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$F_w=830N$

Giancoli 6th Edition, Chapter 7, Problem 9

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Chapter 7, Problem 9 is solved.

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Transcript for this Giancoli solution

To answer this question we’ll consider the momentum way of expressing Newton’s second law which is that the force equals change in momentum over change in time. This is the force of the wind that we’re concerned with here. So to answer this question we’ll need to know what is the change in momentum of the wind, that’s going to be its mass times its change in velocity, the mass of the wind will stay constant so it’s just the change in velocity of the wind that's going to affect the change in momentum. There's going to be some work involved in figuring out what is the mass of the wind because we’re told that there's forty kilograms of wind for every second passing through every square meter, so first we’ll suppose: let the amount of time that we’ll consider be one second for convenience, and then based on that we’ll say that the amount of mass of air will be forty kilograms every second for every square meter times one second which gives us forty kilograms passing through every square meter in a second. Then we’ll have to turn that into kilograms by saying that the mass of air is going to be forty kilogram for every square meter times the number of square meters in a person. So the area of a person is: a person is one point five meters high and half a meter wide, so the area of a person is three quarters of a meter squared, so we’ll substitute that and have: forty kilograms per meter squared times three quarter square meters and now we finally have kilograms, the amount of mass of air that actually been stopped by a person, thirty kilograms. The force then exerted by the wind is going to be: ‘m’ times ‘Δv’ over ‘Δt’ and the mass we’ve established is thirty kilograms times one hundred kilometers per hour times one hour for every three thousand six hundred seconds times one thousand meters for every kilometer divided by the amount of time which we said is one second, working that out gives us that the force is eight hundred and thirty three newtons. This is based on estimates so in two significant figures it’s eight hundred and thirty newtons is the force exerted by this wind during a Chicago storm. Interestingly, the textbook asks us to compare that with then friction between a person and the ground and if this air force is more than the friction, it means that the person would be sliding around. That would be quite dramatic. So the friction between a person and the ground is going to be: ‘µ’ times ‘FN’, the normal force being equal to the weight of the person, ‘m’ times ‘g’, so we have ‘µ’ is one point zero times a typical seventy kilogram mass times nine point eight newtons per kilogram which gives us the friction between a person’s feet and the ground as six hundred and eighty six newtons. So this means the person will slide because the force exerted by the wind exceeds the friction between their feet and the ground.