# Homework Help: Cable Car Power

1. Nov 20, 2013

### hk4491

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. Nov 20, 2013

### mic*

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?

3. Nov 20, 2013

### mic*

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?

4. Nov 20, 2013

### hk4491

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.

5. Nov 20, 2013

### mic*

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.

6. Nov 20, 2013

### mic*

I hope that is some sort of help. Best of luck