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# Mechanics 1- Connected Particles Watch

1. A train consists of an engine of mass 50,000kg coupled to two trucks A and B of masses 10,000kg and 6,000kg. The train moves along the track horizontally at constant deceleration. The tension in the coupling between truck A and truck B is zero.

i) By applying newtons second law to truck B, show that the deceleration is 0.25ms^-2

ii) Find the tension in the coupling between the engine and A

iii) Determine whether the engine exerts a driving force or a braking force and find its magnitude.

No idea how to go about this, any ideas on a method to work this out, all questions on my assignment are simular to this one.
2. F=ma
the mass is 6000 and the force is -1500
rearrange to find the deceleration

Can't remember how to do ii or iii atm, I'll edit this if I remember
3. If this is a university assignment, we can't really help you.

Mark T the tension in the towbar between the engine and A. It points both ways into the centre of the towbar.

Mark X the driving force at the front of the engine (remember it could be a braking force but it will just come out with a negative magnitude if this is the case).

Apply F = ma on whatever sections you like.
4. (Original post by Mr M)

Mark T the tension in the towbar between the engine and A. It points both ways into the centre of the twobar.

Mark X the driving force at the front of the engine (remember it could be a braking force but it will just come out with a negative magnitude if this is the case).

Apply F = ma on whatever sections you like.
Not a university assignment! It's M1 as level, okay I'll try your method be back in 10
5. (Original post by Mr M)

Mark T the tension in the towbar between the engine and A. It points both ways into the centre of the towbar.

Mark X the driving force at the front of the engine (remember it could be a braking force but it will just come out with a negative magnitude if this is the case).

Apply F = ma on whatever sections you like.
I get a positive tension and a negative X for this question.

I took X in the right direction, but due to the negative answer it is acting at left, so its a braking force. But why is the tension positive, shouldn't it be negative, because won't the braking force cause thrust?
6. (Original post by raheem94)
I get a positive tension and a negative X for this question.

I took X in the right direction, but due to the negative answer it is acting at left, so its a braking force. But why is the tension positive, shouldn't it be negative, because won't the braking force cause thrust?
I think you can answer the question yourself by looking at the system as a whole or, if you have already done that, look at the forces on the engine. You will find the equations are not consistent.

Updated: April 15, 2012
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