physics given me the answer if you guys can .read the text below.?

Question

During a chicago storm, winds can whip horizontally at speeds of 100km/h. If the air strikes a person at the rate of 40kg/s per square meter and is brought to rest.
estimate the force of the wind on a person.
Assume the person is 1.50m high and 0.5m wide. Compare to the typical maximum forces of friction (u=1.0) between the person and the ground, if the person has a mass of 70kg.


help me tell the formula , its a problem in my book i cound find it out , but in the back of the book there is an answer for it and the answer is : 8x10^2 N, Fwin> ffr= 7x10^2 N. but i dont know how to get this answer.

Answer 1

If your area A=1.5*0.5m^2, then the air strikes you with rate of R=40*A= 30kg/s; the constant wind with speed v=100*1000/3600 = 250/9 m/s will drive you with force Fwin=R*v = 30*250/9 = 833.3 N, while to drag you lying on the ground one needs the force Ffric=u*mg= 1*70*9.81 = 686.7 N; thus as Fwin>Ffric the wind would pick you up and carry in the air – cool!

Answer 2

I think the answer in the book forgot a factor of 1/2.
Here's what I believe is the correct way to solve this problem:
the wind pressure is the stagnation pressure of the flow, or
Pw = 1/2 (rho) v^2. The force on the person is then Pw * A, where area is the effective area of the person.
v: is 100 km/hr or 27.8 m/sec
rho: you are told that the mass flow per unit area is 40 kg/(s*m^2)
Mass flow per unit area is (rho)*v.

Therefore Fw = Pw * A = 1/2 (rho)*v^2*A = 1/2 (rho * v) * v * A=
1/2*40*27.8*.75 = 417 N.
The book's answer of 800 N appears to be wrong. Or else I am going to be embarrassed!