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Math Help - Mechanics - Train

  1. #1
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    Mechanics - Train

    Please help

    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
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  2. #2
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    Quote Originally Posted by Natasha1
    Please help

    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:

    <br />
m.g.\sin(10^{\circ})\approx 20000 \times 9.81 \times 0.1736\approx 34060 <br />
newtons

    This is in balance with the two slowing forces.

    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
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  3. #3
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    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.
    Last edited by Natasha1; May 4th 2006 at 01:37 PM.
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  4. #4
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    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:

    <br />
-34730 = ma<br />

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

    so:

    <br />
a=\dot v =-34730/20000=-1.7365<br />

    which may be integrated to give:

    <br />
v(t)=-1.7365 t+c<br />

    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:

    <br />
v(t)=\dot s=-1.7365 t+15<br />
,

    which we may integrate again to give:

    <br />
s(t)=-\frac{1.7365 t^2}{2}+15t+const<br />

    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:

    <br />
s(8.64)=-\frac{1.7365\times 8.64^2}{2}+15\times 8.64=64.8 m

    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
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  5. #5
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    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 ?
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  6. #6
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    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.

    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 ?
    yes this is all right. The only problem I had with your solution was that
    I could not follow your reasoning to arrive at the 64.8m.

    RonL
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  7. #7
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    what does the graph P against x look like?
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  8. #8
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    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
    Attached Thumbnails Attached Thumbnails Mechanics - Train-new-picture-1-.jpg  
    Last edited by CaptainBlack; May 8th 2006 at 05:40 AM.
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  9. #9
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    Quote Originally Posted by CaptainBlack
    With a bit of luck it looks like the attachment

    RonL
    Merci Monsieur ;-)
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