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Found in: Page 346

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Question: Figure 12-29 shows a diver of weight 580 N standing at the end of a diving board with a length of L =4.5 m and negligible v mass. The board is fixed to two pedestals (supports) that are separated by distance d = 1 .5 m . Of the forces acting on the board, what are the (a) magnitude and (b) direction (up or down) of the force from the left pedestal and the (c) magnitude and (d) direction (up or down) of the force from the right pedestal? (e) Which pedestal (left or right) is being stretched, and (f) which pedestal is being compressed?

1. The magnitude of force from the left pedestal, ${F}_{1}=1.2×{10}^{3}\text{N}$ .
2. The direction of force from the left pedestal is downward.
3. The magnitude of force from the right pedestal, ${F}_{2}=1.7×{10}^{3}\text{N}$ .
4. The direction of force from the right pedestal is upward.
5. The left pedestal is being stretched.
6. The right pedestal is being compressed.
See the step by step solution

## Step 1: Understanding the given information

$L=4.5\text{m}\phantom{\rule{0ex}{0ex}}d=1.5\text{m}\phantom{\rule{0ex}{0ex}}W=580\text{N}\phantom{\rule{0ex}{0ex}}$

## Step 2: Concept and formula used in the given question

Using the concept of static equilibrium, you can write the equation for the force and torque. From these equations, you can find forces and their directions. The equations are given below.

Static Equilibrium conditions:

$\sum {F}_{x}=0\phantom{\rule{0ex}{0ex}}\sum {F}_{y}=0\phantom{\rule{0ex}{0ex}}\sum \tau =0\phantom{\rule{0ex}{0ex}}$

## Step 3: (a) Calculation for the magnitude of the force from the left pedestal

FBD of the diving board:

Applying equilibrium conditions to FBD:

$\sum {F}_{y}=0\phantom{\rule{0ex}{0ex}}{F}_{1}+{F}_{2}-W=0·······\left(1\right)\phantom{\rule{0ex}{0ex}}$

Taking torque at point

$\sum \tau =0\phantom{\rule{0ex}{0ex}}{F}_{1}×d+W×\left(L-d\right)=0······\left(2\right)\phantom{\rule{0ex}{0ex}}{F}_{1}×1.5+580\left(4.5-1.5\right)=0\phantom{\rule{0ex}{0ex}}$

By solving for F1 :

${F}_{1}=-1160\text{N}$

By rounding off to appropriate significant figures:

${F}_{1}=1200\text{N}\phantom{\rule{0ex}{0ex}}$

The magnitude of force from the left pedestal is,

.${F}_{1}=1200\text{N}\phantom{\rule{0ex}{0ex}}=1.2×{10}^{3}\text{N}\phantom{\rule{0ex}{0ex}}$

## Step 4: (b) Calculation for the direction (up or down) of the force from the left pedestal

The negative value of F1 indicates that it is directed downward.

## Step 5: (c) Calculation for the magnitude of the force from the right pedestal

Now plugging in the value of F1 into equation no. :

$-1200+{F}_{2}-580=0$

By solving for F2 :

${F}_{2}=1780\text{N}$

By rounding off to appropriate significant figures:

${F}_{2}=1700\text{N}$

The magnitude of force from the right pedestal is ${F}_{2}=1.7×{10}^{3}\text{N}$ .

## Step 6: (d) Calculation for the direction (up or down) of the force from the right pedestal

The positive value of F2 indicates that it is directed upward.

## Step 7: (e) Calculation for which pedestal (left or right) is being stretched

The force on the left pedestal is in an upward direction, so the left pedestal is being stretched.

## Step 8: (f) Calculation for which pedestal is being compressed

The force on right pedestal is in a downward direction, so the right pedestal is being compressed.