- #1

Guillem_dlc

- 184

- 15

- Homework Statement
- The bench-guide of a machine (see figure) has the cross-section shown in the figure and an extension ##L=60\, \textrm{cm}##. The force exerted by the weight supported by the bench is ##1500\, \textrm{N}## and the bench moves (in a direction perpendicular to the figure) at a constant speed ##c=6000\, \textrm{mm}/\textrm{min}##. The clearance between the bench and the guide is ##h=0,1\, \textrm{mm}## and it is assumed to be filled with a lubricating oil of viscosity ##\nu =5\, \textrm{stoke}## and relative density ##\rho_r=0,9##. Calculate:

a) the shear stress ##\tau##, at ##\textrm{N}/\textrm{m}^2##. Solution: ##\tau =450\, \textrm{N}/\textrm{m}^2##

b) The power ##P##, in ##\textrm{W}##, required to overcome the frictional resistance of the fluid. Solution: ##P=6,361\, \textrm{W}##

c) The dynamic viscosity ##\mu## of the lubricant, in \textrm{decapoise} if, if it is desired to reduce the above power by ##5\%##. Solution: ##\mu =0,4274\, \textrm{kg}/(\textrm{m}\cdot \textrm{s})##

- Relevant Equations
- ##v=\dfrac{\mu}{\rho}##, ##\tau =\mu \dfrac{c}{h}##

Figure: bench (bancada), guide (guide), oil (aceite).

My attempt at a solution:

$$\textrm{extension } L=0,6\, \textrm{m}$$

$$F=1500\, \textrm{N},\,\, c=0,1\, \textrm{m}/\textrm{s},\,\, h=0,0001\, \textrm{m}$$

$$v=5\, \textrm{stoke}=0,0005\, \textrm{m}^2/\textrm{s},\,\, \rho_r=0,9$$

a) $$\rho_r=\dfrac{\rho}{\rho_{\textrm{H2O}}}\rightarrow \rho =900\, \textrm{kg}/\textrm{m}^3$$

We calculate the area of the bench:

$$v=\dfrac{\mu}{\rho}\rightarrow \mu =v\rho =0,45\, \textrm{Pa}\cdot \textrm{s}$$

$$\tau =\mu \dfrac{c}{h}\rightarrow \boxed{\tau =450\, \textrm{N}/\textrm{m}^2}$$

I have tried to do this. And b) and c) I have no idea how to approach them.

My attempt at a solution:

$$\textrm{extension } L=0,6\, \textrm{m}$$

$$F=1500\, \textrm{N},\,\, c=0,1\, \textrm{m}/\textrm{s},\,\, h=0,0001\, \textrm{m}$$

$$v=5\, \textrm{stoke}=0,0005\, \textrm{m}^2/\textrm{s},\,\, \rho_r=0,9$$

a) $$\rho_r=\dfrac{\rho}{\rho_{\textrm{H2O}}}\rightarrow \rho =900\, \textrm{kg}/\textrm{m}^3$$

We calculate the area of the bench:

$$v=\dfrac{\mu}{\rho}\rightarrow \mu =v\rho =0,45\, \textrm{Pa}\cdot \textrm{s}$$

$$\tau =\mu \dfrac{c}{h}\rightarrow \boxed{\tau =450\, \textrm{N}/\textrm{m}^2}$$

I have tried to do this. And b) and c) I have no idea how to approach them.