# M1: Tension between a car and a towed trailer

Watch
Announcements
#1
I need help on the following question:

I have made force diagrams for both the car and the trailer:
Attachment 620592

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?
1
3 years ago
#2
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
2
#3
(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.

0
3 years ago
#4
(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 no acceleration

No acceleration - net force is 0

Net force is 0 forces are equal
2
3 years ago
#5
the car must be providing a propulsive force equal to the two resistances.
1
3 years ago
#6
Out of interest, where did this question come from? The wording is odd - the phrase 'the force in' is, at best, clumsy.
0
#7
(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.
0
#8
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?
0
3 years ago
#9
(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.
0
3 years ago
#10
(Original post by W. A. Mozart)
Cambridge University Press
Ta. Must not comment .. 1
3 years ago
#11
(Original post by RogerOxon)
Ta. Must not comment .. 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
1
3 years ago
#12
(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)
0
3 years ago
#13
(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
0
3 years ago
#14
0
3 years ago
#15
0
3 years ago
#16
(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).
0
3 years ago
#17
(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
0
3 years ago
#18
The acceleration is negative - the car is slowing.
0
3 years ago
#19
(Original post by RogerOxon)
The acceleration is negative - the car is slowing.
what is the acceleration?
0
3 years ago
#20
(Original post by cosford2)
what is the acceleration?
You wrote pretty-much the right equation earlier. 0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (135)
14.35%
I'm not sure (41)
4.36%
No, I'm going to stick it out for now (284)
30.18%
I have already dropped out (24)
2.55%
I'm not a current university student (457)
48.57%