Once again: Could you please not double post? It is annoying having to edit to answer you posts.
(Changed to red, because this is an actual forum rule which should be followed by everyone anyway. - gmalivuk)
cobra bubbles wrote:
By this logic the forces between a wire and a truck it drags does no work because the relative velocity is zero.
correct. a wire pulling an object does no work, because there is no relative motion. the force acting on the wire does the work. the wire simply transfers the force. static friction simply transfers force also.
Or what about a car accelerating or braking without the wheels slipping?
cars can brake without the wheels slipping. kinetic energy is removed via kinetic friction between the brake pads and the discs, not by the static friction between the tire and the road. this is why you don't want to brake so hard that the tires skid. kinetic friction is typically less than static friction between two surfaces. so you get more braking acceleration by *not* slipping.
Fine so the static friction makes it possible to extract energy from the motion of the car so stop ignoring it.
cobra bubbles wrote:accelerating cars are different. static friction is defined as F=< uN, the magnitude of friction is less than or equal to the coefficient of friction times the normal force. static friction can change in response to the force opposing it, up to the maximum value, when slippage starts, and the kinetic friction regime is entered. to accelerate a car, the engine applies torque to the tire, which interacts with the road via static friction, up to some max value when the wheels start to slip. since the tire is pushing on the road backward, the road pushes the tire forward by newton's third law, causing a net force in the forward direction. this does not mean that the static friction force did work on the car. it does not. the torque of the engine turning the tires through an angle is the source of the cars kinetic energy.
Thanks for this nice breakdown of how a car manages to accelerate. Al I was saying was that just because a force produces no work, you can't just ignore its presence when summing up the resultant force at maximum speed.
cobra bubbles wrote:
You are switching back and forth between refference frames, could we please maybe choose one and stick to it? In the cars reference frame the interface between ground and wheels are indeed moving.
i am not switching reference frames. i am keeping in the frame of the ground. as the cart moves by the ground at speed v, the rim of the wheel is at speed -v, so the net speed of the wheel with respect to the ground is v + (-v) = 0. there for displacement is zero and work is zero. i did not change reference frames.
Oh no! You claimed that the air moves faster after it has passed the propeller than before. That is true in the car frame, not in the ground frame.
cobra bubbles wrote:so yes, we *can* ignore static friction. it is simply a mechanism to transmit force. it does not transmit energy.
Not when summing up the resultant force.
cobra bubbles wrote:
Where did you get that from? The propeller is rotating!
yes, the propeller is rotating. but it is a necessary consequence of newton's second law. at terminal velocity, the propeller is unable to transmit energy to the cart.
Sure a force of magnitude zero cant do work. But it is not zero at maximum velocity, it is still counteracting the braking on the wheels.
cobra bubbles wrote:sure, iceboats and sailboats can acclerate to a speed faster than that of the wind. they do it by exploiting the forces involved in a different manner. for those boats, the final speed is determined by a constant wind force and a slowly increasing resistive force. if the resistive forces increase slow enough as a function of velocity, they will balance the force due to the wind, giving net force zero at a higher terminal velocity.
the dwfttw wind cart operates on a different principle. the friction forces are very small, and don't effect the final velocity much. the main effect is that the propellers thrust drops to zero, preventing any more acceleration.
ice boat: constant wind force, speed limited by resistance.
dwfftw cart: decreasing propeller force, speed limited by when thrust goes to zero.
so, the ice boat/sailboat comparison is really not a good one. the situations are subtly different.
No. Not a constant wind force. It is tacking down wind
, beating the wind to its goal. The component of the force going forward diminishes as it goes faster, but it can still move with a downwind velocity component faster than the wind. The propeller cart is completely analogous, the blades going around in a circle just removes the need to change direction all the time.
cobra bubbles wrote:one thing i would like to have proponents for the dwfttw travel answer is:
what limits the final speed of the cart?
Under perfect conditions the final speed of the cart is given by the gearing ratios. It is when the thrust of the propeller equals the braking on the wheels. It could in theory be made as high as one wanted, just like the "under the ruler" video. But the faster you gear it to go, the harder it will be to overcome the real friction whci hcan never be fully avoided.
cobra bubbles wrote:the carts, by design, have small profiles, so air resistance is negligible. the gears and axles are low friction, so the forces involved there are small. rolling friction is small. static friction between the wheels and ground cannot affect the motion. these dissipative forces all waste energy, mainly through generation of heat, leaving less energy from the wind to end up as kinetic energy of the cart. a free body diagram of the cart will only have the thrust of the propeller acting in the horizontal direction that can possibly affect the motion.
Once agian you are forgetting the braking on the wheels in your free body diagram.
cobra bubbles wrote:since the only kinetic energy left to exploit for motion is the kinetic energy of the wind, the thrust of the propeller must depend on it.
then, as the cart speeds up, the relative amount of kinetic energy available decreases, since the air is moving slower relative to the cart as the carts speed goes up.
Now you again switched back to car frame. In the ground frame the kinetic eneregy of the air is constant.
cobra bubbles wrote:thus, it is natural to suppose the magnitude of the thrust goes as some power of the difference between the wind and cart speeds. it may go as the difference squared, or cube or something. (actually, i think my calcs show it goes as the cube of the difference) but in any case, as the cart goes faster, the thrust will drop, until wind speed = cart speed and thrust available from the wind decreases to zero.
No. Again the propeller is rotating. It pushes back on the air, even if that air is stagnant relative to the car.
cobra bubbles wrote:in order to have a speed faster than the wind, the thrust must then be something like T= f(w-v) + g(v) + C.
boundary conditions dictate that C=O, since when the wind is zero, and the cart is not moving, there is no thrust. i.e. the cart does not start to move from rest if wind is zero.
When the cart is at rest in no wind there is no rotation and no thrust. It obviously cant work without wind, nobody said it could. So yes, of course, C=0
cobra bubbles wrote:so T=f(w-v) + g(v). as the carts speed goes up, the f(w-v) portion goes to zero, leaving T=f(v). the thrust is then only dependent as some function of velocity, which is eventually counteracted by the resistive forces. this could allow the cart to go faster than the wind.
this is the point that i am puzzling over now. is f(v)=0 or not? my gut says it is zero, which is why i think the cart will not go faster than the wind.
i am trying to think of some mechanism by which to show what this extra component of thrust could be.
No f(v) will not be zero. Because the faster the car goes, the faster the propeller is forced to rotate because of the coupling to the wheels.
Now I think we are getting there