# Power to suspend a mass in air

How much power is needed to suspend a mass in the air?

The four parts below deal with the same problem. I post them all here, should there be some subtle things I didn't see.

## The textbook problem

A 9th edition college physics textbook has this as an example problem in its energy chapter (details cleaned-up):

An elevator car ("a lift") and its passengers has a mass of 1800 kg and negligible friction force in all its moving part. How much power must a motor deliver to lift the elevator car and its passengers at a constant speed of 3.00 m/s? Answer: Speed is constant, so $a=0$. Now let $F$ be the total force and $T$ be the force exerted by the motor. Then, $$\sum F = T-Mg = 0 \qquad \Rightarrow \qquad T = Mg$$ with power = force x velocity, we then have $$P = T\cdot v = Mg \cdot v$$ $$P = 1800 \,kg\cdot9.8\,m/{s^2}\cdot 3.00\,m/s=5.29\times10^4\,{\textrm W}$$ And, what power must the motor deliver at the instant the speed of the elevator is $v$ if the motor is designed to provide the elevator car with an upward acceleration of 1.00 m/s per second? Answer: Now $a=1.00 \,m/{s^2}$, and then $$\sum F = T-Mg = Ma \qquad \Rightarrow \qquad T = M(a+g).$$ Proceeding as before, we have $$P = T\cdot v = M(a+g) \cdot v$$ $$P = 1800 \,kg\cdot (1.00+9.8)\,m/{s^2}\cdot 3.00\,m/s= 5.83 \times10^4\,{\textrm W}.$$ That's the end of the example.

My questions. We might as well ask these:

1. How much power must the motor deliver to suspend the elevator and its passengers so it's kept at the constant height? That is, zero velocity? Certainly it can't be zero right? Because otherwise the elevator car will free-fall. But how much power is required?
2. What power must the motor deliver at the instant the speed of the elevator is $v=3.00$ (as before) if the motor is designed to provide the elevator car with a downward acceleration of 1.00 m/s per second?

I've been looking for an answer but to no avail.

## The helicopter problem

This is the real question that motivates this post.

I've been asked by a friend who's going to build a model-helicopter. "Ok, I have a 2000 W engine for my 50 kg model-helicopter. Assuming 100% efficiency, will it able to float 1 m above ground for at least 1 minute?"

"Ok then, what is minimum power required for the engine if I want it to float 1 m above ground, assuming 100% efficiency?"

## The pulley problem

This is a simpler version of the above two problems. Using a rope and a pulley, what is the minimum power required for a man to suspend a 1 kg mass in the air 1 m above ground for 1 second?

## The table or string problem

There's not much problem here. Suspending a 1 kg mass 1 m above the ground? Simple. Put it on a table, or hang it with a string. No movement. No power required.

What's happening here?

• is this homework? – Nikos M. Aug 22 '14 at 19:34
• No, it's not a homework. – fajar Aug 23 '14 at 1:39

When an elevator is moving, you have to do work against gravity. You are changing the potential energy of the system. The faster the elevator moves, the more work per unit time is needed (because power = work / time = force * velocity ). If you are changing the velocity of an object, you are changing its kinetic energy: if it's slowing down, it gives energy back to you; if it's speeding up, you need to give it more energy.

If the elevator car is not moving, no WORK needs to be done. You still need a FORCE - but you could tie a knot in the cable and turn off the power: the elevator car will stay in place, without electricity, without heat being generated...

The helicopter example is different. The only reason a helicopter can hover is because it is pushing air down. Every second that it hovers, it needs to move a volume of air at a certain speed. In this case, the helpful equation is

$$F\Delta t = \Delta p$$

The change in momentum of the air tells you the force that you can get. This can be done by moving a large volume of air a little bit, or moving a little bit of air by a lot. Both situations will give you the same momentum, but since energy goes as velocity squared, the larger blades will be more efficient (up to the point where the drag of the blades becomes an important factor).

To solve the problem you stated, you need to know the size of the blades of the helicopter. Making some really simplified assumptions (there is at least a factor 2 missing in this - but just to get the idea): if you have a helicopter blade that sweeps an area $A$ and pushes air of density $\rho$ down with velocity $v$ then the force is

$$F = m \cdot v = (A\rho v)\cdot v = A\rho v^2$$

and the power needed (kinetic energy of the air pushed down per second) is

$$E = \frac12 m v^2 = \frac12 (A \rho v) v^2 = \frac12 A \rho v^3$$

If we assume $A=3m^2$ (roughly 1 m radius), and $\rho=1 kg/m^3$, then for a force of $500 N$ we need a velocity

$$v = \sqrt{\frac{F}{A\rho}} = \sqrt{\frac{500}{3}} = 13 m/s$$

and the power required is

$$E = \frac12 A \rho v^3 = \frac12\cdot 3 \cdot 13^3 = 3.3 kW$$

This is a bit higher than the 2kW you have available. The solution would be to increase the size of the rotors - the larger the area, the lower the velocity of the air, and the better off you are.

As for the pulley and string, or fixed string - see the comments I made about the elevator. When nothing moves, no work is done. In the case of the helicopter, although the helicopter doesn't move, the wings (blades of the rotor) do - and so does the air that is being moved (and whose motion provides the force needed to keep the helicopter in the air).

I hope that clears up your understanding.

• it does clear up my understanding in the helicopter case. I understand that when nothing moves, no work is done. However: the textbook problem was not intuitively framed. One can take the limit of the velocity to zero, and the power does go to zero. (Taking the limit also means one cannot just tie a knot to halt it). While, of course, the motor still does some [big!] work just to sustain the elevator's position. A more practical question would be: how much electrical power is needed (in Watts) to move the elevator 0.001 m/s upward. Is there any engineering trick I missed here? – fajar Aug 23 '14 at 1:38
• If your motor is running at "zero velocity" it will draw a current to keep from rotating - but most elevators etc would have a brake that engages when there is no motion so there is no current required to maintain "zero". Going very slowly, you would have significant "holding" losses just as holding a weight in your hands will make you tired even though you are "not doing work". (In the physics sense) Usually this problem would be addressed with gearing: a motor could drive the elevator slowly with relatively little torque and a big gear ratio. – Floris Aug 23 '14 at 3:19
• Your units don't work out... $A\rho v$ gives you mass per time not force (mass distance per time time) Then there are similar problems for your energy equation. – Rick May 14 '15 at 16:52
• @Rick - you are absolutely right, that was sloppy of me. I have corrected the equation (I think). Thanks for pointing it out! – Floris May 14 '15 at 18:20
• Nice, your answer now agrees with mine to a factor of $\sqrt{2}$ – Rick May 14 '15 at 18:35

As for the helicopter problem, theoretically, arbitrarily low power can be sufficient to float a load, if you push down a lot of air with very low speed. However, you need longer and lighter (and maybe wider) blades for that, so the problem you'll have to solve is that of structural strength. Let me note that recently a muscle-driven helicopter (http://en.wikipedia.org/wiki/Human-powered_helicopter ) was demonstrated.

• Awesome link - I did not know that the Sikorsky prize had finally been won. Wow. – Floris Aug 22 '14 at 20:05
• Yes, awesome link! It gives me new light. – fajar Aug 23 '14 at 1:40
• Note the huge area swept out by the blades :) – Rick May 14 '15 at 18:36

Energy is required to push something over a distance. In the case of the elevator, when it's not moving a brake can be engaged and the power removed and the elevator will sit just fine. That's because it doesn't take any energy to keep something still.

But wait, if all I want to do is keep the helicopter still, then it doesn't require any energy?

Sort of. Not only do you have to conserve energy but also momentum. Gravity will giver your helicopter downward momentum at a constant rate that it must shed in order to remain stationary. In the case of the elevator, the car gave the momentum to the cable which gave the momentum to the building which gave it to the ground, which gave it to the ground which balanced out with a bunch of other momentum transfers to result in no net movement.

In the case of the helicopter it must transfer this momentum to the air. This is actually a pretty complicated process but can be approximated by the following:

Assume the blades cover an area $A$ and that air passes downward at a velocity $v_1$ given a pressure boost of $\Delta p$.

Now the ambient air is all at the same pressure so this pressure difference must transition to a difference in velocity via the Bernoulli formula. So let us call atmospheric pressure $p_0$, the pressure directly above the blades $p_1$ and the pressure below the blades $p_2$

Now is we assume the air coming in from the top of the helicopter starts with negligible velocity then in order to accelerate to $v$ in must undergo a pressure drop of

$$p_0-p_1=\frac12 \rho v^2$$

And the exiting air, in the process of returning to atmospheric pressures must change velocity according to the equation

$$p_2+\frac12 \rho v^2=p_0 + \frac12 \rho {v_f}^2$$

Where $v_f$ is the final velocity of the air moving downward.

Combining these equations give:

$$\Delta p = \frac12 \rho {v_f}^2$$

By conservation of momentum we know that the lift $L$ must equal the mass rate times the change in velocity:

$$\dot m = A\,\rho\,v$$ $$\Delta v = v_f$$

$$L= \dot m \Delta v = A\,\rho\,v\,v_f$$

We also know that the lift must be equal to the pressure difference times the area:

$$L = \Delta P \, A$$

equating

$$A\,\rho\,v\,v_f=\Delta P \, A$$

$$\rho\,v\,v_f=\Delta P$$

and from earlier

$$\Delta p = \frac12 \rho {v_f}^2$$

$$\rho\,v\,v_f = \frac12 \rho {v_f}^2$$

$$v = \frac12 v_f$$

Are power requirements $W$ are given as the force on the air by the velocity.

$$W= L \, v$$

$$W= L\,\frac12 v_f$$

$$W= L\,\frac12 \sqrt{\frac{2\Delta P}{\rho}}$$

$$W= L\,\sqrt{\frac{\Delta P}{2\rho}}$$

$$W= L\,\sqrt{\frac{L}{2\rho A}}$$

$$W= \frac{L^\frac32}{\sqrt{2\rho A}}$$

Note this analysis assumes perfect efficiency while in reality I would guess that helicopter blades are probably around 30% efficient. But the scaling should at least be right. The power requirements will grow with weight to the 1.5 power and will be inversely proportional to the length of the blades.

## Plugging in numbers

Since you didn't include an area in order for a 50kg helicopter powered with a 2000W engine to fly it would need blades sweeping a diameter $D$

$$4 A=\pi D^2$$

$$D= \frac{L^\frac32}{W}\sqrt{\frac2{\pi\,\rho}}$$

$$D= \frac{(g \,50 kg)^\frac32}{2000W}\sqrt{\frac2{\pi\, 1.2 \frac{kg}{m^3}}}\approx 4 m$$

So my guess would be no, the model helicopter would not fly, as its blades are probably not that long.