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Deal with tension problems in a systematic way?

  • #1
Something about tensions throws me off. I can never seem to get tension problems.
A freight train consists of 250 cars each of mass 64 metric tons. The acceleration of the train is a=0.043m/s2. What is the tension in the coupling that holds the first car to the locomotive? the last car? Ignore friction.

Here's a little diagram.

[___]---[___]---[___]---[___]---[___]\ ----->a


Let me treat the last car first. The free body diagram looks like this:
O-------> T
T=ma=2.75x10^3 N
Correct?

On the first car we have (ignoring normal force and weight, which cancel out),

<-----O----->
Right? Since the car pulls on the car behind it, the car behind it pulls on it. And the locomotive pulls on the first car is the forward direction.
If you treat the train as a whole, the net force is 6.88x10^5 N in the direction of a.
Considering the front boxcar,
T1 - T2=Fnet=ma
a is known, m is known, Fnet is known
T2 must be sufficiently large to accerelate the remainder of the train of mass M by a. So
T1=a(m+M)
T1=6.88x10^5 N since m+M is the total mass of the train

Is my work ok? Can you give me some advice on how to deal with tension problems in a systematic way?
Thanks.
 
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Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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You say "A freight train consists of 250 cars each of mass 64 metric tons." Can we assume that does NOT include the locomotive?

Assuming that, the cars behind the locomotive have a mass of
250(64)= 16000 metric tons= 1.6 x 106kg. Since "F= ma", in order to accelerate them at a=0.043m/s2, we must have
F= 1.6 x106* 4.3 x 10-2= 6.88 x 104 Newtons. The only way to do that is for the locomotive to pull them so that is the tension in the first coupling.

Now, can you calculate the tension in the coupling between the first and second cars?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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And if you can do that, can you calculate the tension on the coupling between the first and second cars?

How about between the last and next to last cars?
 
  • #4
Yes. It does not include the locomotive.

Sure.

[___]---[___]---[___]---[___]---[___]\ ----->a

I prefer general cases.
So I consider the nth car on a train of N cars. I want to find the tension in the cable connecting the nth car with the (n-1)th car.

Force diagram:

<------[__n__]---------> T2
T1

The mass of the car is m.
a is the acceleration of the car. It is also the acceleration of the train.
Fnet on n=ma=T2-T1
-T1 provides the acceleration of the N-n cars of mass m(N-n) behind n.
So T1=ma(N-n)
So T2=ma(N-n+1).
This result gives me the same answers I obtained before when I use N=250, n={250,1}.
 
Last edited:
  • #5
That one was ok.
How about this one?

Two blocks slide down a ramp making an angle P with the horizontal. They are tied together by a string. They have coefficients of friction uk1 and uk2. What is a? What is T?

|\
|.\[m2]
|..\ \
|...\ \
|....\ \
|.....\[m1]
|......\
|.......\

Nice diagram, huh?

Now on m1 we have,

<----- force of frition uk1(m1)gcosP
<----- force of tension T
-----> force of gravity (m1)gsinP

On m2 we have,

<----- force of frition uk2(m2)gcosP
-----> force of tension T
-----> force of gravity (m2)gsinP

(m2)(a2)= -uk2(m2)gcosP + T + (m2)gsinP
(m1)(a1)= -uk1(m1)gcosP - T + (m1)gsinP
I can conclude that T is the same for both equations since there is only one rope. Can I conclude that a1=a2? If so, how do I show that a1=a2? I often get in trouble with my problems because I assume things rather than showing them.
 

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