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Q11P

Expert-verifiedFound in: Page 346

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**

**Answer:**

** **

- The magnitude of force from the left pedestal, ${F}_{1}=1.2\times {10}^{3}\text{N}$ .
- The direction of force from the left pedestal is downward.
- The magnitude of force from the right pedestal, ${F}_{2}=1.7\times {10}^{3}\text{N}$ .
- The direction of force from the right pedestal is upward.
- The left pedestal is being stretched.
- The right pedestal is being compressed.

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

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:

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

** **

FBD of the diving board:

Applying equilibrium conditions to FBD:

$\sum {F}_{y}=0\phantom{\rule{0ex}{0ex}}{F}_{1}+{F}_{2}-W=0\xb7\xb7\xb7\xb7\xb7\xb7\xb7\left(1\right)\phantom{\rule{0ex}{0ex}}$

Taking torque at point

$\sum \tau =0\phantom{\rule{0ex}{0ex}}{F}_{1}\times d+W\times \left(L-d\right)=0\xb7\xb7\xb7\xb7\xb7\xb7\left(2\right)\phantom{\rule{0ex}{0ex}}{F}_{1}\times 1.5+580\left(4.5-1.5\right)=0\phantom{\rule{0ex}{0ex}}$

By solving for F_{1} :

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

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,

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

** **

The negative value of F_{1} indicates that it is directed downward.

Now plugging in the value of F_{1} into equation no. :

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

By solving for F_{2} :

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

By rounding off to appropriate significant figures:

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

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

The positive value of F_{2} indicates that it is directed upward.

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

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

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