Trust and Tension - again I am stuck with the forces directions...

I am lost in Forces signs again...

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Here is the problem:
-------------------------

A car of mass 800Kg is towing a caravan of mass 300Kg along a horizontal road. the resistance forces (assumed constant) on the car and the caravan are 700N and 1200N respectively.
a) the car exerts a driving force of 3000N. Find the acceleration of the system and the tension in the coupling.
b) Find the force in the coupling when the system is travelling at a constant speed of 50 km/h.
c) Find the force in the coupling when the car exerts a braking force of 2000N.

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In fact, this problem is from one of the threads on this forum and it's solved there. I can follow the solution there OK, but I am getting confused with the forces signs. I shall repeat here the solution for (c) to show where I get confused. So...

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I draw the picture like so:

Caravan to the left.
Car to the right.
A rod between them.
I choose the positive direction to the right
----------------------------> +

(c) Breaking:

ma ---------->

CARAVAN ----> T T<---- CAR <-------Breaking force

<--------Resistance CARAVAN <------- Resistance CAR

(BTW - I was confused with directions of the resistance forces at first, too!

Probably this my confusion stemmed from my limited English language?
I feel I misintepreted "resistance forces oppose the motion"....

Intuitively, I thought as the system is breaking, the resistance forces should be opposite to the ACCELERATION, i.e. to the right!

Now I believe that the resistance forces are opposite to the VELOCITY. So, although the car & caravan are deccelerating, they still "phisically move" to the right (velocity > 0), hence resistance forces are to the left - is this my contemplation correct?

I feel now that it is, but your confirmation/correction will be really appreciated...

Thanks!)

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OK.
And now to my "main" question....

Writing the equations of motion gives:

(1) CAR AND CARAVAN

- Braking force - Resitance car - Resistance caravan = (M car+ M caravan) x a

(2) CARAVAN

T - Resistance caravan = M caravan x a

(3) CAR

- Braking force - Resitance car - T = M car x a

From (1)

a = - 39/11 m s-2 (negative as it shoud be when breaking)

From (2)

T = + 136.36 N (as in the textbook)

(Equation (3) is not really needed here....)

However, although I do get the correct (as in the textbook) answer, I still feel confused about the following....

MY QUESTION
-------------------
Why do I get the forces in the Rod T as positive?

Positive T means that the directions I had initially chosen in the picture are correct:

CARAVAN ----> T T<---- CAR <-------Breaking force

But, when the car is breaking, it should be THRUST in a rod, not TENSION?

I am confused ....

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Again, below is a quote form a (different) thread on this forum:

"Thrust is the opposite of tension. If a rod (e.g. towbar) is being squashed, as when a car and caravan are braking, the towbar will push on both vehicles rather than pulling on them. So the arrow for the force at each end of the towbar will be towards the vehicles (rather than away from them if it were a tension)."

It all makes sense to me, but I can see a contradiction between what this quote says and the answer to the above problem!

Where do I go wrong in my contemplations?

Thank you in advance for your help...

Reply 1

Absent Agent

21

Original post by ahsstudent

I am lost in Forces signs again...

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Here is the problem:
-------------------------

A car of mass 800Kg is towing a caravan of mass 300Kg along a horizontal road. the resistance forces (assumed constant) on the car and the caravan are 700N and 1200N respectively.
a) the car exerts a driving force of 3000N. Find the acceleration of the system and the tension in the coupling.
b) Find the force in the coupling when the system is travelling at a constant speed of 50 km/h.
c) Find the force in the coupling when the car exerts a braking force of 2000N.

Again, below is a quote form a (different) thread on this forum:

"Thrust is the opposite of tension. If a rod (e.g. towbar) is being squashed, as when a car and caravan are braking, the towbar will push on both vehicles rather than pulling on them. So the arrow for the force at each end of the towbar will be towards the vehicles (rather than away from them if it were a tension)."

It all makes sense to me, but I can see a contradiction between what this quote says and the answer to the above problem!

Where do I go wrong in my contemplations?

Thank you in advance for your help...

Sorry, I have a bit of difficulty following your solution but I will try to explain something here. First, always it's better to think about a problem logically rather than with rules or signs in this case. You know that the car is pulling a caravan in the direction we take as positive to the right. Since the caravan with its resistance forces resit the motion of the car, the forces must act in the opposite direction to the velocity (if the car has a constant velocity) or acceleration of the car. Therefore, the quote is highly twisted (but I will explain that if you would like to).

Caravan--<--------T-------->>--Car

That is,

F_{total} - T- R_{car}= m_{total} a

, where

T

is sum of the resistance forces acting on the caravan only (as the resistance forces on the car don't affect the tension in the couple), and

R_{car}

the resistance forces on the car.

Edit: @ahsstudent: Looking at the conventional force diagrams, I should admit that the directions of the tensions in the couple are not correct. They should be pointing towards the middle of the couple because essentially the tension in the couple is the reaction force resulting from the force pulling the couple.

(edited 8 years ago)

Reply 2

ahsstudent

OP

4

Thank you for coming back, but I am still confused!

You write:

"You know that a caravan is pulling a car in the direction we take as positive to the right."

What do you mean by "a caravan is pulling"? Is it not a car that is pulling a caravan?

What have you denoted by F_{total}?

"Therefore, the quote is highly twisted (but I will explain that if you would like to)."
And yes, please explain how I have twisted the quote?
Thank you :-)

Reply 3

Absent Agent

21

Original post by ahsstudent

Thank you for coming back, but I am still confused!

You write:

"You know that a caravan is pulling a car in the direction we take as positive to the right."

What do you mean by "a caravan is pulling"? Is it not a car that is pulling a caravan?

What have you denoted by F_{total}?

"Therefore, the quote is highly twisted (but I will explain that if you would like to)."
And yes, please explain how I have twisted the quote?
Thank you :-)

Sorry, I edited the part when I said the caravan is pulling the car. It should have been otherwise.
F total is the total thrust of the car.

Reply 4

Absent Agent

21

Original post by ahsstudent

...

"Thrust is the opposite of tension. If a rod (e.g. towbar) is being squashed, as when a car and caravan are braking, the towbar will push on both vehicles rather than pulling on them. So the arrow for the force at each end of the towbar will be towards the vehicles (rather than away from them if it were a tension)."

I don't think the highlighted word in red is not correct to be mentioned here. The only case that the towbar can be squashed (when the both vehicles break) is when the breaking force of the car is greater than the breaking force of the caravan.

edited.

(edited 8 years ago)

Reply 5

ahsstudent

OP

4

I am compeltely lost now. But thank you for trying.

Reply 6

Absent Agent

21

Original post by ahsstudent

I am compeltely lost now. But thank you for trying.

I believe it would be better if you could ask questions step by step to see where we are. What are you particularly first stuck on? Do you have any difficulties with my first post?