Extra force on a train problem

In summary, two long trains carrying coal are traveling in the same direction at different speeds. The rate of coal transfer between the trains is 4 tons/min for each 100 feet of train length. Using the momentum equation, the extra force on each train per unit length caused by this mechanism can be calculated. The slower train experiences an extra force of +13.3 lb*ft/s^2 per unit length, while the faster train experiences an extra force of -13.3 lb*ft/s^2 per unit length. This is due to the 10 ft/s difference in speed between the two trains. By exchanging coal, the slower train gains the same amount of momentum that the faster train loses.
  • #1
UFeng
27
0

Homework Statement



Two long trains carrying coal are traveling in the same direction side by side on separate tracks. One train is moving at 40 ft/sec and the other at 50 ft/sec. In each coal car a man is shoveling coal and pitching it across to the neighboring train. The rate of coal transfer is 4 tons/min for each 100 feet of train length. This rate is the same for both trains. Find the extra force on each train per unit length caused by this mechanism.

Homework Equations



I think I will need: sum of Forces = Mass * dv/dt but I'm not certain



The Attempt at a Solution


I'm not sure exactly what to do. Would you just find the weight of coal added to each train or is there more to it. Does the momentum equation have any use here. I'm confused on how to start this. Any help would be appreciated!
 
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  • #2
Yes, I think momentum is the best approach. The fast train will lose momentum as it exchanges fast-moving coal for slow-moving coal. If you can find out how much momentum is lost per second you can use
dp/dt = m*dv/dt = F
 
  • #3
So this is what I've come up with...

using F = d(mv)/dt = m*dv/dt + v*dm/dt => where dv/dt = 0, so F = v*dm/dt

and dm/dt = 4 tons/min

F(slower moving train)=d(mv)/dt = (4 tons/min)*(1 min/60sec)*(40ft/sec)/(100 ft of length)
= 0.0267 tons-ft/s^2 per unit length = 53.4 lb ft/s^2 per unit length

similarly for the faster moving train...

with v = 50 ft/s => F(faster moving train) = 66.67 lb ft/s^2 per unit length

How does this look? Did I miss something?
 
  • #4
My calcs are often wrong, but here is what I did for the fast train:
dp = mv - mv = m(v1 - v2) since the mass doesn't change.
= 4*2000/60 lb/s*(50-40) ft/s
= 1333 lb*ft/s^2
This is per 100 ft of train, so per foot it is 13.3 lb*ft/s^2.
 
  • #5
Delphi51 said:
My calcs are often wrong, but here is what I did for the fast train:
dp = mv - mv = m(v1 - v2) since the mass doesn't change.
= 4*2000/60 lb/s*(50-40) ft/s
= 1333 lb*ft/s^2
This is per 100 ft of train, so per foot it is 13.3 lb*ft/s^2.

If that is for the fast train would the force on the slow moving train be the same( for example: the slower train has an "extra force" of +13.3 while the faster train has an "extra force" of -13.3)? Also, when it asked for the force on "each train," would they both be the same but have opposite signs? It seems like the momentum change would be the same with the slower train gaining the same amount of momentum that the fast train is losing. I'm not sure why my solution is wrong.
Why wouldn't you just calculate the force for each train rather than find the difference (4*2000/60 lb/s*(50-40) ft/s). Sorry if I'm missing something very basic. Thanks very much for the help!
 
Last edited:
  • #6
I think I get it. If they were both traveling at the same velocity and all other things being the same, the "extra force" would be zero, but since the faster train is moving at 50 ft/s (10 ft/s faster than the slower one), the "extra force" is from this 10ft/s difference. For example, the coal from the fast train being thrown into the slow train creates an "extra" force of +13.3 lb*ft/s^2 and addes momentum to the slower train. If this is true then the coal from the slower train being thrown into the fast train creates an "extra" force of -13.3 lb*ft/s^2 and the faster train loses the same amount of momentum that the slower train gains.

Does that sound correct? I think I might understand now.
 
  • #7
Sounds great!
 
  • #8
Thanks for the help!
 

1. What is the extra force on a train problem?

The extra force on a train problem is a physics problem that involves calculating the additional force needed to move a train on a track with varying slopes or curves.

2. How is extra force on a train calculated?

The extra force on a train is calculated by using the formula F = ma, where F represents force, m represents mass, and a represents acceleration. The mass of the train and the angle of the slope or curve are also taken into consideration in the calculation.

3. What factors affect the extra force on a train?

The main factors that affect the extra force on a train are the mass of the train, the angle of the slope or curve, and the coefficient of friction between the train and the track. Other factors such as air resistance and the shape of the train can also have an impact.

4. How does the extra force on a train affect its speed?

The extra force on a train can either increase or decrease its speed, depending on the direction and magnitude of the force. If the force is in the same direction as the train's motion, it will increase its speed. If the force is in the opposite direction, it will decrease its speed.

5. Why is it important to consider the extra force on a train?

It is important to consider the extra force on a train because it can impact the train's ability to move and stop safely. Calculating the extra force allows engineers to design trains and tracks that can handle different slopes and curves, ensuring the safety and efficiency of train travel.

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