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A train of 3000 KN weight starts from rest and reached a velocity of 54 km/h after o

by manal950
Tags: 3000, km or h, reached, rest, starts, train, velocity, weight
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manal950
#1
Jun10-12, 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
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rock.freak667
#2
Jun10-12, 09:54 AM
HW Helper
P: 6,202
Quote Quote by manal950 View Post
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
Right a = 0.25 m/s2 (Always remember to put your units).

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).
manal950
#3
Jun10-12, 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

manal950
#4
Jun12-12, 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 ..
rock.freak667
#5
Jun12-12, 04:43 PM
HW Helper
P: 6,202
Quote Quote by manal950 View Post
force on the train
1 weight
2 frictional resistance
3 tractive force

now How I can find the tractive force
In which direction do these forces act?

How would you find the resultant horizontal force?
manal950
#6
Jun12-12, 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
rock.freak667
#7
Jun12-12, 06:14 PM
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P: 6,202
Quote Quote by manal950 View Post
I don't know how to find resultant horizontal force?

anyway the direction for each one weight dawn , frictional resistance opposite of tractive force
Right so let's say frictional force is going left and the tractive force is going right. The train is moving to the right. How would you calculate the value of the force that the train is moving with from these two forces?

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?
manal950
#8
Jun14-12, 08:32 AM
P: 177
Resultant force = tractive force - frictional force
rock.freak667
#9
Jun14-12, 09:56 AM
HW Helper
P: 6,202
Quote Quote by manal950 View Post
Resultant force = tractive force - frictional force
Right!

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.
manal950
#10
Jun15-12, 12:59 AM
P: 177
tractive force = Resultant force - frictional force
= 76452.5 - 3000
now this is correct ??
rock.freak667
#11
Jun15-12, 07:22 AM
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P: 6,202
Quote Quote by manal950 View Post
tractive force = Resultant force - frictional force
= 76452.5 - 3000
now this is correct ??
Check the signs in your equation.
manal950
#12
Jun15-12, 08:35 AM
P: 177
76452.5 + 3000 ... ?
rock.freak667
#13
Jun15-12, 04:14 PM
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P: 6,202
Quote Quote by manal950 View Post
76452.5 + 3000 ... ?
That should work.
HallsofIvy
#14
Jun15-12, 04:22 PM
Math
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Thanks
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Quote Quote by manal950 View Post
tractive force = Resultant force - frictional force
= 76452.5 - 3000
now this is correct ??
Remember that the "tractive force", the force applied to the track by the locomotive, must be larger then the resultant force as it has to overcome the frictional force.

You can write it as "Resultant force- frictional force" if you remember to treat "frictional force" as negative.
manal950
#15
Jun16-12, 09:47 AM
P: 177
now is my answer if fully correct ... ?
rock.freak667
#16
Jun16-12, 12:43 PM
HW Helper
P: 6,202
It should be.
manal950
#17
Jun23-12, 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
azizlwl
#18
Jun23-12, 09:20 AM
P: 963
Quote Quote by manal950 View Post
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 ?
frictional resistance of the track is 5N/KN of the train's weight


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