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22P

Expert-verifiedFound in: Page 326

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**A straight ladder is leaning against the wall of a house. The ladder has rails 4.90 m long, joined by rungs 0.410 m long. Its bottom end is on solid but sloping ground so that the top of the ladder is 0.690 m to the left of where it should be, and the ladder is unsafe to climb. You want to put a flat rock under one foot of the ladder to compensate for the slope of the ground. (a) What should be the thickness of the rock? (b) Does using ideas from this chapter make it easier to explain the solution to part (a)? Explain your answer.**

The thickness of rock is 0.0574 m.

The length of rails of ladder is $L=4.90m$

The length of rungs joined is $l=0.410m$

The position for the top of ladder is $d=0.690m$

**For small angles, the angular displacement is described as the ratio of arc length to the radius of the circular path.**

The angle for the top of the ladder is given as:

$\theta =\frac{d}{L}$

For $d=0.690m$ and L=4.90 m , the above equation gives-

$\theta =\frac{0.690m}{4.90m}\phantom{\rule{0ex}{0ex}}\theta =0.14rad$

The thickness of rock is given as:

$t=l\xb7\theta $

For $\theta =0.14rad$ and $l=0.410m$, the above equation becomes-

$t=\left(0.410m\right)\left(0.14rad\right)\phantom{\rule{0ex}{0ex}}t=0.0574m$

Therefore, the thickness of rock is 0.0574 m .

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