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Expert-verified Found in: Page 230 ### Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922 ## (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) $$x' = 2.5\;{\rm{m}}$$.

See the step by step solution

## 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 FP 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. ### Want to see more solutions like these? 