# M1 Question

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#1

truck Z truck Y truck X Engine
100N<---------- 100N<---------- 300N<---------500N<---------->37000N
........ l 20000kg l ----- l 20000kg l ----- l40000kg l --------l 40000kg l
........ ---------- .........-----------.......--------- .......... -----------

A train consists of an engine and three trucks with the masses and resistance to motion as shown in the diagram (best that I could do on here...) There is also a driving force of 37000N. All the couplings are light, rigid and horizontal.

i) show the acceleration of the train is 0.3ms^-2.
ii)Draw a diagra showing all the forces acting on the truck Z inthe line of its motion. Calculate the force in the coupling between trucks Y and Z.

With the driving force removed, brakes applied, so adding a further resistance of 1100N to the totl of the resistances shown in the diagram.

iii)calculate the new acceleration of the train.
iv) calculate the new force in the coupling between trucks Y and Z if the brakes are applied
(A) to the engine
(B) to truck Z

In each case state whether the force is a tension or a thrust.

I have tried all of the questions apart from iv) which I cant work out which equation to use for.

i) i got 0.225ms^2 not 0.3ms^2 i did total force(27000) / total mass (120000)

ii) I got the total to be 392000N

iii) 0.1s^-2

any help on the last question would be very helpful indeed, and if possible and corrections on my previous answers...

Thanks
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#2
sorry it did it twice, I did say it didnt work the first time, but it apparently did...
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11 years ago
#3
Is the driving force 3 700 N or 37 000 N? The diagram says one figure, and the text another.
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11 years ago
#4
I notice that if it is 37 000 N, then the acceleration comes out with the right answer of .3 ms^-2, so I think, you've used the wrong value.
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#5
Its 37 000 N.

But doing 27000/120000 still only gets 0.225

Im I doing something wrong by force/mass=acceleration ?
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11 years ago
#6
(Original post by CaroBoo)
Its 37 000 N.

But doing 27000/120000 still only gets 0.225

Im I doing something wrong by force/mass=acceleration ?
You're not working out the resultant force correctly.

37000 - 500 -300 - 100 -100 does not equal 27000.
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#7
ahh, thank you.
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11 years ago
#8
Good. Now part ii.

Doesn't it strike you as odd that the force in the coupling is more than 10 times the driving force of the train?

If you've not got a better answer, post some working we can look at.
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#9
I did think it was a bit high yeh...
I did mg-T=ma for the tension truck Z -> truck Y
20 000(9.8) - T =20 000(0.3)
so T=190 000 N

Then T-mg=ma for truck Y -> truck Z
T-20 000(9.8) = 20 000(0.3)
so T=202 000 N

I then added the two T's equalling 392 000 N

Thanks for the help by the way
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11 years ago
#10
(Original post by CaroBoo)
I did think it was a bit high yeh...
I did mg-T=ma for the tension truck Z -> truck Y
20 000(9.8) - T =20 000(0.3)
so T=190 000 N

Then T-mg=ma for truck Y -> truck Z
T-20 000(9.8) = 20 000(0.3)
so T=202 000 N

I then added the two T's equalling 392 000 N

Thanks for the help by the way
The weight of the truck is acting at right angles to the direction of the motion of the train, so it will have no effect!

You don't need to do anything with truck Y.

Your diagram, should just have the truck (Z), the resistive force, and the force in the coupling. Those are the only forces directly effecting truck Z, in line with the direction the train is moving.

Have a go with that.
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#11
so R=100 N
a=0.3ms^2
t=?

I have no equations with just these in, The only equations with t inall have the mass in too.
Sorry to be a pain.
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11 years ago
#12
(Original post by CaroBoo)
so R=100 N
a=0.3ms^2
t=?

I have no equations with just these in, The only equations with t inall have the mass in too.
Sorry to be a pain.
You will need the mass, as it is part of the F=ma formula. You may be confusing mass with weight. Weight is mg and is a force acting vertically downwards, which is what you had, although it is not relevant to this question.

As you've correctly ascertained you will be using F=ma.

In this case the resultant force is going to be the pulling force, via the couple, and lets call that T, and that is what we are trying to find out, and the resisting force of 100 N.

So we get T-100 = mass times acceleration.

I'm afraid that's all I have time for today.

Have a look at part iii again afterwards, the answer is somewhat less than you have, but perhaps after these first two parts, you can see why.
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#13
Thank you vey much fo your help tonight.
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4 years ago
#14
(Original post by ghostwalker)
You will need the mass, as it is part of the F=ma formula. You may be confusing mass with weight. Weight is mg and is a force acting vertically downwards, which is what you had, although it is not relevant to this question.

As you've correctly ascertained you will be using F=ma.

In this case the resultant force is going to be the pulling force, via the couple, and lets call that T, and that is what we are trying to find out, and the resisting force of 100 N.

So we get T-100 = mass times acceleration.

I'm afraid that's all I have time for today.

Have a look at part iii again afterwards, the answer is somewhat less than you have, but perhaps after these first two parts, you can see why.
why T-100 what about the 37000 driving force?
wouldnt it be....37000-100-T=ma
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4 years ago
#15
(Original post by Hard work)
why T-100 what about the 37000 driving force?
wouldnt it be....37000-100-T=ma
That has more than one error.

What's your diagram for the forces acting on truck Z?
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