- What is the acceleration of two falling sky divers (total mass = 132 kg including parachute) when the upward force of air resistance is equal to one-fourth of their weight?
- After opening the parachute, the divers descend leisurely to the ground at constant speed. What now is the force of air resistance on the sky divers and their parachute? See Fig. 4–44.
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This is Giancoli Answers with Mr. Dychko. Here's a free body diagram of the skydivers. We have the force of air friction upwards equal to 0.25 times their weight in part A, and their weight is also acting on them downwards, mg. And air resistance force upwards is 0.25 times mg, where put mg in place of the weight, Fg. And we can say that the force up minus the force down equals ma. And solve for a by dividing both sides by m so, we get a is air friction force minus force of gravity divided by m, and that's 0.25 mg minus mg divided by m and the m's cancel giving the acceleration is 0.25 g minus g which is negative 3/4 g. And g is 9.8 times 0.75 gives negative 7.35 m/s squared. And then when there's zero net force on the skydivers, that means the upward force of air resistance equals the downward force of their weight. And so, their weight is 132 kilograms times 9.8 which means the air friction force must be 1.29x10^3 Newtons upwards.
why does it become -.75g? Why isn't it .25(9.8) - 9.8? Is it because its technically countering the downward motion, and causing minor upwards acceleration?
Because when you have .25g-g is the same thing as .25g-1 which = -.75g then you substitute the 9.8 and get -.75(9.8) = -7.35
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