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Tough M2 Question Helpage Please Happy Xmas too

This be the question:

A car of total mass 1200kg is travelling along a straight horizontal road at 40ms-1 when the driver suddenly applies the brakes. The brakes exert a constant force and the car comes to rest after travelling 80m. The other resistances on the car total 500N.

a) Find the magnitude of the force exerted by the brakes (I did this and got 11500N)

A trailer, with no brakes, is now attached to the car by means of a tow-bar. The mass of the trailer is 600kg, and when the trailer is moving, it experiences a constant resistance of magnitude 420N. The tow-bar is modeled as a light rod which remains parallel to the road at all times. The car and trailer come to a straight hill, which is inclined to the horizontal at an angle x (sinx = 1/14). They move together down the hill. The driver suddenly applies the brakes, which exert a force of same magnitude as before. The other resistances on the car remain at 500N

b) Find the deceleration of the car and trailer when brakes applies (I got 6.2ms-2)

c) Find the magnitude of the force exerted on the car by the trailer when the brakes are applied (STUCK)

I really would like to do this question myself, but I’m having trouble picturing the forces in the last section. What about the tension in the tow-bar?

Any help is much appreciated and I’ll rep if you want.

Reply 1

Rich L

The key is to realise that the car and the trailer form a newton 3rd law pair (tension in towbar is the same in either direction), so the force of the car slowing down the trailer is the same as the force of the trailer on the car. You should be able to calculate the force slowing down the trailer...

Reply 2

fidel69

this hsa made me realise how boring mechanics is,
why am i doing a mechanics centred degree a uni?

Reply 3

Rich L

Ha yep this is why i'm gonna do chemistry. Just can't bring myself to face 3.5 more years of physics...

Reply 4

firefist_ace

Since I'm doing this exam in jan I thought I might as well have a go. I got the same answers for the first two parts. Part c, I'm not sure how to go about it really, is the tension in the bar 420 and add 420 from the mass of trailer so the force is 840n? I'm not sure this is right so if someone explain :smile:

Reply 5

gcseeeman

Lou Reed

I shall leave the Maths forum to mull this over the early hours of the morning. It has never failed me yet!!

Firstly, people have been talking of the 'tension' in the tow-bar after braking. It is actually 'thrust' (it pushes out towards the car and the trailer). This may seem a pointless thing to know (and it doesn't really affect the answer) but this terminology apparently came up in a recent mechanics exam and confused many people.

As for the answer, just look at the car on its own (or the trailer as the thrust is equal). Note all the force acting on it, and the net force of this is equal to the car's mass times acceleration. The excess force needed to make the net force equal this is the answer.

You said you wanted to do it by yourself so I'll leave it there, but if you want more help ask :smile:

Reply 6

gcseeeman

Lou Reed

Ah...so..

On the car, forces:

<--(component of weight)
-->(braking force 11500N + Thrust)

That right? If it is indeed thrust then surely it would matter because the force will be in the opposite direction?

Thanks for the help.

Not exactly. It's:

Acting down the slope (forwards) - thrust from the tow-bar (or as they call it, the force from the trailer) and the weight.

Acting back up the slope - braking force (+ orignal 500N resistance)

(total of this is equal to mass of the car x acceleration)

I said thrust doesn't really matter as it is the same (they called it something different. Basically I'm just saying there is no tension)

i.e. if you look at the trailer, thrust acts backwards on it, along with resistance, and the weight pushes it down.

Does that help?

Reply 7

gcseeeman

Lou Reed