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A physics question

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LAN_403

dmg

Mon Mar 07 2011, 04:26AM

Registered Member #2628 Joined: Fri Jan 15 2010, 12:23AM
Location:
Posts: 627

There is something that I dont quite understand,

Here is the situation.
Assume there is a body of water (like a tank maybe), and from edge to edge is point A and point B.

You have a paddle in hand, or some other object that has significant surface area and will resist the flow of water as you move from point A to B.

In the first senario you will go as fast as you can, exerting the most ammount of force you are capable off to cover the distance.

In the second senario you do it once again, but this time you go as slow as you want, using as least force as possible to cover the same distance, but at a longer time.

My question is, assuming everything is constant but the speed of your hand and the time it takes, will the net force you excert be the same in the two senarios? or will differ? If so can it be mathematicly explained as to why?

Ash Small

Mon Mar 07 2011, 01:23PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Every action has an equal and opposite reaction.

Sulaiman

Mon Mar 07 2011, 01:28PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

IF there were no friction/viscosity etc. then a single push/paddle will get you across at constant speed.
Since there are several loss mechanisms involved, proportional to v, v^2 and v^3 (v=speed)
the faster you go the more energy expended on the journey.

coillah

Mon Mar 07 2011, 07:51PM

Registered Member #1517 Joined: Wed Jun 04 2008, 06:55AM
Location: Chico CA
Posts: 304

Ash Small wrote ...

Every action has an equal and opposite reaction.

This is not a very helpful reply. Although, it does set you on the right track.

A couple of things to consider are, the speed of the boat when you reach the finish, the drag force on the boat at certain speeds (as was already mentioned), and the total energy expended during the entire process.

If you are speeding along as fast as you can, the drag force on the boat will provide an acceleration against the direction of travel. To compensate for this negative acceleration you must push harder with the paddle to keep the same speed. This amounts to a larger amount of energy expended because E = F*d where E is energy, F is the force, and d is the stroke length of the rowing motion. Your F variable is large, and so is the d, so your E term is large as well.

If you are going slowly, frictional forces are smaller, so to maintain the same speed you only need to push less. Again this amounts to less expended energy.

Ash Small

Tue Mar 08 2011, 05:21AM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

The law of conservation of energy obviously still applies here.

As has been pointed out, the velocity of the boat at the end is relevant. As far as the viscosity/drag issue is concerned, though, energy lost due to friction will be converted to heat. The faster you paddle, the more heat will be generated due to frictional losses.

The most efficient way to paddle from point A to point B will be the method that produces the least turbulence.

The formula for calculating the most efficient velocity for any displacement hull is roughly the square root of the waterline length in feet knots (one knot is 2000 yards per hour)(nautical miles).

ie the most efficient velocity for a nine foot boat is three nautical miles per hour.

Above this speed energy is wasted because the boat is trying to climb it's own bow wave. This remains the case until the boat reaches sufficient velocity to overtake it's bow wave, at which point the boat starts planing. There is then a different formula that comes into play which I won't go into as it is not relevant in this case.

Renesis

Tue Mar 08 2011, 04:08PM

Registered Member #2028 Joined: Mon Mar 16 2009, 08:13PM
Location: Norway
Posts: 319

Ash Small wrote ...

The faster you paddle, the more heat will be generated due to frictional losses.

The frictional loss (in watts) is clearly larger when you paddle fast. But then again, the time this loss is exerted is less, because you reach your destination faster. So it sounds reasonable that the total energy spendt is somewhat identical in both cases.

But as Ash mentions, in the case of boats, this is also a question of hull design. A better example would be a sawmill, sawing through a thick log. The total amont of energy consumed by the motor would be relatively identical at different sawing speeds (not considering heat losses in the motor and its circuitry)

Steve Conner

Tue Mar 08 2011, 04:34PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

No, it's not. The frictional loss due to turbulence goes up as the square of the speed, or the cube or whatever. But the time taken to reach the destination only goes down linearly. So the faster you want to get there, the more energy you must spend in total.

Frictional losses for a vehicle moving through air obey similar laws, so this also explains why driving at a sensible speed saves gas. Well, it would if anyone ever did it.

In the case of a sawmill, it probably takes the same energy to cut X feet of timber no matter what speed you do it.

klugesmith

Tue Mar 08 2011, 05:25PM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

OP:
>>My question is, assuming everything is constant but the speed of your hand and the time it takes, will the net force you excert be the same in the two senarios?

"Net force" is probably the wrong word for what you want to know. The harder you paddle, the more propulsive force.

One of several correct replies:
>>The frictional loss due to turbulence goes up as the square of the speed, or the cube or whatever. But the time taken to reach the destination only goes down linearly.

We have to be careful here to define our term "loss". The speed exponent for power is higher, by one, than the speed exponent for drag force. They both go up sharply (as someone else pointed out) when a boat exceeds its critical hull speed.

Back to original question, interpreted to mean: at what speed do you cross the pond with the least work (energy)? We can forget about the time, since energy is force x distance, and the distance is constant.
So the slower the better.
Suppose, instead of paddling, you tow the boat using a rope that goes over some pulleys, to a falling weight. A large weight or a small one will fall the same distance, equal to length of the pond. The large weight will pull the boat faster, of course, but in theory an infinitesimal weight will eventually get the boat across.

Dr. Slack

Tue Mar 08 2011, 08:26PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

A good thing to do with physics questions is to look for upper and lower bounds.

What happens if you approach an infinite time getting across?

Speed approaches zero, so viscous force retarding the paddle's speed (for a Newtonian liquid like water) approaches zero. We haven't said what time we are going to allocate to accelerate the paddle, but if we acclerate it for 50% of the time, then slow it down for the second half, then that force for that will also tend to zero.

So the force isn't zero, but we can always pick a long enough time such that the force will be below any arbitrarily small force we care to specify.

The working is similar for trying to cross in a time approaching zero, the force needed will approach infinity

If instead of being filled with water, it's filled with fine sand, then the situation is different. Sand, which for some purposes can be regarded a fluid, is definitely non-Newtonian. When you push the paddle through slowly, it will pile up in front then trickle round the sides to the back, and resist the paddle's movement with probably a fairly constant force, regardless of speed (at a high enough speed the situation will change as dynamic effects become significant).

Really, the answer to your question is inherent in defining the force/speed behaviour for the paddle in your particular fluid-filled tank.

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