Connected Particles Question (ENGAA 2019)

[TomK]

Homework Statement

ENGAA 2019

Relevant Equations

Equation of motion

The correct answer is 'C'. Why is my working wrong?

 

[haruspex]

The image of your working is rather faint, and it would benefit from some explanation of what you are doing.
You seem to be considering forces on the car (ma, D₁), but that is rather pointless since you have no given info on the tractive force on the car. Yes, you can deduce it, but it is more direct to consider forces on the caravan.

 

[TomK]

The image of your working is rather faint, and it would benefit from some explanation of what you are doing.
You seem to be considering forces on the car (ma, D₁), but that is rather pointless since you have no given info on the tractive force on the car. Yes, you can deduce it, but it is more direct to consider forces on the caravan.

I have started the question again. This is the working I have so far:

E = (l . 'resultant force on bar') / (A . x)
x = extension of bar

(caravan equation of motion): T - D2 = Ma
T= tension in bar

(car equation of motion): F - T - D1 = ma
F = driving force of car

I need to re-arrange the top equation for 'x', but I don't understand what the resultant force on the bar should be.

If you pull a bar from both ends, what counts as the resultant force? Do I need to add the 'resultant force on the car' and the 'resultant force on the caravan' together? Or, is the resultant force on the bar just the sum of the forces which act left on the caravan and act right on the car?

This is very confusing.

 

[haruspex]

If you pull a bar from both ends, what counts as the resultant force?

If we take the bar as massless, ΣF=ma tells us that the net force on it must be zero.
Tension is not quite the same as a force; it is more a pair of equal and opposite forces at each point along the bar.
See section 2 at https://www.physicsforums.com/insights/frequently-made-errors-mechanics-forces/

 

[TomK]

If we take the bar as massless, ΣF=ma tells us that the net force on it must be zero.
Tension is not quite the same as a force; it is more a pair of equal and opposite forces at each point along the bar.
See section 2 at https://www.physicsforums.com/insights/frequently-made-errors-mechanics-forces/

If the car has a resistive force of D1, does that compress the bar? If the car has driving force of F, does that extend the bar? Likewise for the caravan, does D2 extend the bar?

How can the resultant force on the bar be zero if it extends? I don't know if I'm meant to be adding or subtracting forces? I can't visualise it.

 

[haruspex]

How can the resultant force on the bar be zero if it extends?

Because it is being pulled equally in opposite directions by the car and the caravan. That's tension.

 

[TomK]

Because it is being pulled equally in opposite directions by the car and the caravan. That's tension.

I still don't understand what the total force is. If F = D1 + D2, how can it be pulled equally? D2 is pulling left, and F is pulling right.

 

[haruspex]

I still don't understand what the total force is. If F = D1 + D2, how can it be pulled equally? D2 is pulling left, and F is pulling right.

Separate the system into three bodies, car, bar and caravan.
In general, the car pulls on the bar with force T₁, and by Newton's laws the bar pulls on the car with an equal and opposite force.
Similarly a force T₂ between bar and caravan.
Net force on the bar is T₁-T₂=m_bara. Since m_bar is negligible here, the difference between T₁ and T₂ is negligible, so call them both T (but in opposite directions).
For the purposes of finding the extension of the bar, we need T.

If F is the total driving force on the system then F-D₁-D₂=(M+m)a. But that is not helpful since it sheds no light on T. You can combine it with the equation for the car, F-D₁-T=ma, to obtain T-D₂=Ma, but you can get that directly by considering the forces on the caravan.

 

[TomK]

Separate the system into three bodies, car, bar and caravan.
In general, the car pulls on the bar with force T₁, and by Newton's laws the bar pulls on the car with an equal and opposite force.
Similarly a force T₂ between bar and caravan.
Net force on the bar is T₁-T₂=m_bara. Since m_bar is negligible here, the difference between T₁ and T₂ is negligible, so call them both T (but in opposite directions).
For the purposes of finding the extension of the bar, we need T.

If F is the total driving force on the system then F-D₁-D₂=(M+m)a. But that is not helpful since it sheds no light on T. You can combine it with the equation for the car, F-D₁-T=ma, to obtain T-D₂=Ma, but you can get that directly by considering the forces on the caravan.

Thank you. It is clear to me now.