Constant speed question

[RichardTM]

Hi

I have a question about how to calculate constant speed, but im only given mass and a few forces and I know speed is distance over time. So now im not quite sure how I would calculate this.

 

Given:

A train is accelerating on a horizontal track.

mass -> 120 Mg (I assume Mg is an error and making it tons)

forces -> Resistance to movement is 5N per kN of the mass

-> In draw-bar of the locomotive is 80kN

(If needed I can type out the full question iv been qiven)

 

Thank you

Richard

Edited May 26, 2015 by RichardTM

[swansont]

Mg is tons (1000 kg = 1 Mg)

 

You should have seen an equation or two relating speed and acceleration.

 

v = v₀ + at

 

is one of them, if you know the time (it can be derived from the definition of acceleration, a = dv/dt)

 

Another is

 

v_f² = v₀² + 2a.s

 

That's a dot product between acceleration (a) and displacement (s), so the direction of the force related to the displacement matters. (It's also the result of applying the definition of "work" to the problem)

 

Once you know the forces, once of those equations should be applicable

[RichardTM]

Thank you

I never knew Mg was a ton, no one ever told me that.

 

I don't have time or acceleration(other than gravity) or distance

 

Here is the exact question.

 

1.3 A train with a mass of 120 Mg accelerates uniformly on a horizontal track. The resistance to movement is 5 N per kN of the mass of the train. The force in the draw-bar of the locomotive is 80 kN.

 

Calculate the following:

1.3.1 The force required for the constant speed.

 

Constant speed = distance / time .But I don't have distance or time.

[Klaynos]

You don't need time or distance.

 

Draw a force diagram, what must be true if the locomotive is at a constant velocity?

[studiot]

 

Given:

A train is accelerating on a horizontal track.

 

I have a question about how to calculate constant speed,

 

 

 

Would you perhaps like to explain these two statements first?

 

Is the track curved?

Edited May 26, 2015 by studiot

[RichardTM]

@studio If you wouldn't mind please seeing my 2nd post that should explain everything for me.

Thank you

Richard

[studiot]

OK I see the question, what I don't see is any strategy for solving it.

 

If you can come up with one we can fill in the detail.

 

 

Do you understand what the question is asking?

 

Hint

It gives you details what happens when the train is accelerating and Force = mass times acceleration.

 

It implies that even when acceleration = 0 (Hence that force is zero), other forces must be acting if there is still a force in the coupling.

Can you think what these forces might be?

[RichardTM]

You don't see a strategy for solving this question because I don't know how to.

 

Iv always been taught speed= distance/time no one has ever told or taught me a different way to work it out, that is why im so confused, Iv tried googling it, but it keeps telling me speed=distance/time.

 

To my knowledge the question is asking for the force that need to be applied for the train to run or drive at a constant speed.

Yes F=m.a

Im understanding that I need to work out constant speed 1st to get/help work out my applied force some how, once iv got that applied force I can take away the friction or resistance force?

[Klaynos]

You do not need to know the speed.

 

F=ma

 

If speed (velocity) is constant what does a=?

 

Have you drawn a diagram showing the forces?

[RichardTM]

a= 0 because the question says constant speed.

Or are we assuming a= 9.8 (gravity)?

 

So would you be saying

 

F= 120000x0

= 0

Is my answer ? Or just part ?

[Klaynos]

Draw the diagram.

 

You've just said that the overall (net) force is 0. The question gives you some other forces. Draw them on.

Draw the diagram.

 

You've just said that the overall (net) force is 0. The question gives you some other forces. Draw them on.

[studiot]

 

Thank you

I never knew Mg was a ton, no one ever told me that.

 

 

 

Actually, not quite.

 

Mg is the correct abreviation for mega grammes or grammes times 1 million.

 

So yes 1Mg - 1000kilogrammes or 1000kg

 

and yes this is a metric-ton, the correct term fot this is tonne, not ton.

 

This is note quite the same as an imperial ton = 2240 pounds.

 

 

Im understanding that I need to work out constant speed 1st to get/help work out my applied force some how, once iv got that applied force I can take away the friction or resistance force?

 

Yes you seem to have a strategy that only needs slight amendment, really to look at it form a slight different angle.

 

Whether the train is accelerating or travelling at constant speed, you can assume the friction is the same, since you are not told any different.

 

So when the train is accelerating the loco is pulling for two reasons.

 

1) To apply the acceleration Force

 

2) To apply the pull to overcome friction.

 

These two are additive. To give a total force F₁ = F_a + F_r = the force in the drawbar under acceleration

 

When the train is travelling at constant speed F_a is zero so F₂ = 0 + F_r = the force in the drawbar at constant speed

 

You are told what F₁ is and what F_r is and asked to calculate F₂

 

How is your stategy coming on?

[RichardTM]

@klaynos I cant attach the image the image button is broken (the ok and cancel buttons don't work) and it wont allow me to paste links on to here

 

@studiot

5N per kN of the mass:

 

120 000kg to N or kN

120 000x 9.8

= 1176000kN

 

5N to kN

5/1000

=0.005kN

 

So 1176000/0.005

=235200000kN

 

So F2=Fa+Ffr

F2=0+235200 000

F2= 235200 000kN ? I don't think this is right

 

(I was going to do a simultaneous equation with the 2 equations but I don't have Fa for the 1st equation.)

Edited May 27, 2015 by RichardTM

[Fuzzwood]

Why so difficult? You get 5 N of resistance for each kN of weight (not mass, mass implies a number in grams, weight a number in Newtons). You have 1176000 kN, so that totals to 5 * 1176000 kN = 5880000 N of resistance

[Klaynos]

5N to kN

5/1000

=0.005kN

 

This is where you start to go wrong.

 

If it was written as 5N friction for every 1000N of weight, does that make or clearer?

Why so difficult? You get 5 N of resistance for each kN of weight (not mass, mass implies a number in grams, weight a number in Newtons). You have 1176000 kN, so that totals to 5 * 1176000 kN = 5880000 N of resistance

My back of the envelope doesn't agree...

 

120 000 Kg

 

Goes to 1 200 000 N for the weight.

 

Which is then 1200 *5 for the friction.

[RichardTM]

1200 x 5 = 6000

 

But if I were to use gravity at 9,8 and not 10

id get 1 176 000N

 

So 1176 x 5 = 5880kN

 

F2=Fa+Ffr

F2=0+5880kN

F2= 5880kN

Edited May 27, 2015 by RichardTM

[Klaynos]

I think you're still missing a term reread the question.

[RichardTM]

+80 for the loco ?

 

5880+80

=5960kN