Unicycling

Originally posted by DeepThought
ΣF_x' = F_fr - W*sinΦ = m*a_x' = 0 (again, a_x' = 0; velocity of body is constant).

This line is incorrect. You are trying to balance the static frictional force with the component of weight along the slope. Consider a block on a level plain, with coefficient of friction 1. The static friction is just mg. Since there are no other forces act ...[text shortened]... eed in order to overcome gravity and the second the term to take into account frictional losses.

Thank you for the block on a flat surface explanation, I don't think about pure Physics problems like this all that often. Either I just forgot about this frictional caveat, or I never actually made the connection, and hence was easily mislead.

I believe I am following along now.

Originally posted by DeepThought
The problem is that the wrong frictional force was being assumed,

No, that is not the problem I am talking about. I am ignoring friction altogether.
It is a fact that the force required to push an object up a slope at a constant speed is proportional to the speed at which you want it pushed. If that were not the case you would be violating some fundamental laws of energy conservation.

Originally posted by joe shmo
I've riden bikes up and down things you wouldn't dare imagine...but that is besides the point.

No, it is not besides the point. If you had ridden up hills, you would know perfectly well that the force required to pedal was proportional to the speed you ride.

In the idealized (absense of mechanical friction) physical model, provided it is not accelerating (you left that bit out) is exactly...Yes
Physics is real, not some imaginary concept. I can assure you that it is not friction that causes you to sweat when cycling at 40mph up a steep hill. It is another law of physics to blame. You are creating potential energy per unit time, therefore the input energy must also be per unit time. The faster you go, the more energy you need per unit time and the stronger the force you will need.

Originally posted by joe shmo
This problem is designed to show what the minimum torque needed is in the ideal case neglecting ALL forms of mechanical friction.

That's easy. Just work out the force of gravity parallel to the slope. Anything greater than that is your required force up the slope, which you can convert to torque. How much greater determines the velocity.

Originally posted by twhitehead
No, it is not besides the point. If you had ridden up hills, you would know perfectly well that the force required to pedal was proportional to the speed you ride.

[b]In the idealized (absense of mechanical friction) physical model, provided it is not accelerating (you left that bit out) is exactly...Yes
Physics is real, not some ima ...[text shortened]... faster you go, the more energy you need per unit time and the stronger the force you will need.[/b]

Power is not force. The work done is independent of the velocity. W = F•r. I don't care what velocity you climb a hill, the Work is the same in any case. However, the Power is inherently dependent on velocity. P = d(F•r)/dt . Power and Force are fundamentally different quantities.

Originally posted by twhitehead
That's easy. Just work out the force of gravity parallel to the slope. Anything greater than that is your required force up the slope, which you can convert to torque. How much greater determines the velocity.

Your almost correct. Anything more than that determines the acceleration. It does not determine the velocity.

Originally posted by twhitehead
No, that is not the problem I am talking about. I am ignoring friction altogether.
It is a fact that the force required to push an object up a slope at a constant speed is proportional to the speed at which you want it pushed. If that were not the case you would be violating some fundamental laws of energy conservation.

This is wrong. F = ma. If the net force on an object is constant then it will accelerate. If the net force is zero it will continue at the same rate. To maintain a constant speed the net force must be zero. Since none of the forces depend on velocity (we are ignoring frictional losses) the cyclist needs to apply the same torque independently of speed. If you don't believe me go through my energy argument a page ago.

I'm wondering about your point about cycling uphill though. Initially a cyclist has to accelerate, so that they need a greater force than to maintain a steady velocity and they have to overcome the rotational inertia of the wheel (which isn't much). Ignoring friction again the deceleration caused by gravity is constant which means that at 1 m/s and assuming a gentle slope with a 1m/s^2 component of acceleration downhill a cyclist who stops pedalling will find themselves stationary after a second. If the cyclist is travelling at 10 m/s (=36 km/h ~22.5mph) then they will still have 90% of their speed after a second. So possibly it's just that it seems easier as one loses a smaller proportion of one's speed.

The only factors I can think of which would change my conclusion are the moment of inertial of the wheels or something to do with gearing. In the former case I think it's too small an effect. The gearing seems likely to be the root of the difference. I think it's just that once you're up to speed you're working in the zone where your body finds it easiest to supply power, in the same way that a car engine works best at some range of engine speeds. So I'm thinking the reason for the effect you're describing is due to physiology rather than physics.

Originally posted by DeepThought
So I'm thinking the reason for the effect you're describing is due to physiology rather than physics.

No. I am thinking my understanding of force may be wrong. One needs to apply the same force to the pedals, but because they are rotating faster, it takes more energy overall.
So the torque will be the same even if you are going down the slope at a constant speed and using torque for braking, or even if you are stationary.

Originally posted by twhitehead
No. I am thinking my understanding of force may be wrong. One needs to apply the same force to the pedals, but because they are rotating faster, it takes more energy overall.
So the torque will be the same even if you are going down the slope at a constant speed and using torque for braking, or even if you are stationary.

Let's power the bike with a motor and give the poor cyclist a rest - this way we don't need to worry about physiology. To travel some distance L at constant speed the cyclist has to supply a torque T and turn the pedals a fixed number of times, suppose θ ∈ [0, ∞ ) is the angle the wheel has turned through and L = θa, where a is the radius of the wheel. The total energy one has to supply is Tθ = TL/a and just depends on the mass of the bike and rider and slope of the incline via T/a and the total distance travelled which is L. It is independent of speed. The power one has to supply is P = Tθ/t (t being the time for the journey) = T(L/t)/a = Tv/a. Where v is the velocity of the bike. So one has to deliver the same energy but more power. The journey time is proportionately shorter for the faster cyclist. An ideal engine (one whose efficiency is always 100% whatever power it's asked to supply - it's electric before anyone starts complaining about thermodynamics) would deliver the same amount of energy for the same trip no matter how fast it had to go, humans on the other hand are kind of lossy.

Originally posted by joe shmo
Yes, that is what I am saying. The body is not being accelerated, the net force acting on the body is zero.

If you are lifting something straight up, you are accelerating it, against gravitational acceleration so there has to be a force action on the body to continue dragging it uphill against gravity.

Originally posted by sonhouse
If you are lifting something straight up, you are accelerating it, against gravitational acceleration so there has to be a force action on the body to continue dragging it uphill against gravity.

It's moving at constant speed, there is no acceleration.

Originally posted by sonhouse
If you are lifting something straight up, you are accelerating it, against gravitational acceleration so there has to be a force action on the body to continue dragging it uphill against gravity.

You are confusing the concepts of "Force" and "Net Force". In the scenario your questioning there is a "Force" acting on the body ( actually two). However, the body is moving up with constant velocity ( i.e. It is not accelerating). Consequently, the two forces acting on the body are equal and opposite by Newtons Second Law. Which states in this case: F_net = m*a = 0

Originally posted by joe shmo
You are confusing the concepts of "Force" and "Net Force". In the scenario your questioning there is a "Force" acting on the body ( actually two). However, the body is moving up with constant velocity ( i.e. It is not accelerating). Consequently, the two forces acting on the body are equal and opposite by Newtons Second Law. Which states in this case: F_net = m*a = 0

Well, lets take the case of a rocket accelerating a mass from zero relative velocity. 1 G of acceleration is the same as lifting a body up off ground level, it is the law of equivalence that Einstein proposed, you can't tell the difference between a rocket accelerating you in space or a vertical lift on Earth. Both are in an accelerating environment. So F=MA and pulling up against Earth is just negative acceleration equivalence.

Originally posted by sonhouse
Well, lets take the case of a rocket accelerating a mass from zero relative velocity. 1 G of acceleration is the same as lifting a body up off ground level, it is the law of equivalence that Einstein proposed, you can't tell the difference between a rocket accelerating you in space or a vertical lift on Earth. Both are in an accelerating environment. So F=MA and pulling up against Earth is just negative acceleration equivalence.

Firstly, This is not the same scenario. The rocket is accelerating at 1g. The Net Force on the rocket in your scenario is greater than zero. The velocity of the rocket is constantly changing explicitly because of this condition. The velocity of the box in the previous scenario was not changing at all...it had no acceleration. There are forces doing the work in both cases. In the case of the rocket the net force is greater than zero, and in the case of the box the net force was equal to zero.

Originally posted by joe shmo
Firstly, This is not the same scenario. The rocket is accelerating at 1g. The [b]Net Force on the rocket in your scenario is greater than zero. The velocity of the rocket is constantly changing explicitly because of this condition. The velocity of the box in the previous scenario was not changing at all...it had no acceleration. There ...[text shortened]... the net force is greater than zero, and in the case of the box the net force was equal to zero.[/b]

The only thing holding back the box from being accelerated is the ground in the way. If it were a hundred meters in midair it would be accelerating so it has at least potential kinetic energy.