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M1: Tension between a car and a towed trailer

I need help on the following question:
asset (1).jpg

I have made force diagrams for both the car and the trailer:
Attachment not found


I used N2L to try and work out the tension, but I ended up getting two different values somehow.

Can somebody please show me how they would apply N2L to each body?
(edited 7 years ago)

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You can't apply N2L as there is no acceleration
Draw a free body diagram of the Trailer & resolve
P.S don't forget about the unknown thrust of the car
Original post by EwanWest
You can't apply N2L as there is no acceleration
Draw a free body diagram of the Trailer & resolve
P.S don't forget about the unknown thrust of the car


I think I understand it now.

I forgot about the driving force because I saw 'constant speed' and thought it was 0, but now I realise that it's the net force which is zero.

asset (2).jpg
Original post by W. A. Mozart
I need help on the following question:

I have made force diagrams for both the car and the trailer:

I used N2L to try and work out the tension, but I ended up getting two different values somehow.

Can somebody please show me how they would apply N2L to each body?


Constant speed \Rightarrow no acceleration

No acceleration F=ma=0\Rightarrow F=ma=0 - net force is 0

Net force is 0 \Rightarrow forces are equal
(edited 7 years ago)
the car must be providing a propulsive force equal to the two resistances.
Out of interest, where did this question come from? The wording is odd - the phrase 'the force in' is, at best, clumsy.
Original post by RogerOxon
Out of interest, where did this question come from? The wording is odd - the phrase 'the force in' is, at best, clumsy.


'Mechanics 1 for AQA' by Cambridge University Press.
Whilst I am here, to save me from making a new thread...

When calculating the time using s = ut + 0.5at^2, do you have to show that the solution can be either positive or negative, and then state that the time can not take a negative value?
Original post by W. A. Mozart
When calculating the time using s = ut + 0.5at^2, do you have to show that the solution can be either positive or negative, and then state that the time can not take a negative value?

I would state that, 'for t>=0 the solution is', but I don't know how this gets marked.
Original post by W. A. Mozart
Cambridge University Press

Ta. Must not comment .. :wink:
Original post by RogerOxon
Ta. Must not comment .. :wink:


A car moves along a straight road. When it passes a set of traffic lights the car is travelling at a speed of 8ms-1 the car then moves with a constant acceleration for 10 seconds and travels 200metres. A) show that the acceleration is 2.4ms-2 B) find the speed of the car at the end of the 10 seconds C) the road is horizontal and the car has mass 1200kg. A constant resistance force of 1800n acts on the car while it is moving. (I) find the magnitude of the driving force that acts on the car while it is moving (Ii) at the end of the 10 second period the driving force is removed the car then moves subject to the resistance force of 1800n until it stops . Find the distance that the car travels while it is slowing down
Can some help with c)ii please
Original post by cosford2
C) the road is horizontal and the car has mass 1200kg. A constant resistance force of 1800n acts on the car while it is moving. (I) find the magnitude of the driving force that acts on the car while it is moving (Ii) at the end of the 10 second period the driving force is removed the car then moves subject to the resistance force of 1800n until it stops . Find the distance that the car travels while it is slowing down
Can some help with c)ii please

You have the initial speed from B, and the (constant) acceleration (negative) from F=ma. What formulas are you expected to have? (If needed, you can easily calculate the time taken to stop)
Original post by RogerOxon
You have the initial speed from B, and the (constant) acceleration (negative) from F=ma. What formulas are you expected to have? (If needed, you can easily calculate the time taken to stop)
it is in the Newton's second law section so f=am I think
I was thinking more of v2=u2+2asv^2=u^2+2as, with aa from F=maF=ma, so v2=u2+2Fmsv^2=u^2+2\frac{F}{m}s. Rearrange to get ss.
Original post by RogerOxon
I was thinking more of v2=u2+2asv^2=u^2+2as, with aa from F=maF=ma, so v2=u2+2Fmsv^2=u^2+2\frac{F}{m}s. Rearrange to get ss.


Is a 1800=1200a?
Original post by cosford2
Is a 1800=1200a?

That would get you the deceleration.

You have the initial speed (u), end speed (v=0) and the deceleration. You should be able to use the formula above to calculate the distance traveled (s).
Original post by RogerOxon
That would get you the deceleration.

You have the initial speed (u), end speed (v=0) and the deceleration. You should be able to use the formula above to calculate the distance traveled (s).

Ok so would it be 0^2=32^2+1800/1200s
The acceleration is negative - the car is slowing.
Original post by RogerOxon
The acceleration is negative - the car is slowing.
what is the acceleration?
Original post by cosford2
what is the acceleration?

You wrote pretty-much the right equation earlier.
a=Fm=18001200=1.5ms2a=\frac{F}{m}=\frac{-1800}{1200}=-1.5 ms^2
(edited 7 years ago)

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