Find Tractive Force of 3000KN Weight Train in 1 min

[manal950]

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

 

[rock.freak667]

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]

force on the train
1 weight
2 frictional resistance
3 tractive force

now How I can find the tractive force

 

[manal950]

I don't know where is the other helper ..

 

[rock.freak667]

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]

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]

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]

Resultant force = tractive force - frictional force

 

[rock.freak667]

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]

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

 

[rock.freak667]

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

Check the signs in your equation.

 

[manal950]

76452.5 + 3000 ... ?

 

[rock.freak667]

76452.5 + 3000 ... ?

That should work.

 

[HallsofIvy]

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]

now is my answer if fully correct ... ?

 

[rock.freak667]

It should be.

 

[manal950]

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]

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

 

[manal950]

5 X 3000

unit is Kn or N

and my other questions is the Resultant force
ma + tractive force

or only F = ma