# Shear stress

## Homework Statement

how to make μ = ρ v as in 8-41 ?

## The Attempt at a Solution

since μ = viscocity , it has unit =( kgms^-2)s m^-2 , = kg(m^-1)(s^-1)
ρ v = (kgm^-3)(ms^-1 ) = kg(m^-2)(s^-1) , the unit for LHS and RHS are not the same , can someone help pls ?

#### Attachments

• 88.1 KB Views: 279

## Answers and Replies

Related Introductory Physics Homework Help News on Phys.org
SteamKing
Staff Emeritus
Homework Helper

## Homework Statement

how to make μ = ρ v as in 8-41 ?

## The Attempt at a Solution

since μ = viscocity , it has unit =( kgms^-2)s m^-2 , = kg(m^-1)(s^-1)
ρ v = (kgm^-3)(ms^-1 ) = kg(m^-2)(s^-1) , the unit for LHS and RHS are not the same , can someone help pls ?
You've mixed up lower case v for velocity with the lower case Greek letter nu (##\nu##), which is used to represent the kinematic viscosity of a fluid.

The kinematic viscosity is defined ##\nu = \frac{\mu}{\rho}##

See this article for a discussion of the units:

https://en.wikipedia.org/wiki/Viscosity

You've mixed up lower case v for velocity with the lower case Greek letter nu (##\nu##), which is used to represent the kinematic viscosity of a fluid.

The kinematic viscosity is defined ##\nu = \frac{\mu}{\rho}##

See this article for a discussion of the units:

https://en.wikipedia.org/wiki/Viscosity
ok , can you pls explain how the author get the formula of viscous sublayer ? after substitute u= sqrt rt (τ / ρ ) into 8-41 , I have u = ( v / y ) sqrt rt (τ / ρ ) , hwo to get the same form as in 8-42 ?

#### Attachments

• 67.4 KB Views: 273
TSny
Homework Helper
Gold Member
Note that ##\sqrt{\tau_w/\rho}## is defined to be a fictitious velocity ##u##- (called "friction velocity"). (I'm not sure I got the notation right. It looks like there is a subscript "-" on the ##u## for the fictitious velocity. But it is hard for me to read.) See discussion below equation 8-41.
##u##- is not the same as the actual velocity ##u##.

Last edited:
Note that ##\sqrt{\tau_w/\rho}## is defined to be the "fictitious velocity" ##u##-. (I'm not sure I got the notation right. It looks like there is a subscript "-" on the ##u## for the fictitious velocity. But it is hard for me to read.) See discussion below equation 8-41.
##u##- is not the same as the actual velocity ##u##.
u / ##u##- , i haver u [ sqrt rt ( p / т ) ] , how to get the same form as in the 8-42 ?

TSny
Homework Helper
Gold Member
Solve ##u##- = ##\sqrt{\tau/\rho}## for ##\tau##. Then sustitute this expression for ##\tau## into equation 8-41. You should then be able to rearrange it to get 8-42.

Solve ##u##- = ##\sqrt{\tau/\rho}## for ##\tau##. Then sustitute this expression for ##\tau## into equation 8-41. You should then be able to rearrange it to get 8-42.
hwo to do that ? I have tried , but didn't get the same form ,

#### Attachments

• 31.2 KB Views: 292
haruspex