# M1: Tension between a car and a towed trailer

Watch
Announcements

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?

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

reply

Report

#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

Draw a free body diagram of the Trailer & resolve

P.S don't forget about the unknown thrust of the car

2

reply

(Original post by

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

**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 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

reply

Report

#4

(Original post by

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?

**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?

No acceleration - net force is 0

Net force is 0 forces are equal

2

reply

Report

#6

Out of interest, where did this question come from? The wording is odd - the phrase 'the force in' is, at best, clumsy.

0

reply

(Original post by

Out of interest, where did this question come from? The wording is odd - the phrase 'the force in' is, at best, clumsy.

**RogerOxon**)Out of interest, where did this question come from? The wording is odd - the phrase 'the force in' is, at best, clumsy.

0

reply

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?

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

reply

Report

#9

(Original post by

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?

**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?

0

reply

Report

#10

(Original post by

Cambridge University Press

**W. A. Mozart**)Cambridge University Press

1

reply

Report

#11

(Original post by

Ta. Must not comment ..

**RogerOxon**)Ta. Must not comment ..

Can some help with c)ii please

1

reply

Report

#12

(Original post by

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

**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

0

reply

Report

#13

(Original post by

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)

**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)

0

reply

Report

#16

(Original post by

Is a 1800=1200a?

**cosford2**)Is a 1800=1200a?

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

reply

Report

#17

(Original post by

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).

**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).

0

reply

Report

#19

(Original post by

The acceleration is negative - the car is slowing.

**RogerOxon**)The acceleration is negative - the car is slowing.

0

reply

Report

#20

(Original post by

what is the acceleration?

**cosford2**)what is the acceleration?

0

reply

X

### Quick Reply

Back

to top

to top