track is designed so that \( y=x^{2} / 28 \), where \( x \) and \( y \) are in feet. Calculate the magnitude of the total acceleration of the skip as it reaches a level of \( 2.9 \mathrm{ft} \) below the top. Neglect the dimensions of the skip compared with those of the path. Recall that the radius of curvature is given by Answer: \( a= \) \( \mathrm{ft} / \mathrm{sec}^{2} \)

ZF1UOT The Asker · Mechanical Engineering

Transcribed Image Text: The mine skip is being hauled to the surface over the curved track by the cable wound around the 30 -in. drum, which turns at the constant clockwise speed of \( 115 \mathrm{rev} / \mathrm{min} \). The shape of the track is designed so that \( y=x^{2} / 28 \), where \( x \) and \( y \) are in feet. Calculate the magnitude of the total acceleration of the skip as it reaches a level of \( 2.9 \mathrm{ft} \) below the top. Neglect the dimensions of the skip compared with those of the path. Recall that the radius of curvature is given by Answer: \( a= \) \( \mathrm{ft} / \mathrm{sec}^{2} \)
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Transcribed Image Text: The mine skip is being hauled to the surface over the curved track by the cable wound around the 30 -in. drum, which turns at the constant clockwise speed of \( 115 \mathrm{rev} / \mathrm{min} \). The shape of the track is designed so that \( y=x^{2} / 28 \), where \( x \) and \( y \) are in feet. Calculate the magnitude of the total acceleration of the skip as it reaches a level of \( 2.9 \mathrm{ft} \) below the top. Neglect the dimensions of the skip compared with those of the path. Recall that the radius of curvature is given by Answer: \( a= \) \( \mathrm{ft} / \mathrm{sec}^{2} \)
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S3JUUF

【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/2At \( \mathrm{{y}={2.9}{f}{t},} \) the value of x:\( \begin{align*} \mathrm{{y}} &= \mathrm{\frac{{x}^{{2}}}{{28}}}\\[3pt]\mathrm{{2.9}} &= \mathrm{\frac{{x}^{{2}}}{{28}}}\\[3pt]\mathrm{{x}} &= \mathrm{{9.011}{f}{t}} \end{align*} \)Radius of curvature at \( \mathrm{{x}={9.011}{f}{t}:} \)\( \begin{align*} \mathrm{\rho} &= \mathrm{\frac{{{\left[{1}+{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}^{{2}}\right]}^{{\frac{{3}}{{2}}}}}}{{\frac{{{d}^{{2}}{y}}}{{{\left.{d}{x}\right.}^{{2}}}}}}}\\[3pt]\mathrm{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}} &= \mathrm{\frac{{x}}{{14}}}\\[3pt]\mathrm{\frac{{{d}^{{2}}{y}}}{{{\left.{d}{x}\right.}^{{2}}}}} &= \mathrm{\frac{{1}}{{14}}}\\[3pt]\mathrm{\rho} &= \mathrm{\frac{{{\left[{1}+{\left(\frac{{x}}{{14}}\right)}^{{2}}\right]}^{{\frac{{3}}{{2}}}}}}{{\frac{{1}}{{14}}}}}\\[3pt] &= \mathrm{{14}{\left[{1}+{\left(\frac{{x}}{{14}}\right)}^{{2}}\right]}^{{\frac{{3}}{{2}}}}}\\[3pt] &= \mathrm{{14}{\left[{1}+{\left(\frac{{9.011}}{{ ... See the full answer

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