Radial Force on Railway Engine of 120 Tons at 25mph

[John O' Meara]

A railway engine of mass 120 tons travels round a curve of radius 220 yards at a speed of 25 miles per hour. Find in tons weight the radial force acting on the engine. If the resultant of the radial force and the weight of the engine acts in a direction normal to the plane of the track, and the gauge is 56.5 ins. how must the outer rail be banked up above the level of the inner rail.
66mph=88 ft/s, =>25mph=36.63 ft/s. F=ma = mv^2/r, F=244.44 tons, F= 244.44 divided by 32ft/s/s = 7.638 tons wt. where F= radial force.
Question: why divide by 32ft/s/s, I thought you multiplied the mass by the acceleration due to gravity to get the weight.
2nd part: first find the angle "theta" with which the resultant R is inclined to the vertical.
tan(theta) = 7.64/120 which give the wrong answer for theta: because 56.5"sin(theta) doesn't give you 3.6inches.Thanks guys.

 

[gulsen]

I'm too lazy to make unit transformations. What I've understood from your post, this should solve the problem analitically:

$$Fcos(\theta) = mg$$
$$Fsin(\theta) = mv^2/r$$
$$\frac{v^2/r}{g} = tan(\theta)$$
$$atan(\frac{v^2/r}{g}) = \theta$$
$$(gauge)sin(\theta) = (height)$$