Question 8.15: For the following thrust bearing (Fig. 8.21), show that the ...

It is given that the velocity profile is

and load

where $n=h_{1} / h_{2}$

Force on the slider in the x-direction is

Now, $\tau_{s}=-\mu\left(\frac{\partial u}{\partial y}\right)_{y=h}$

and $u=U\left(1-\frac{y}{h}\right)\left[1-3 \frac{y}{h}\left(1-\frac{2}{n+1} \cdot \frac{h_{1}}{h}\right)\right], \text { so we get }$

Also, $h=h_{1}-\left(h_{1}-h_{2}\right) \frac{x}{L}$

Thus, $\int_{0}^{L} \tau_{s} d x=\frac{L}{h_{1}-h_{2}} \int_{h_{2}}^{h_{1}} \tau_{s} d h$

Also, load $P=\int_{0}^{L}\left(p-p_{0}\right) \frac{ d x \cos \alpha}{\cos \alpha} \quad \begin{array}{l}\text { (neglecting contribution of } \\\tau_{s} \text { to load; } \alpha \text { is small) }\end{array}$

∴ $F_{s}=\int_{0}^{L} \tau_{s} d x+\tan \alpha(P)$

or $F_{s}=\int_{0}^{L} \tau_{s} d x+\tan \alpha(P)$$F_{s}=\frac{\mu U L}{h_{2}(n-1)}\left(-2 \ln n+\frac{6(n-1)}{n+1}\right)+\left[\frac{h_{1}-h_{2}}{L}\right] \times$

Now the force on the guide is

But $\tau_{G}=-\mu\left(\frac{\partial u}{\partial y}\right)_{y=0}$

The expression is $\tau_{G}=\mu U\left[\frac{4}{h}-\frac{6}{n+1} \cdot \frac{h_{1}}{h^{2}}\right]$

∴ $F_{G}=\int_{0}^{L} \tau_{G} d x$

$=\frac{L}{h_{1}-h_{2}} \int_{h_{2}}^{h_{1}} \tau_{G} d h$ (as before)

which is the same as $F_{s}$.