Q11P (II) A 75-kg adult sits at one e... [FREE SOLUTION]

Book edition 7th

Author(s) Douglas C. Giancoli

Pages 978 pages

ISBN 978-0321625922

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Short Answer

^{(II) A 75-kg adult sits at one end of a 9.0-m-long board. His 25-kg child sits on the other end. (a) Where should the pivot be placed so that the board is balanced, ignoring the board’s mass? (b) Find the pivot point if the board is uniform and has a mass of 15 kg.}

The results of the values of the positions of the pivot point are (a) \(x = 2.25\;{\rm{m}}\) and (b) \(x' = 2.5\;{\rm{m}}\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

The mass of the adult is \(M = 75\;{\rm{kg}}\).

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

The mass of the board is \({m_{\rm{B}}} = 15\;{\rm{kg}}\).

The length of the board is \(l = 9\;{\rm{m}}\).

Step 2: Understanding the torque exerted on the board

In this problem, the upward force F_P does not exert any torque about the pivot point. Also, the pivot must be kept in such a place where the net torque on the board is equal to zero.

Step 3: Free body diagram and calculation of the position of the pivot point

The following is the free body diagram.

The relation to calculate the net torque can be written as:

\(\begin{array}{c}\sum \tau = 0\\\left( {Mg \times x} \right) - \left( {mg \times \left( {l - x} \right)} \right) = 0\end{array}\)

Here, \(g\) is the gravitational acceleration and \(x\) is the distance from the left end point to point P.

On plugging the values in the above relation, you get:

\(\begin{array}{c}\left( {75\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^2} \times x} \right) - \left( {25\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^2} \times \left( {9\;{\rm{m}} - x} \right)} \right) = 0\\x = 2.25\;{\rm{m}}\end{array}\)

Thus, \(x = 2.25\;{\rm{m}}\) is the position of the pivot point from the adult.

Step 4: Calculation of pivot point if the board is uniform

The relation to calculate the pivot point can be written as:

\(\begin{array}{c}\sum \tau = 0\\\left( {Mg \times x'} \right) - \left( {mg \times \left( {l - x'} \right)} \right) - \left( {{m_{\rm{B}}}g \times \left( {\frac{l}{2} - x'} \right)} \right) = 0\end{array}\)

On plugging the values in the above relation, you get:

\(\begin{array}{c}\left[ \begin{array}{l}\left( {75\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^2} \times x'} \right) - \\\left( {25\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^2} \times \left( {9\;{\rm{m}} - x'} \right)} \right) - \left( {15\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^2} \times \left( {\frac{{9\;{\rm{m}}}}{2} - x'} \right)} \right)\end{array} \right] = 0\\x' = 2.5\;{\rm{m}}\end{array}\)

Thus, \(x' = 2.5\;{\rm{m}}\) is the distance of the pivot point from the adult.

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Physics Principles with Applications

Answers without the blur.

Short Answer

^{(II) A 75-kg adult sits at one end of a 9.0-m-long board. His 25-kg child sits on the other end. (a) Where should the pivot be placed so that the board is balanced, ignoring the board’s mass? (b) Find the pivot point if the board is uniform and has a mass of 15 kg.}

Step by Step Solution

Step 1: Given data

Step 2: Understanding the torque exerted on the board

Step 3: Free body diagram and calculation of the position of the pivot point

Step 4: Calculation of pivot point if the board is uniform

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Physics Principles with Applications

Answers without the blur.

Short Answer

(II) A 75-kg adult sits at one end of a 9.0-m-long board. His 25-kg child sits on the other end. (a) Where should the pivot be placed so that the board is balanced, ignoring the board’s mass? (b) Find the pivot point if the board is uniform and has a mass of 15 kg.

Step by Step Solution

Step 1: Given data

Step 2: Understanding the torque exerted on the board

Step 3: Free body diagram and calculation of the position of the pivot point

Step 4: Calculation of pivot point if the board is uniform

Most popular questions for Physics Textbooks

Question: The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.2 m above the ground, how steep a slope can the truck be parked on without tipping over (Fig. 9–81)?

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^{(II) A 75-kg adult sits at one end of a 9.0-m-long board. His 25-kg child sits on the other end. (a) Where should the pivot be placed so that the board is balanced, ignoring the board’s mass? (b) Find the pivot point if the board is uniform and has a mass of 15 kg.}

^{Question: The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.2 m above the ground, how steep a slope can the truck be parked on without tipping over (Fig. 9–81)?}