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A train of 3000 KN weight starts from rest and reached a velocity of 54 km/h after o 
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#1
Jun1012, 09:49 AM

P: 177

Hi all
1 ) A train of 3000 KN weight starts from rest and reached a velocity of 54 km/h after one min . if the frictional resistance of the track is 5N/KN of the train's weight . Find the tractive force developed by the engine ? I don't understand what is mean "tractive force " ?? anyway I start solve the question w = 3000 kN mass = 305810 u = 0 v = 54 km/h v = u + at a = 0.25 F = m X a F = 305810 X 0.25 76452.5 N 


#2
Jun1012, 09:54 AM

HW Helper
P: 6,202

In the formula F=ma, a is the resultant acceleration of the mass m (Train in this case). So F is the resultant force. If you draw a free body diagram, what are the forces on the train? (The tractive force is essentially the force generated by the train itself). 


#3
Jun1012, 10:35 AM

P: 177

force on the train
1 weight 2 frictional resistance 3 tractive force now How I can find the tractive force 


#4
Jun1212, 09:35 AM

P: 177

A train of 3000 KN weight starts from rest and reached a velocity of 54 km/h after o
I don't know where is the other helper ..



#5
Jun1212, 04:43 PM

HW Helper
P: 6,202

How would you find the resultant horizontal force? 


#6
Jun1212, 04:47 PM

P: 177

I don't know how to find resultant horizontal force?
anyway the direction for each one weight dawn , frictional resistance opposite of tractive force 


#7
Jun1212, 06:14 PM

HW Helper
P: 6,202

in other words, if we are in a tug of war competition and you are pulling with 10 N and I am pulling with 11 N, what would be the resultant force in this case? 


#8
Jun1412, 08:32 AM

P: 177

Resultant force = tractive force  frictional force



#9
Jun1412, 09:56 AM

HW Helper
P: 6,202

So you calculated the resultant force in your first post. And they told you what the frictional force was. So you can rearrange to get the tractive force. 


#10
Jun1512, 12:59 AM

P: 177

tractive force = Resultant force  frictional force
= 76452.5  3000 now this is correct ?? 


#11
Jun1512, 07:22 AM

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P: 6,202




#12
Jun1512, 08:35 AM

P: 177

76452.5 + 3000 ... ?



#13
Jun1512, 04:14 PM

HW Helper
P: 6,202




#14
Jun1512, 04:22 PM

Math
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P: 39,682

You can write it as "Resultant force frictional force" if you remember to treat "frictional force" as negative. 


#15
Jun1612, 09:47 AM

P: 177

now is my answer if fully correct ... ?



#16
Jun1612, 12:43 PM

HW Helper
P: 6,202

It should be.



#17
Jun2312, 09:11 AM

P: 177

my teacher said the answer is not correct
and he said why you don't multiply 5 X 3000 I don't know what he mean ? and my other questions is the Resultant force ma + tractive force or only F = ma 


#18
Jun2312, 09:20 AM

P: 963




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