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# M1 help watch

1. A car of mass 900kg is towing a trailer of mass 300kg along a straight horizontal road. The car and the trailer are connected by a light inextensible tow-bar and when the speed of the car is 20ms^-1 the brakes are applied. This produces a braking force of 2400N. Find:
a. deceleration of the car
b. the magnitude of the force in the tow-bar.

My answer: a. consider the whole system: -2400=1200a
a= -2ms^2

b. T= 300a
T= -600N

I don’t get why i keep getting the force on the tow-bar as a negative number.
2. (Original post by Mme_Bonii)
I don’t get why i keep getting the force on the tow-bar negative. Is there something wrong with my diagram?
Have you got the picture? Or the question?
3. (Original post by Daniel100499)
Have you got the picture? Or the question?
Yes i do! I'm just having a bit of trouble uploading the pictures
4. (Original post by Mme_Bonii)
Yes i do! I'm just having a bit of trouble uploading the pictures
5. It's possibly due to the fact that you're not considering consistently which direction you take as negative and which direction you take as positive.

You've taken the braking force to be negative, therefore the acceleration (deceleration) must also be negative. The tension is therefore positive as it is acting in the opposite direction as the braking force.

Draw a body diagram with all forces acting on the trailer (or car, should be the same).
6. Hello
choose the direction of the vectors carefully. check my attachment.

(Original post by Mme_Bonii)
A car of mass 900kg is towing a trailer of mass 300kg along a straight horizontal road. The car and the trailer are connected by a light inextensible tow-bar and when the speed of the car is 20ms^-1 the brakes are applied. This produces a braking force of 2400N. Find:
a. deceleration of the car
b. the magnitude of the force in the tow-bar.

My answer: a. consider the whole system: -2400=1200a
a= -2ms^2

b. T= 300a
T= -600N

I don’t get why i keep getting the force on the tow-bar as a negative number.
(Original post by Mme_Bonii)
A car of mass 900kg is towing a trailer of mass 300kg along a straight horizontal road. The car and the trailer are connected by a light inextensible tow-bar and when the speed of the car is 20ms^-1 the brakes are applied. This produces a braking force of 2400N. Find:
a. deceleration of the car
b. the magnitude of the force in the tow-bar.

My answer: a. consider the whole system: -2400=1200a
a= -2ms^2

b. T= 300a
T= -600N

I don’t get why i keep getting the force on the tow-bar as a negative number.
Attached Images

7. (Original post by gurungg48)
Hello
choose the direction of the vectors carefully. check my attachment.
what if i took the right direction a positive?
8. (Original post by BTAnonymous)
It's possibly due to the fact that you're not considering consistently which direction you take as negative and which direction you take as positive.

You've taken the braking force to be negative, therefore the acceleration (deceleration) must also be negative. The tension is therefore positive as it is acting in the opposite direction as the braking force.

Draw a body diagram with all forces acting on the trailer (or car, should be the same).

This is how I drew the diagram. I took the right direction as positive
9. (Original post by Mme_Bonii)

This is how I drew the diagram. I took the right direction as positive
When the braking is taking place, the force in the towbar is a thrust, i.e. it is pushing outwards on the two objects. Since you've drawn the force as a tension, i.e. pulling on each object, it will come out as negative.

Think about it: there has to be a force on the trailer pushing backwards on it to make it slow down.
10. (Original post by tiny hobbit)
When the braking is taking place, the force in the towbar is a thrust, i.e. it is pushing outwards on the two objects. Since you've drawn the force as a tension, i.e. pulling on each object, it will come out as negative.

Think about it: there has to be a force on the trailer pushing backwards on it to make it slow down.
Oh, makes sense. What about the the tension opposite to the one of the trailer, is it just there to cancel it out? Or is there another reason?
11. (Original post by Mme_Bonii)
Oh, makes sense. What about the the tension opposite to the one of the trailer, is it just there to cancel it out? Or is there another reason?
As well as the towbar pushing the trailer in a backwards direction, to slow it down, the towbar pushes the car forwards which cancels out with some of the braking force. If you walk into the back of someone, you exert a forwards push on them while they exert a backwards push on you (try it!).
12. (Original post by tiny hobbit)
As well as the towbar pushing the trailer in a backwards direction, to slow it down, the towbar pushes the car forwards which cancels out with some of the braking force. If you walk into the back of someone, you exert a forwards push on them while they exert a backwards push on you (try it!).
I’m sorry if my question sounds stupid, but do you have an explanation as to why it’ll push the car forward and cancel out with some of the braking force?
13. (Original post by Mme_Bonii)
I’m sorry if my question sounds stupid, but do you have an explanation as to why it’ll push the car forward and cancel out with some of the braking force?
We know that the towbar is pushing backwards on the trailer, so it is pushing outwards, not pulling inwards (this is the only option for a string). A rigid thing such as the towbar can either push outwards or pull inwards, but it must do the same thing at each end. It is either being stretched (tension pulling inwards at both ends) or being squashed (thrust pushing outwards at both ends).
14. Please look at the attachment

(Original post by Mme_Bonii)
I’m sorry if my question sounds stupid, but do you have an explanation as to why it’ll push the car forward and cancel out with some of the braking force?
Attached Images

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