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Chapter 18: Q. 44 (page 512)

A 15 $掳$ C, 2.0-cm-diameter aluminum bar just barely slips between two rigid steel walls 10.0 cm apart. If the bar is warmed to 25 $掳$ C, how much force does it exert on each wall?

Short Answer

Expert verified

The accurate answer is $14.5 k n$

Step by step solution

01

Description

A 15 $掳$ C, 2.0-cm-diameter aluminum bar just barely slips between two rigid steel walls 10.0 cm apart.

The young module depicts

$p = \frac{F}{A} = Y \frac{鈭� L}{L}$

02

Finding the result

The description provided that

$F = A Y \frac{鈭� L}{L} = \frac{蟺}{4} D^{2} Y \frac{鈭� L}{L}$

From the expression for linear thermal expansion is:

$\frac{鈭� L}{L} = 伪鈭� T$

Therefore, the final expansion will be:

$F = \frac{{蟺顿}^{2}}{4} Y 伪鈭� T$

The table of 18.4 depicts that the coefficient of linear thermal expansion of Aluminium is almost $23 . 10^{- 5} 1 / k$ .

The table of 14.3 depicts that Aluminium's young modules is $Y = 20 . 10^{10} N / m^{2}$ .

Thus, $F = \frac{蟺路 0 . 02^{2}}{4} 路 20 路 10^{10} 路 2.3 路 10^{- 5} 路 10 = 14500 N$

Therefore, If the bar is warmed to 25 $掳$ C, the force of $14.5 k N$ , it exert on each wall.

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