Question

A 15$°$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$°$C, how much force does it exert on each wall?

Short answer

The accurate answer is$14.5kn$

Step-by-step solution

Description 

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

The young module depicts

$p=\frac{F}{A}=Y\frac{∆L}{L}$

Finding the result 

The description provided that

$F=AY\frac{∆L}{L}=\frac{\mathrm{\pi }}{4}{D}^{2}Y\frac{∆L}{L}$

From the expression for linear thermal expansion is:

$\frac{∆L}{L}=\alpha ∆T$

Therefore, the final expansion will be:

$F=\frac{{\mathrm{\pi D}}^2}{4}Y\alpha ∆T$

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, $F=\frac{\mathrm{\pi }·0.{02}^2}{4}·20·{10}^{10}·2.3·{10}^{-5}·10=14500N$

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