Drag Force and Terminal Velocity Problems
Fishbane Chp 5
Q33. Consider Eq. (5–32) for the drag force, which has two terms. Which term, if any,
dominates at low speeds and which, if any, dominates at high speeds? How would you
decide the size of the domains for which one term is much larger than the other?
The first term is proportional to v while the second one is proportional to v2; so at low
speeds the first term dominates (as v is small and v2 is even smaller), while at high speeds
the second term dominates (as v is large but v2 is even larger). At a certain speed, v1, the
two terms become the same: bv1 = cv12, or v1 = b/c. When v << v1 it’s the low-speed
regime, where the first term dominates; and when v >> v1 it’s the high-speed regime, where
the second one dominates.
P37. Estimate the drag force on an automobile cruising at 65 mi h . Assume that the drag
2
coefficient CD is 0.45 and that the car’s cross-sectional area is 4 m . Take air to
. kg m3 .
have a density of 125
We convert the speed: (65 mi/h)(1.61103 m/mi)/(3.6103 s/h) = 29 m/s.
The drag force is
FD  21 ACD v 2  21 (125
. kg/m3 )(4 m2 )(0.45)(29 m/s) 2  9.5  102 N .
P38. A parachute is rigged so that a parachutist of total mass 116 kg with full gear
reaches the ground at a terminal speed of 4.9 m s . Assuming that the drag force on
2
the parachutist, moving with speed v, has a magnitude equal to kv , what is the
value of k?
At terminal velocity, a = 0 and the weight is balanced by the drag force:
Fd = mg = kv2, so
k = mg /v2 = 116 kg(9.8 m/s2)/(4.9 m/s)2 = 47 N · s2/m2 = 47 kg/m.
2
P40. A race car of mass 800 kg has a maximum acceleration from rest of 4.8 m s .
Assume that the car’s engine is such that the force on the tires is constant and that the
. m2 . If the car’s top speed is
car’s effective cross-sectional area into the air is 18
observed to be 90 m s , what is its drag coefficient CD ? Take the density of air as
125
. kg m3 .
The maximum acceleration is achieved when there is no drag (low speed).
Thus Fengine = ma = (800 kg)(4.8 m/s2) = 3.8103 N.
At a top speed of vmax = 90 m/s, the acceleration is zero and
FD  21 ACD v 2  Fengine ,
so
CD = 2Fengine/Av2 = 2 (3.8103 N)/(1.25 kg/m3)(1.8 m2)(90 m/s)2 = 0.42.
P44. Assume that the drag force on a parachute is given by Eq. (5–30) in the text. The
3
. kg m . If a 90effective area of the parachute is 30 m and the density of air is 125
kg parachutist finds that his terminal speed is 6.0 m s , what is the drag coefficient?
2
Because FD  21 ACD v 2 , at terminal speed we have mg  21 ACD v 2 , so
CD = 2mg/Av2
= 2(90 kg)(9.8 m/s2)/(1.25 kg/m3)(30 m2)(6 m/s)2 = 1.3.
P45. A parachutist of 60-kg mass uses a parachute similar to the one used by the
parachutist of the previous problem. Determine the terminal speed of this lighter
person.
Similar to the previous problem, set mg  21 ACD v 2 , which gives v = constant m1/2.
With m1 = 90 kg, v1 = 6 m/s, and m2 = 60 kg, the terminal speed v2 of the 60-kg
person must satisfy
v1/ v2 = (m1/ m2)1/2 , which gives
v2 = (m2/ m1)1/2 v1 = (60 kg/90 kg)1/2 (6 m/s) = 5 m/s.