windy physics?

Question

During a Chicago storm, winds can whip horizontally at speeds of 100 km/h. if the air strikes a person at the rate of 40 kg/s per square meter and is brought to rest, estimate the force of the wind on a person. assume the person a 1.50 m high and .50 m wide. compare to the typical maximum force of friction ( mu=1.0) between the person and the ground, if the person has a mass of 70kg.

Answer 1

It was along time ago that I studied Physics in School but I here you are with my try:

1) Speed of wind = 100 km/hr = 27.7 m/s

Value of air striking 40 Kg/s m² is not very clear to me. I would interpret as "stress rate" due to employed units. Let´s call it "S" and we assume that these "kg" are kg-force, so :

S= 40 Kg/sm² = (40 x 9.81 N) = 392 N/sm²

Surface of person = A = 1.50 m x 0.50 m = 0.75 m²

Knowing that stress is defined as:

S = F/A we solve for F:

F = S*A = (392 N/sm²)(0.75 m²) = 294 N/s

2) Friction force between the person and the ground is computed as follows:

f = mu*W where: W = weight of person = mg
f = (1.0)(70 kg)(9.81 m/s²) = 686.7 N

3) If we compare both forces F = 294 N and f = 686.7 N we can see that friction force is larger than the force exerted on the person by wind.

We may conclude that this person (well grounded) will not be taken away by the wind due to higher force of friction.

Hope it helps!

Good luck!