Do I need to calculate in car and trailer tension separately?

  • Thread starter Thread starter lemon
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves a car towing a trailer, with specific masses and an acceleration provided. Participants are exploring the calculation of the driving force and the tension in the coupling between the car and the trailer, while questioning the interpretation of external forces.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the definition of external forces and whether the driving force includes friction. There are attempts to calculate the tension in the coupling, with some expressing uncertainty about the need to calculate tensions separately for the car and trailer.

Discussion Status

Some participants have provided calculations for the driving force and tension, while others have questioned the assumptions made in those calculations. There is a recognition of the need to consider forces on individual components, but no consensus has been reached on the correct approach to calculating tension.

Contextual Notes

Participants are navigating the constraints of the problem, including the definitions of forces involved and the implications of Newton's laws. There is a mention of potential errors in mass assignments in calculations, which may affect the results.

lemon
Messages
199
Reaction score
0
1. A car of mass 1000kg is towing a trailer of mass 650kg and the two have an acceleration of 2,3m/s^2. Assuming that the only external force is the driving force between the wheels and the road, calculate:
1: the value of this force
2: the tension in the coupling between the car and the trailer.

I'm a little unsure about this question. When it says 'external force', the driving force between the wheels and the road, it does just mean the driving force, right? And not the frictional force?
Also, my calculations for tension have come out at 0N. But I think there must still be a Tension/resultant force between the two vehicles as they are not moving at a constant velocity. Do I need to calculate in car and trailer tension separately?

2. F=ma
F-T=ma

3.
(a) F=(1000kg+650kg)2.3m/s^2
=2500N (2sf)
(b) F-T=ma
T=F-ma
T=2500-2500=0
 
Last edited:
Physics news on Phys.org
Welcome to PF!

Hi lemon! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)
lemon said:
I'm a little unsure about this question. When it says 'external force', the driving force between the wheels and the road, it does just mean the driving force, right? And not the frictional force?

They're the same …

by good ol' Newton's third law, action is equal and opposite to reaction, so the (drive) force of the car on the road equals the (friction) force of the road on the car. :wink:
Do I need to calculate in car and trailer tension separately?

(b) F-T=ma
T=F-ma
T=2500-2500=0

Yes, you can only find T if it's an external force, so that means that you must consider only the forces on the car (or only the forces on the trailer).

(So your m in that last equation is wrong.)
 
Welcome to PF!

Hi lemon! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)
lemon said:
I'm a little unsure about this question. When it says 'external force', the driving force between the wheels and the road, it does just mean the driving force, right? And not the frictional force?

They're the same …

by good ol' Newton's third law, action is equal and opposite to reaction, so the (drive) force of the car on the road equals the (friction) force of the road on the car. :wink:
Do I need to calculate in car and trailer tension separately?

(b) F-T=ma
T=F-ma
T=2500-2500=0

Yes, you can only find T if it's an external force, so that means that you must consider only the forces on the car (or only the forces on the trailer).

(So your m in that last equation is wrong.)
 
Welcome to PF!

Hi lemon! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)
lemon said:
I'm a little unsure about this question. When it says 'external force', the driving force between the wheels and the road, it does just mean the driving force, right? And not the frictional force?

They're the same …

by good ol' Newton's third law, action is equal and opposite to reaction, so the (drive) force of the car on the road equals the (friction) force of the road on the car. :wink:
Do I need to calculate in car and trailer tension separately?

(b) F-T=ma
T=F-ma
T=2500-2500=0

Yes, you can only find T if it's an external force, so that means that you must consider only the forces on the car (or only the forces on the trailer).

(So your m in that last equation is wrong.)
 
Hi:
So, to answer part b correctly:

Car: 2500N-T=1000kg x 2.3ms
T=2500N-2300N=200N

Trailer: 2500N-T=650kg x 2.3m
T=2500N-1495N=1005N

Total T=Tcar+Ttratiler=200N+1005N=1205N
 
test
 
ms-2
 
lemon said:
Hi:
So, to answer part b correctly:

Car: 2500N-T=1000kg x 2.3ms
T=2500N-2300N=200N

Trailer: 2500N-T=650kg x 2.3m
T=2500N-1495N=1005N

Total T=Tcar+Ttratiler=200N+1005N=1205N

lemon said:
test

lemon said:
ms-2

Hi lemon! :smile:

I see you had the same problem as I did, with the server! :biggrin:

Your car equation is completely correct …

and you should have stopped there! :wink: :rolleyes:

Your trailer equation is wrong, because there's only one force on the trailer, and that's T.

F is a force on the car, not on the trailer … all the trailer gets is what comes through the coupling.

So you could have said T = mtrailera, which would give you the same result as F - T = mcara … either will do, but the first one is shorter. :wink:
 
You rock! Roll! Rave!
Whatever.
Thanx
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
1K
Replies
3
Views
1K
Replies
15
Views
2K
  • · Replies 25 ·
Replies
25
Views
6K
Replies
8
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
11
Views
3K
Replies
6
Views
4K