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Cable Car Power

  1. Nov 20, 2013 #1
    1. The problem statement, all variables and given/known data

    A cable railway in Austria has a length of 5.6 km. The cars on the cable need 60 minutes for a whole trip along the cable. Assume that 12 cars with a payload of 550 kg are travelling upwards, and 12 empty cars are travelling downwards. The angle between the cable and the ground in 30°. What is the power generated by the motor operating the railway?
    Hint: Against which outer forces does the motor have to bring the power, so that the cars start moving?


    2. Relevant equations

    P=W/t

    W= F*d

    3. The attempt at a solution

    Can someone please tell me if the following method I used is correct?


    Assuming the mass of an empty car is "m", the mass of the cars going upwards will be (550 + m)

    The only relevant force acting on the cars is the weight.

    For the upwards travelling cars: (where Fc means the vector component on the cable)

    Fc = -(550 + m)*g*cos30*12

    For the downwards travelling cars:

    Fc = m*g*cos30*12

    After adding the two the total force would be: F= -550*g*cos30*12

    Then calculating the work: W = -550*g*cos30*5,600

    The power would be: P= (-550*g*cos30*5,600)/3,600 = 87,133 W = 87.1 kW
     
  2. jcsd
  3. Nov 20, 2013 #2
    Working from the hint given, what force does the motor act against, and then, applying the formula for work you have given, over what distance?
     
  4. Nov 20, 2013 #3
    Also, since friction appears to be neglected in this problem, if ALL the cars were empty - up and down - how much power would the motor need to start the cable moving?
     
  5. Nov 20, 2013 #4
    I included in my calculations that the primary force being acted against is the weight of the cable cars, over the distance of the entire cable which is 5.6 km, with a time of 60 minutes.
     
  6. Nov 20, 2013 #5
    If F and s are in the same direction, why would this line;

    After adding the two the total force would be: F= -550*g*cos30*12

    ...show a negative value for 550?
    Also, if the force (of the cart going upward) acts against gravity, why would you apply a factor of cos30 in this equation?


    This line;

    Then calculating the work: W = -550*g*cos30*5,600

    ...will return a result for work done in the horizontal direction. The work being done which is of interest in this problem appears to be the vertical component.
     
  7. Nov 20, 2013 #6
    I hope that is some sort of help. Best of luck :smile:
     
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