Mechanics/ Maths No 2!

dmg24

Now I am feeling really blonde, and you guys are just far to good at these things (creep, creep!).

Next topic, Force and Motion, and I don't know where to start with it!

One horse pulls with a force of x N a cart of mass 800 kg along a horizontal road at constant speed. Three horses, each pulling with a force of x N give the cart an acceleration of 0.8 ms^-2. Find the time it would take two horses to increase the speed of the cart from 2 ms^-1 to 5 ms^-1 given that each horse pulls with a force of x N and that the resistance to motion has the same constant value at all times.

Thank you for any ideas! x

melancholly

F = ma (hopefully right?)

Then you can use the data from the 3 horses to get
3x = 800 * 0.8 and then get x (640/3? sounds a bit of an ugly number!)

Then you have the force being applied..... does that help?


at work so have to run off..... will ponder horses and carts while i sit through a dull meeting!

EDIT - just had a quick look at the other thread - have left out all units because it's just too confusing when typing equations like this!

melancholly

ooooh - the friction will be the same all the time (at least that much is clear because they tell us in the question!), but in one case it will be split 3 ways and the other time between 2 horses........... that sounds like it might be relevant (or not!)

dmg24

I was hoping you would be about!

I got the f = ma bit, but what concerned me was how the one horse information came into it?

I am hoping that my tutor may 'forget' about this module ... give me Pure anyday! x

m277

You're right to be concerned about the info on the one horse.

What it tells you is that the resistance from the cart is exactly equal to the pulling force of one horse. You get to this conclusion because the horse and cart are travelling at a constant speed (no acceleration) which means using
net Force = mass*acceleration
x - Resistance= 800*0
so the resistance is x

From the three horse info we have a forward force of 3x and a resistance of x (constant resistance).
So the net F= ma equation becomes:
3x - x= 800*0.8 giving x= 320

Now we use this info for the two horses. Again the resistance is x which we now know is 320N so for net F=ma:
2x-x = 800*acceleration
giving the acceleration=2/5 (or 0.4)

Now we can use suvat:
s= unknown
u= 2
v= 5
a= 2/5
t= want to know
I think you can work it out from here.

dmg24

Fantastic, thanks ever so much m277, I still don't understand mechanics, but I can see where you got the numbers from! x

m277

My advice would be to always draw a diagram and list any formulas and values you are going to use.

Like in this case, draw a cart with the force of the horse going in one direction and the resistance in the other. Then list the values you want to find like I've done in my working above. Everything becomes a lot clearer on paper than in your head because you could accidentally miss some info somewhere.