Mathametician required

[allover]

Any help would be appreciated
My question to any one capable of working this out

I have a 100 tonne truck on a gradient of 17.5 degrees with no brakes descending at 1 meter/second constant,(don't worry about friction) I have another truck tied by a tow rope trying to prevent it from moving
Question, how much power/ energy/? do i require to make the broken down truck stationary
Ditto at 11.5 degrees
Reference?

[trash]

Oh dear ... kinetics HSC question. Blow in my ear to dust out the cobwebs.

Lets see. Start with the truck. 100 tonnes. Shall we use 9.8ms^-2 for acceleration due to gravity which gives us 9800000 Newtons force
17.5 degrees. Without being able to draw the vectors, you'll just have to take my word for it that we want 9800000N Opposite/Hypotenuse which means we want a SIN function.

SIN(17.5)=X/9800000N
SIN(17.5)=0.3 x 9800000 = X = 2947000 Newtons for force is required to stop to it from rolling down the hill.

If the rope broke, then the Truck would accelerate down the hill at 2947000/100000 = 2.947ms^-2

[gordon_s1942]

If the rope broke, then the Truck would accelerate down the hill at 2947000/100000 = 2.947ms^-2

Then all the Kings Horses and all the Kings Men wouldnt be able to put allovers truck back together again.

Perhaps Trash can complete the equation and tell us how far the truck would have to travel before hitting the inevitable immovable object to render it into a pile of scrap ?

[allover]

Thanks Trash, i was hoping you would be able to answer it, what if the truck was travelling at 1.6666 m/s ie 100 m per minute. What would that increase to?
regards

[lsemmens]

You getting people to do your homework for you Allover?

[allover]

You getting people to do your homework for you Allover?

Isemmens, i will take that as you don't have a clue to the answer
Mass, 100 tonnes, speed 1.66m/s
KE=0.5 x m x v2
KE= 0.5 x 100,000kg x 1.66m/s x 1.66m/s
KE=138,000kg (138 tonnes)

As i did accountancy not physics etc, i really haven't got a clue or else i would not be asking, but what i am trying to determine is how much affect does the incline have on decreasing the above figure, ie 17.5 to 11.5 degrees
The answer could be life changing, but every great idea i have had there turns out a valid reason that it has not been done before

[lsemmens]

I did do physics in high school, but it was so long ago, that I forgot most of it. It was more a light hearted comment, than anything else. Had I the time, I probably could have worked it out but trash was way ahead of me. Sorry, should have added a smiley.

[allover]

Thanks Isemmens, no probs, the smiley button does make a big diff

[trash]

sorry for the delay, I've been on holidays. But it looks like it got answered.

I was always good at accounting, commerce and economics, but I hated them.
I wasn't that good at maths in high school. I was ok, if it was applied maths like physics, but I didn't have enough brain power at the time for 4U maths that I would have liked to have had and well maths teachers are not very good at teaching maths.

When I got into university level applied engineering maths, I was disappointed that I probably could have done a lot better in the HSC because all that pointless maths had a purpose and made sense.


The maths used in this question is just trigonometry. A complex word used to confuse and scare maths students.

Really it's nothing but triangles. All you need to know is how to work out lengths and angles.
The trick is to remember the three functions SIN COS and TAN.

The three sides are named in relation to the angle you are working with.
<b>Opposite</b> side to the angle = O
<b>Adjacent</b> side to the angle = A
<b>Hypotoneuse</b> The longest side of the triangle = H

TAN = Opposite/Adjacent. TAN is the easiest to remember. It's the odd one of the three used with right angle triangles.
SIN = Opposite/Hypotoneuse. This is the one to remember SIN "OH!".

If you can remember the first two, then the third is easy to work out.
COS = A/H

Finally once you realise you can cut up any triangle into two right angle triangles, everything else falls into place as you pick up the missing pieces.

a=angle
A=the length of the side opposite.

A/SIN(a) = B/SIN(b) = C/SIN(c)

[allover]

Thanks Trash, believe it or not i stumbled onto a tutorial as you have mentioned and it started to come back to me. I had done it at high school but that was the last time and a long time ago

[mol666]

Any help would be appreciated
My question to any one capable of working this out

I have a 100 tonne truck on a gradient of 17.5 degrees with no brakes descending at 1 meter/second constant,(don't worry about friction) I have another truck tied by a tow rope trying to prevent it from moving
Question, how much power/ energy/? do i require to make the broken down truck stationary
Ditto at 11.5 degrees
Reference?

Just Get a bigger Truck to Tow it


Haha