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AS Level Mechanics Question

A car, of mass 1200kg tows a trailer, of mass 300kg. They move with a constant acceleration in a straight line on a horizontal road. The trailer is connected to the car by a horizontal tow bar.
A resistance force of magnitude 40N acts on the trailer.
A resistance force of magnitude 200N acts on the car.
The speed of the car and trailer increases from 4ms-1 to 6ms-1 as they travel 50 metres.
A tractive force of magnitude P newtons acts on the car.

a) I've calculated this as 0.2 ms^-2
b) Find the tension in the tow bar.
c) Find P.

So i'm struggling to understand how to calculate tension in general. And also how to get P. Thanks a lot :smile:
Reply 1
Original post by Sh1vam.naik
A car, of mass 1200kg tows a trailer, of mass 300kg. They move with a constant acceleration in a straight line on a horizontal road. The trailer is connected to the car by a horizontal tow bar.
A resistance force of magnitude 40N acts on the trailer.
A resistance force of magnitude 200N acts on the car.
The speed of the car and trailer increases from 4ms-1 to 6ms-1 as they travel 50 metres.
A tractive force of magnitude P newtons acts on the car.

a) I've calculated this as 0.2 ms^-2
b) Find the tension in the tow bar.
c) Find P.

So i'm struggling to understand how to calculate tension in general. And also how to get P. Thanks a lot :smile:

Think about the trailer only. You know its mass and its acceleration, so you can work out the resultant force acting on it. What is this resultant force made up of? It might help to draw a diagram showing all of the forces acting on the trailer along with their directions.
Original post by Pangol
Think about the trailer only. You know its mass and its acceleration, so you can work out the resultant force acting on it. What is this resultant force made up of? It might help to draw a diagram showing all of the forces acting on the trailer along with their directions.

Yeah I just wanted to check that I was doing it right? I got 20N as the tension because (F=ma of the trailer = 60N and taking away resistive force gives you 20N? Is that right?)
Reply 3
Original post by Sh1vam.naik
Yeah I just wanted to check that I was doing it right? I got 20N as the tension because (F=ma of the trailer = 60N and taking away resistive force gives you 20N? Is that right?)

Nearly. You need to be careful with the directions of the forces. If there was only 20 N pulling the trailer forwards but a resistive force of 40 N, then the backwards force would be bigger than the forwards force and the trailer would accelerate backwards!
Original post by Sh1vam.naik
a) I've calculated this as 0.2 ms^-2
b) Find the tension in the tow bar.
c) Find P.

So i'm struggling to understand how to calculate tension in general. And also how to get P. Thanks a lot :smile:

a. Yes, correct.

Original post by Sh1vam.naik
Yeah I just wanted to check that I was doing it right? I got 20N as the tension because (F=ma of the trailer = 60N and taking away resistive force gives you 20N? Is that right?)

No. The resistance on the trailer is 40N, so to accelerate, you need the tension to be more than that. You should be adding the resistive force.

Draw a diagram and label one direction as the positive one. Keep all your signs consistent with this. For the trailer:

T20=maT-20=ma

T=ma+20\therefore T=ma+20
(edited 3 years ago)
Original post by Pangol
Nearly. You need to be careful with the directions of the forces. If there was only 20 N pulling the trailer forwards but a resistive force of 40 N, then the backwards force would be bigger than the forwards force and the trailer would accelerate backwards!

No I meant the force of the trailer going forward is 60N but there is a 40N resistive force so the resultant is 20N forward for the trailer?
Reply 6
Original post by Sh1vam.naik
No I meant the force of the trailer going forward is 60N but there is a 40N resistive force so the resultant is 20N forward for the trailer?

The resultant forward force on the trailer is the 60 N you calculated using F = ma. This is made up of the tension and the restive force. The tension points forwards, the resistive force backwards. You need to take their directions into account as well as their magnitudes.
Original post by Pangol
The resultant forward force on the trailer is the 60 N you calculated using F = ma. This is made up of the tension and the restive force. The tension points forwards, the resistive force backwards. You need to take their directions into account as well as their magnitudes.

Thanks very much for your help so far but I'm still very confused. So at the moment we just need to focus on the trailer? The forward (right) force is
(F=ma which is m x a which is 300 x 0.2 = 60N) but you said this is the resultant force? But there is a resistive force on the trailer (left) of 40N. So overall 60-40 = 20N going forward (to the right). What do I do from here? Sorry I'm just having a very dim day today it seems
Reply 8
Original post by Sh1vam.naik
Thanks very much for your help so far but I'm still very confused. So at the moment we just need to focus on the trailer? The forward (right) force is
(F=ma which is m x a which is 300 x 0.2 = 60N) but you said this is the resultant force? But there is a resistive force on the trailer (left) of 40N. So overall 60-40 = 20N going forward (to the right). What do I do from here? Sorry I'm just having a very dim day today it seems

No - the 60 N force is the total (i.e. resultant) force going forward. This force is made up from the tension T (forwards) and the resistive force of 40 N (backwards). Together they make up the resultant forwards force of 60 N. So T - 40 = 60.
Original post by Pangol
No - the 60 N force is the total (i.e. resultant) force going forward. This force is made up from the tension T (forwards) and the resistive force of 40 N (backwards). Together they make up the resultant forwards force of 60 N. So T - 40 = 60.

Makes much more sense now thank you very much! :biggrin:
what paper is this from

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