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Back to subcatalog.CATAGORY: Mechanics QUESTION: A 5000 LB AIRPLANE LOOPS AT A SPEED OF 200 MI/HR. fIND THE RADIUS OF THE LARGEST CIRCULAR LOOP POSSIBLE, AND THE FORCE ON THE PLANE AT THE BOTTOM OF THE LOOP.

There are alot of variations on the airplane loop problem. The usual idea is that at the top of the loop the wings of the plane provide no lift so that the acceleration is given by gravity only. Since the path is circular the magnitude of the acceleration is also given by v^2/r (v^2 means v squared). Equating the two expressions gives (1) v^2/r = g where g = acceleration due to gravity v = speed of the airplane r = radius of the loop. Solving for r gives (2) r = v^2/g In this problem you are given nasty English units :-( !!!!! Even the English do not use English units! I will use mks units. Here is the speed conversion. v = 200mph X (5280 ft/mi) X (12 in/ft) X (1m/39.37in) X (1/3600 s/hr) = 89.4 m/s In mks g = 9.8 m/s^2 substitution of these into (2) gives r = (89.4 m/s)^2/9.8 m/s^2 = 816 m OK. Ok back to English units (yeech!) 816m X (39.37 in/m) X ( 1 ft/12 in) = 2680 ft ****************** * r = 2680 ft * ****************** What is meant by "force on the plane at the bottom of the loop" ? The question itself is ambiguous. I will take it to mean the lift on the plane required to maintain it on the circular path at the bottom of the loop. At the bottom of the loop gravity is pulling down with force mg. The lift force ,F(lift), is up. These two forces compete ans the result a force is that up and equal in magnitude to the centrepital force m[v^2/r]. Thus (3) m[v^2/r] = F(lift) - mg Solving (3) for F(lift) gives (4) F(lift) = m[v^2/r] + mg Now lets be smart here. We know from the first part of the problem that v^2/r = g. If we substitute into (4) we get (5) F(lift) = m[g] + mg = 2mg But mg is just the weight of the plane which is 5000 lb. the lift force is then 10,000 lb ********************** * F(lift) =10,000 lb * **********************