# Power, Tension, Incline ?

1. Jan 18, 2013

### davekardle

1. The problem statement, all variables and given/known data

Hi guys can you help me with this problem?

I have to find the power required to lift a boat up an incline (24.25 degrees to the horizontal). The boat lock is attached to a counterweight which is 25T heavier when the boat is ascending the incline.

The mass of a boat and lock is 1020T so the counterweight Weight is 1045T.

What the tension on the chord and the power required to lift the boat at a constant velocity?

The height of incline is 55m
Linear incline length is 134m

2. Relevant equations

Tension = F(push by the counterweight) - F(pull by the boat)

? and then for power

(T x displacement )/time ?

3. The attempt at a solution

Tension on cable= Fpush - Fpull
T= (50T)(9.81)sin24.25 - (25T)9.81sin24.25
T=100.8KN

Work= Tension x displacement
Work= 100.8KN x 133

POWER= (100.8KN x 266m)/8(60)s (power ascending and descending)

Power= 55.6KW

Is this correct ?

2. Jan 18, 2013

### Staff: Mentor

I don't see where your values 50T and 25T come from, you had 1020 and 1045 in the problem statement.

Tension depends on the setup - tension where? Tension in a cable holding the counterweight has to be more, for example.

Do you mean $P=\frac{100.8kN \cdot 266m}{8\cdot 60s}$?
Where do the 266 and 8 come from?

3. Jan 18, 2013

### davekardle

266 = ( 133m x 2 )

Ascending and descending the incline.
is it correct ?

When ascending the boat lock is 25t lighter than the counterweight
when descending the boat lock is 25 Heavier than the counterweight.

it takes 4 min to lift the lock to the max height
so a full cycle down the incline is 8 min.

Should I include all this for the power ? or just the lift.

4. Jan 18, 2013

### Staff: Mentor

133 != 134, and why do you use two times that value now? I think the boat is just ascending.

That is different from your initial problem statement.

It is not useful to add forces like that in this case - consider ascending and descending boat as separate processes, until you see that they are the same.

Please add the full problem statement in the first post, not just some parts of it. It is hard to understand what you are doing that way, and asking for everything one by one is annoying.
The position/connections of the chord where you want to calculate torque are still unclear, for example.

5. Jan 18, 2013

### davekardle

I used the two values because I also need to know the power the required to control the speed of both lock and counterweight when descending.
I need to charge each boat lift for each complete cycle.

Here is the full problem I edited. ( sorry for the mix up)

1. The problem statement, all variables and given/known data

Hi guys can you help me with this problem?

I have to find the power required to lift a boat up an incline (24.25 degrees to the horizontal).
The boat is 25T lighter than the counterweight when it ascends the incline
and 25T heavier than the counterweight when it descends the incline.

The mass of a boat and lock is 1020T so the counterweight Weight is 1045T when ascending.
When descending only lock weight is considered which is 300T hence the counterweight must weight 275T.

The height of the incline is 55m
Linear incline length is 134m
Time for the life is 4min

The boat lock is attached to the counterweight by two group of 14 cables.

I'm provided with two electric motors to provide 60KW power each.

The motor power the lift of the boat and control the speed of the lock
and the counterweight.

2. Relevant equations

Tension = F(push by the counterweight) - F(pull by the boat)

? and then for power

(T x displacement )/time ?

3. The attempt at a solution

Tension on cable= Fpush - Fpull
T= (50T)(9.81)sin24.25 - (25T)9.81sin24.25
T=100.8KN

Work= Tension x displacement
Work= 100.8KN x 133

POWER= (100.8KN x 266m)/8(60)s (power ascending and descending)

Power= 55.6KW

Is this correct ?

6. Jan 18, 2013

### haruspex

If that's true then I cannot imagine the set-up. I would have thought the cables ran over pulleys to the counterweight. The counterweight presumably slides on the same incline. Any difference between the two loads (object load v. counterweight) must relate to acceleration or friction or both. Taking it to be just friction, it could be on the slope or in the pulleys or both. If it's not in the pulleys then the tension will be the same both sides of the pulleys.
It's not clear where the electric motor comes in. If the actual motion is driven by the difference in weights, the motor would be for getting the difference in the weights back up to the top for next cycle. A more obvious arrangement would be to have constant counterweight with the motor providing the difference, e.g. counterweight = 660T, motor provides the difference of 385T in each half of the cycle.

7. Jan 19, 2013

### Staff: Mentor

I thought the same.

It is strange that the counterweight is heavier when it goes down - you don't need the motor then (neglecting friction), gravity is sufficient to lift the boat.

That might be an explanation.

@davekardle: Please, do not send PMs like that. I see this thread in the subscriptions anyway.