The force, F, of the wind blowing against a building is given by F=CD ρV^2 A / 2, where V is the

The force, F, of the wind blowing against a building is...

The force, $F$, of the wind blowing against a building is given by $F=C_{D} \rho V^{2} A / 2,$ where $V$ is the wind speed, $\rho$ the density of the air, $A$ the cross-sectional area of the building, and $C_{D}$ is a constant termed the drag coefficient. Determine the dimensions of the drag coefficient.

Key Concepts

Dimensional Analysis

Dimensional analysis involves using the fundamental dimensions (typically mass, length, and time) associated with physical quantities to verify the consistency of equations and to derive relationships between variables. It is used to check that the units on both sides of an equation match, and to infer the dimensions of unknown coefficients when the dimensions of other terms are known.

Drag Coefficient

The drag coefficient is a dimensionless parameter in fluid dynamics that quantifies the drag or resistance of an object in a fluid environment. It is used to characterize how the shape, flow conditions, and surface roughness of the object influence the aerodynamic or hydrodynamic drag.

Dimensionless Quantity

A dimensionless quantity is a number without any physical units that results from the ratio of quantities with the same dimensions. In this context, the drag coefficient is dimensionless because it is constructed to ensure that its multiplication with terms having specific physical dimensions results in a force with the appropriate dimensional consistency.

Transcript

00:01 The force f of the wind blowing against a building f is given by cd into rho v squared a by 2.

00:15 Okay, now that's the drag force.

00:19 F is the drag force cd is the drag coefficient or coefficient of drag.

00:42 Rho is the density, the wind, air in this case v is nothing but the velocity of the wind and a is the area exposed cross -sectional area of the building now what they're asking us to do is find the unit of cd the easiest way we can actually find the unit of cd is by substituting the rest of the units apart from cd and then finally finding it.

01:34 So f has the unit, f is nothing but mass times acceleration which would mean in dimensional form, m is the mass acceleration is length divided by t squared.

01:53 Okay that's the dimensions of f that's the right hand side of the problem in the left hand side we've got rho which is the density.

02:06 The density is nothing but mass divided by volume is length cubed so it is going to be nothing but l to the power 3.

02:15 Area, the next thing is velocity v is nothing but length divided by time squared length divided by time rather, mass by basically the meters per second so that would be length by time and finally we've got the area a which is basically length squared because you know normally the units of area is in the form of meter squared centimeter squared so they're all length squared.

02:48 So we've got everything in place so it's just a matter of substituting them in this form here so f equals to, so let's write down the equation once again f equals to cd into rho v squared a by 2 so we're just going to be substituting these dimensions.

03:08 F is ml by t squared.

03:16 That is equal to cd is the one that we are after rho is ml cube so this is ml cube v is nothing but, i'll just spacket it, v is nothing but l by t, the whole squared and area is just the area which is length squared and because 2 is a number it does not have any units so we do not need to consider anything for 2 we're only considering those values that has units because 2 is just a number there are no units associated with 2 so we do not need to consider the units of 2.

04:10 2 does not have units so we can just neglect it now if you look at the whole expression i will try and simplify the right hand side.

04:19 Cd is the one that we are trying to find out the units for so we'll just keep cd as it is...

Please give Ace some feedback