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:

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

 

[kyle8921]

I would like to clarify and provide a response to the content provided. Firstly, the radial force acting on the railway engine of mass 120 tons can be calculated using the formula F=mv^2/r, where m is the mass, v is the velocity, and r is the radius of the curve. Plugging in the values given, we get F=244.44 tons. This is the force acting on the engine in the direction towards the center of the curve.

Now, to answer the question about why we divide by 32ft/s/s, this is because we are using the formula F=ma, where F is the force, m is the mass, and a is the acceleration. The value of 32ft/s/s is the acceleration due to gravity, which is used as a conversion factor to convert the mass from tons to pounds. This is necessary because the units of acceleration used in the formula are in feet per second squared, while the units of mass given are in tons. So, by dividing by 32ft/s/s, we are converting the mass from tons to pounds, which is the unit of force.

Moving on to the second part of the question, we need to find the angle at which the resultant force R is inclined to the vertical. To do this, we can use the trigonometric formula for tangent, tan(theta)=opposite/adjacent. In this case, the opposite side is the radial force F, and the adjacent side is the weight of the engine, which is equal to its mass multiplied by the acceleration due to gravity. So, we get tan(theta)=F/(m*g), where g is the acceleration due to gravity. Plugging in the values, we get tan(theta)=7.64/120=0.0637. Now, to find the angle theta, we can use the inverse tangent function, which gives us theta=3.64 degrees.

Finally, to find the difference in height between the outer rail and the inner rail, we can use the formula h=r*tan(theta), where h is the height difference, r is the radius of the curve, and theta is the angle we just calculated. Plugging in the values, we get h=(220 yards)*(3.64/100)=8.008 feet or approximately 96.1 inches. This means that the outer rail needs to be banked up 96.1 inches above the level of the inner