Question

The EF Express, a train similar to the one shown above, has 4 locomotives (70 tons each) and 75 coal cars (40 tons each). The train is going down a 2.0% incline. Use positive as the direction the train is going.

What is the magnitude of the force in the coupling between the locomotives and the cars while going down the 2.0% incline with a tractive force provided by each locomotive of 20000 lbs?
(include units with answer)

The engines are turned off so there is no tractive force. The train is moving at 15 mph down the incline. What is the required magnitude of the braking force to bring the train to a stop in 2000 ft?
(include units with answer)

Answer 1

Slope=2%

In degrees D=tan^-1 (Slope Percent/100)=tan^-1(0.02)=1.15 deg.

For each of the locomotives,

Sum of the resultant force along y-direction=0

i.e, NL-20000*cos (1.15^o)=0, NL=20000 cos (1.15^o) lbs =19995 lbs

Sum of the resultant force along x-direction=0

i.e, μ*19995 lbs-T-20000*sin (1.15^o)=(20000/32.2)a

μ*19995 lbs-T-401.398=621.11 a

T is the magnitude of force coupling

a=acceleration

Considering both locomotive and coal car,

Sum of the resultant force along y-direction=0

Nc-(40*2000)cos(1.15^o)=0

Nc=79998 lbs

Sum of the resultant force along x-direction=0

i.e, μ*19995 lbs-(3280*2000)*sin (1.15^o)=(6560000/32.2)a

μ*19995 lbs- 131658.79=( 6560000/32.2)a

given, velocity v of the train=15mph d=2000 ft

a can be found from V²=2ad, a=(15* 0.44704)^2/(2*2000* 0.3048)= 0.0368808 m/s2=0.121 ft/s2

so, μ=7.8

Thus,

μ*19995 lbs-T-401.398=621.11 a

T= 155484.4 kip (ans)

Answer 2

To calculate the magnitude of the force in the coupling between the locomotives and the cars while going down the 2.0% incline, we need to consider the weight of the train and the force due to the incline.

1. Coupling Force: The total weight of the train can be calculated by summing the weights of the locomotives and the coal cars. Since there are 4 locomotives, each weighing 70 tons (140,000 lbs), and 75 coal cars, each weighing 40 tons (80,000 lbs), we have: Total Weight = (4 locomotives * 140,000 lbs) + (75 coal cars * 80,000 lbs) Total Weight = 560,000 lbs + 6,000,000 lbs Total Weight = 6,560,000 lbs

The force due to the incline can be calculated using the weight of the train and the incline percentage: Force due to Incline = Total Weight * Incline Percentage Force due to Incline = 6,560,000 lbs * 0.02 Force due to Incline = 131,200 lbs

Therefore, the magnitude of the force in the coupling between the locomotives and the cars while going down the 2.0% incline is 131,200 lbs.

2. Braking Force: To calculate the required magnitude of the braking force to bring the train to a stop, we can use the concepts of kinetic energy and work. The work done by the braking force will be equal to the initial kinetic energy of the train, which will bring it to a stop.

First, we need to calculate the initial kinetic energy of the train using its mass and velocity: Initial Kinetic Energy = (1/2) * Mass * Velocity^2

The mass of the train can be calculated by converting the total weight of the train from pounds to mass (using the conversion factor of 1 lb = 0.453592 kg): Mass = Total Weight / (Gravity * Conversion Factor) Mass = 6,560,000 lbs / (9.8 m/s^2 * 0.453592 kg/lb) Mass ≈ 2,981,195 kg

The velocity of the train can be converted from miles per hour to feet per second (using the conversion factor of 1 mile = 5280 feet and 1 hour = 3600 seconds): Velocity = 15 mph * (5280 ft/mile) / (3600 s/hour) Velocity ≈ 22 ft/s

Now, we can calculate the initial kinetic energy: Initial Kinetic Energy = (1/2) * Mass * Velocity^2 Initial Kinetic Energy = (1/2) * 2,981,195 kg * (22 ft/s)^2

Next, we need to convert the kinetic energy from foot-pounds to joules (using the conversion factor of 1 ft-lb = 1.35582 J): Initial Kinetic Energy = Initial Kinetic Energy * Conversion Factor Initial Kinetic Energy ≈ 1.0105 * 10^8 J

Finally, the required magnitude of the braking force can be calculated by dividing the initial kinetic energy by the stopping distance: Braking Force = Initial Kinetic Energy / Stopping Distance Braking Force = 1.0105 * 10^8 J / (2000 ft * 0.3048 m/ft)

Therefore, the required magnitude of the braking force to bring the train to a stop in 2000 ft is approximately 2.631 * 10^5 N (newtons).