How Do Braking Forces Affect Train Derailment Distances and Coupling Forces?

[Apprentice123]

The composition of meters shown in Figure traffic to 72,4 km/h when the brakes of the wheels A and B were activated, causing the derailment. Determine (a) the distance traveled by the rest and (b) the force in each coupling. The coefficient of kinetic friction between the wheels and rails is 0,30.

Answer:
(a) 98,1 m
(b) FAB = 2,4x10^4 N; FBC = 5,6x10^4 N


My solution:
(a) Kinetic energy:
T = 1/2mv^2 = 1,83x10^7 J
Work:
U = (PT - FT).x
PT (Total weight) = 89x10^4
FT (Total force of friction) = 2,67x10^5
U = 6,23x10^5x

x = T / U = 29,44m Is not the correct answer

 

[Doc Al]

My solution:
(a) Kinetic energy:
T = 1/2mv^2 = 1,83x10^7 J

OK.

Work:
U = (PT - FT).x

I don't understand this equation. You just need the work done by friction.

PT (Total weight) = 89x10^4
FT (Total force of friction) = 2,67x10^5

Only two of the three cars produce friction.

 

[Apprentice123]

OK.

I don't understand this equation. You just need the work done by friction.

Only two of the three cars produce friction.

Yes. Thanks

U = 1,87x10^5
x = T/U = 97,86m

How do I calculate (b)?

 

[Doc Al]

How do I calculate (b)?

Consider each car separately. Apply Newton's 2nd law. (What's the acceleration of each car?)

 

[Apprentice123]

Consider each car separately. Apply Newton's 2nd law. (What's the acceleration of each car?)

Acceleration is the same for all cars?

V^2 = (Vo)^2 +2aX
a = V^2/2X = 2,066 m/s^2

 

[Apprentice123]

I not find the correct answer

FAB = FatA + FatB + (mA+mB)a

FatA = PA . u
FatB = PB . u

FAB = 3,18x10^5 N

 

[Doc Al]

Acceleration is the same for all cars?

V^2 = (Vo)^2 +2aX
a = V^2/2X = 2,066 m/s^2

Looks OK.

I not find the correct answer

FAB = FatA + FatB + (mA+mB)a

I don't understand what you are doing. Analyze each car separately.

For example, consider car A. What horizontal forces act on it? (One force is the coupling force F_B/A from car B.) Apply Newton's 2nd law to solve for that force.

 

[Apprentice123]

Looks OK.


I don't understand what you are doing. Analyze each car separately.

For example, consider car A. What horizontal forces act on it? (One force is the coupling force F_B/A from car B.) Apply Newton's 2nd law to solve for that force.

Yes. But I do not find the answer

Fb/a = (2,67x10^5)/9,81 * 2,066 + 2,67x10^5 * 0,3
Fb/a = 136330,581 N

 

[Doc Al]

Yes. But I do not find the answer

Fb/a = (2,67x10^5)/9,81 * 2,066 + 2,67x10^5 * 0,3
Fb/a = 136330,581 N

ΣF = F_f + F_b/a = m*a

 

[Apprentice123]

ΣF = F_f + F_b/a = m*a

Ok. Thank you very much