# Mechanics - Train

• May 3rd 2006, 08:03 AM
Natasha1
Mechanics - Train

A train of mass 20 000 Kg freewheels down a straight incline of angle 10 degrees at constant speed 15 m/s in that the gravitational effect is cancelled by friction and air resistance.

It then comes to a level stretch, the friction and air resistance remaining the same. Find how far it travels before coming to rest.

The train then starts moving again, by switching on the engine, and after a modest speed (value being immaterial here) comes to an upward incline of x degrees. Friction and air resistance still remain at their previous value.

Find in terms of x, the engine force P needed to maintain a constant speed up the incline. Sketch a graph of P against x. Is any interpretation possible for negative x?

PS: The coefficient of friction 'model' is not used in this question. The frictional force is taken as constant, along with air resistance. g = force du to gravity = 10
• May 3rd 2006, 09:00 AM
CaptainBlack
Quote:

Originally Posted by Natasha1

A train of mass 20 000 Kg freewheels down a straight incline of angle 10 degrees at constant speed 15 m/s in that the gravitational effect is cancelled by friction and air resistance.

Look at the forces acting on the train parallel to the plane, they are the
total force tending to slow the train (friction and air resistence), and the
component of the weight (gravitational force on train) down (parallel to) the
plain.

The component of the weight down the plain is:

$
m.g.\sin(10^{\circ})\approx 20000 \times 9.81 \times 0.1736\approx 34060
$
newtons

This is in balance with the two slowing forces.

Quote:

It then comes to a level stretch, the friction and air resistance remaining the same. Find how far it travels before coming to rest.
Now there is no component of gravity acting in the horizontal direction, but
the slowing forces are the same as before. So this is a problem of a body
being slowed by a constant force. As we know this force and the trains
mass we can solve this fairly easily.

RonL
• May 4th 2006, 01:34 PM
Natasha1
Right then.... please do correct me in my wording or miss calculations. Many thanks in advance :-) (Oh, I use g=10 below, not as 9.81)

Friction + Air resistance = -34730 N

-34730 = ma
-34730 = 20000a
a= - (34730/20000)
a= -1.7365

To find the distance travelled:

u^2 = v^2 + 2as
15^2 = 0 + 2(-1.7365)s
s=64.8 m (I actually get -64.8 m ??? What should I do?)

To find the engine force P (we solve the force along the slope) to get:

P = (Friction + Air resistance) + Weight component
P = 34730 + 20 000*g*sinx
P = 34730 + 20 000*10*sinx

For values of x values smaller than -10 degrees(i.e -11, -12, etc), the driver will have to start using the brakes in order to componsate for the acceleration taking place.
• May 4th 2006, 02:01 PM
CaptainBlack
Quote:

Originally Posted by Natasha1
Right then.... please do correct me in my wording or miss calculations. Many thanks in advance :-) (Oh, I use g=10 below, not as 9.81)

Friction + Air resistance = -34730 N

-34730 = ma
-34730 = 20000a
a= - (34730/20000)
a= -1.7365

To find the distance travelled:

u^2 = v^2 + 2as
15^2 = 0 + 2(-1.7365)s
s=64.8 m (I actually get -64.8 m ??? What should I do?)

When the train reaches the level the slowing forces remain the same, as
we calculated, so as you observe:

$
-34730 = ma
$

(here I am using your value for the slowing forces - need to check this).

so:

$
a=\dot v =-34730/20000=-1.7365
$

which may be integrated to give:

$
v(t)=-1.7365 t+c
$

now if we take $t=0$, as the moment that the train reaches the level part
of the track we know that $v(0)=15$ m/s, so $c=15$.

Now we have:

$
v(t)=\dot s=-1.7365 t+15
$
,

which we may integrate again to give:

$
s(t)=-\frac{1.7365 t^2}{2}+15t+const
$

and if we are measuring distances from the start of the level track $s(0)=0$

Now the train comes to rest when $v(t)=-1.7365 t+15=0$ so
the train stops at $t=15/1.7365\approx 8.64$ s. Now plug this value
of $t$ into the equation for $s$ to find the distance
traveled before stopping on the level section:

$
s(8.64)=-\frac{1.7365\times 8.64^2}{2}+15\times 8.64=64.8$
m

Quote:

To find the engine force P (we solve the force along the slope) to get:

P = (Friction + Air resistance) + Weight component
P = 34730 + 20 000*g*sinx
P = 34730 + 20 000*10*sinx

For values of x values smaller than -10 degrees(i.e -11, -12, etc), the driver will have to start using the brakes in order to componsate for the acceleration taking place.
This last bit looks OK to me.

RonL
• May 4th 2006, 02:11 PM
Natasha1
RoL,

You say (here I am using your value for the slowing forces - need to check this). I'm pretty confident it is -34730 N, as isn't that the friction and air resistance added together. As it is mentioned in the question right at the top that:

A train of mass 20 000 Kg freewheels down a straight incline of angle 10 degrees at constant speed 15 m/s in that the gravitational effect is cancelled by friction and air resistance.

Am I right or wrong :confused: ?
• May 4th 2006, 02:26 PM
CaptainBlack
Quote:

Originally Posted by Natasha1
RoL,

You say (here I am using your value for the slowing forces - need to check this). I'm pretty confident it is -34730 N, as isn't that the friction and air resistance added together. As it is mentioned in the question right at the top that:

I wrote that before it registered that you were using 10 as the value of g.

Quote:

A train of mass 20 000 Kg freewheels down a straight incline of angle 10 degrees at constant speed 15 m/s in that the gravitational effect is cancelled by friction and air resistance.

Am I right or wrong :confused: ?
yes this is all right. The only problem I had with your solution was that

RonL
• May 8th 2006, 03:49 AM
Natasha1
what does the graph P against x look like?
• May 8th 2006, 05:37 AM
CaptainBlack
Quote:

Originally Posted by Natasha1
what does the graph P against x look like?

With a bit of luck it looks like the attachment

RonL
• May 8th 2006, 05:46 AM
Natasha1
Quote:

Originally Posted by CaptainBlack
With a bit of luck it looks like the attachment

RonL

Merci Monsieur ;-)