# Viscosity in lubricating fluid in the bench-guide of a machine

• Guillem_dlc
In summary: The steps seem logical to me.Where does the ## 1500 \rm{N}## force factor in? That is what I am unsure about. I feel like the shear stress should be a function of the load.Where does the ## 1500 \rm{N}## force factor in? That is what I am unsure about. I feel like the shear stress should be a function of the load.We don't care, do we? Because it does not run parallel to this force.I don't know. Does shear stress have dependencies on fluid pressure? Intuitively I feel like it should,
Guillem_dlc
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.

Lnewqban
I'm not sure I fully have been able to factor in all the parameters, but for starters let's see if we can get you closer than "I have no idea" for part (b).

Do a force balance in the direction of motion, for the guide. What do you get for the force of the required push ( in all variables - no numbers)?

Lnewqban
erobz said:
I'm not sure I fully have been able to factor in all the parameters, but for starters let's see if we can get you closer than "I have no idea" for part (b).

Do a force balance in the direction of motion, for the guide. What do you get for the force of the required push ( in all variables - no numbers)?
I've done this, look:

b) Find hypotenuse: ##H^2=c_1^2+c_2^2##
$$A=(0,04+\sqrt{0,015^2+0,03^2}+0,045+\sqrt{0,06^2+0,03^2}+0,05)\cdot 0,6=0,14137\, \textrm{m}^2$$
$$F=\tau \cdot A=63,62\, \textrm{N}$$
Because we have constant velocity:
$$\boxed{P=F\cdot c=6,362\, \textrm{W}}$$

c) We reduce ##5\% \rightarrow P'=6,0439\, \textrm{W}##
$$P'=F'c\rightarrow F'=\dfrac{P'}{c}=\tau' A\rightarrow \tau' =\dfrac{P'}{cA}\rightarrow$$
$$\mu' \dfrac{c}{h}=\dfrac{P'}{cA}\rightarrow \boxed{\mu '=\dfrac{P'h}{c^2A}=0,4275\, \textrm{Pa}\cdot \textrm{s}}$$

Lnewqban
Guillem_dlc said:
I've done this, look:

b) Find hypotenuse: ##H^2=c_1^2+c_2^2##
$$A=(0,04+\sqrt{0,015^2+0,03^2}+0,045+\sqrt{0,06^2+0,03^2}+0,05)\cdot 0,6=0,14137\, \textrm{m}^2$$
$$F=\tau \cdot A=63,62\, \textrm{N}$$
Because we have constant velocity:
$$\boxed{P=F\cdot c=6,362\, \textrm{W}}$$

c) We reduce ##5\% \rightarrow P'=6,0439\, \textrm{W}##
$$P'=F'c\rightarrow F'=\dfrac{P'}{c}=\tau' A\rightarrow \tau' =\dfrac{P'}{cA}\rightarrow$$
$$\mu' \dfrac{c}{h}=\dfrac{P'}{cA}\rightarrow \boxed{\mu '=\dfrac{P'h}{c^2A}=0,4275\, \textrm{Pa}\cdot \textrm{s}}$$
I'm confused. Those look like the answers to me?

erobz said:
I'm confused. Those look like the answers to me?
This is what I have tried to do for (b) and (c).

Guillem_dlc said:
This is what I have tried to do for (b) and (c).
You are saying those are not correct? Do you have the correct answers for part (b) and (c)?

erobz said:
You are saying those are not correct? Do you have the correct answers for part (b) and (c)?
I don't know if the steps I have done are correct or not. This I what I'm asking.

Guillem_dlc said:
I don't know if the steps I have done are correct or not. This I what I'm asking.
The steps seem logical to me.

Where does the ## 1500 \rm{N}## force factor in? That is what I am unsure about. I feel like the shear stress should be a function of the load.

erobz said:
Where does the ## 1500 \rm{N}## force factor in? That is what I am unsure about. I feel like the shear stress should be a function of the load.
We don't care, do we? Because it does not run parallel to this force.

Guillem_dlc said:
We don't care, do we? Because it does not run parallel to this force.
I don't know. Does shear stress have dependencies on fluid pressure? Intuitively I feel like it should, but perhaps it doesn't.

I'm just kind of paralleling the idea from solid mechanics that the frictional force is dependent on the Normal force, and the shear stress might be dependent on pressure in a similar manner.

If there is no dependency (or it's not an issue until extreme pressures), then it seems fine that the ##1500 \rm{N}## is ignorable.

Last edited:
erobz said:
I don't know. Does shear stress have dependencies on fluid pressure? Intuitively I feel like it should, but perhaps it doesn't.

I'm just kind of paralleling the idea from solid mechanics that the frictional force is dependent on the Normal force, and the shear stress might be dependent on pressure in a similar manner.

If there is no dependency (or it's not an issue until extreme pressures), then it seems fine that the ##1500 \rm{N}## is ignorable.
Ok ok

The gauge pressure from the load is pretty small in comparison to atmospheric...there is probably no effect to consider. I'm not trying to steer you into left field, it's just something that I wasn't sure about.

Guillem_dlc
erobz said:
I don't know. Does shear stress have dependencies on fluid pressure? Intuitively I feel like it should, but perhaps it doesn't.

I'm just kind of paralleling the idea from solid mechanics that the frictional force is dependent on the Normal force, and the shear stress might be dependent on pressure in a similar manner.

If there is no dependency (or it's not an issue until extreme pressures), then it seems fine that the ##1500 \rm{N}## is ignorable.
https://www.engineeringtoolbox.com/dynamic-absolute-kinematic-viscosity-d_412.html

Lnewqban said:
Ok perfect now I will take a look at it! Thanks

Lnewqban
Lnewqban said:
"The shear resistance in a fluid is caused by inter-molecular friction exerted when layers of fluid attempt to slide by one another."

Its not clear with this statement from the link why there wouldn't be some pressure dependence in the shear stress in my opinion. That being said, my fluids text doesn't bring up the possible effect of pressure on shear stress either. If there is any effect, it's probably negligible for any reasonable pressures.

erobz said:
"The shear resistance in a fluid is caused by inter-molecular friction exerted when layers of fluid attempt to slide by one another."

Its not clear with this statement from the link why there wouldn't be some pressure dependence in the shear stress in my opinion. That being said, my fluids text doesn't bring up the possible effect of pressure on shear stress either. If there is any effect, it's probably negligible for any reasonable pressures.
Excellent remark!
You may find this lecture interesting:
https://galileo.phys.virginia.edu/classes/152.mf1i.spring02/Viscosity.pdf

Guillem_dlc and erobz

## What is viscosity?

Viscosity is a measure of a fluid's resistance to flow. It is a property that describes how thick or thin a fluid is.

## Why is viscosity important in lubricating fluids?

Viscosity is important in lubricating fluids because it affects the fluid's ability to form a protective film between moving parts. A higher viscosity fluid will provide better lubrication and protect the machine from wear and tear.

## How is viscosity measured?

Viscosity can be measured using a viscometer, which measures the time it takes for a fluid to flow through a small tube under specific conditions. The unit of measurement for viscosity is typically centipoise (cP).

## What factors can affect the viscosity of a lubricating fluid?

The temperature, pressure, and shear rate can all affect the viscosity of a lubricating fluid. As temperature increases, viscosity typically decreases. Pressure can also decrease viscosity, while shear rate can increase or decrease viscosity depending on the type of fluid.

## How does viscosity impact the performance of a machine?

The viscosity of a lubricating fluid can impact the performance of a machine by affecting its ability to reduce friction and wear between moving parts. A lower viscosity fluid may not provide enough lubrication, while a higher viscosity fluid may cause excess drag and decrease efficiency.

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