Originally posted by RamnedA) The repulsive force decreases, and the upper wire falls back down and probably to the side.
follow-ups for 19...
(A) If it is displaced upward on case 1, what happens to the repulsive force and the wire?
(B) In case 2, what happens to the force and the wire if it is displaced downwards?
B) The attraction decreases until the lower wire is no longer under enough force to levitate it.
Originally posted by AThousandYoungB) The wire drops immedeately. Any disturbance downward destroys the equilibrium.
A) The repulsive force decreases, and the upper wire falls back down and probably to the side.
B) The attraction decreases until the lower wire is no longer under enough force to levitate it.
Originally posted by mtthwHaha...
It's done with wires!
I've discovered my initial theories were wrong after reading some of the other posts, so I'll revise:
A: The wires attract each other
In this case, the top wire must be held in place otherwise they both fall. In the equilibrium position, the bottom wire would be perfectly balanced (gravity vs. attraction) directly underneath the top wire.
Effect of perturbations
(i) up: moving the bottom wire up will increase the attractive force, causing it to collide with the top wire - this new arrangement is stable, but it's not the same as the original equilibrium position, so the original equilibrium position is not stable;
(ii) down: moving the bottom wire down will decrease the attractive force, and gravity will pull it down - not stable;
(iii) sideways: moving the bottom wire sideways will reduce the attractive force due to the increase in distance, and induce it to move back so it's directly under the top wire - not stable.
B: The wires repel each other
In this case, the bottom wire must be held otherwise they both fall. In the equilibrium position, the top wire would be perfectly balanced directly on top of the bottom wire.
Effect of perturbations
(i) up: moving the top wire up will decrease the repulsive force, causing it to fall when released, however as it falls the repulsive force will cause it to bounce back up and repeat the cycle again - stable;
(ii) down: moving the top wire down will increase the repulsive force, causing it to bounce back up when released, however as it gets the higher the repulsive force will decrease, allowing it to fall back down and repeat the cycle again - stable;
(iii) sideways: moving the top wire sideways will decrease the repulsive force to decrease due to the increase in distance, but will also cause the top wire to be pushed away from the bottom wire - not stable.
Now, if at least one type of perturbation causes the system to become unstable, then the whole system is unstable. Therefore, it seems clear (now) that neither system is stable.
Originally posted by AThousandYoungCorrect . Next one to follow, last conceptual of this set! Then the second MONSTER problem 😉
A) The repulsive force decreases, and the upper wire falls back down and probably to the side.
B) The attraction decreases until the lower wire is no longer under enough force to levitate it.
Originally posted by HolyTGood. I'm okay with you using resources - it means I'm getting them hard enough!
I didn't remember the formula from so many years back and had to look it up. Is that cheating? Stored energy E = 0.5 * L [inductance] * I [current]. So unless I'm really missing something here, a doubling of current results in a quadrupling of stored energy.
Well, I'll cook up a difficult problem based on the ten questions I just asked. Good Luck.
#10-20 Challenger
One possible means of achieving space flight is to place a perfectly reflecting aluminized sheet into Earth's orbit and to use the light from the Sun to push this solar sail. Suppose such a sail, of area 6.00 X 10^4 m^2 and mass 6,000 kg is placed in orbit facing the sun. (A) What force is exerted on the sail? (B) What is the sail's acceleration? (C) How long does it take for this sail to reach the moon, 3.84 X 10^8 m away? Ignore all gravitational effects, and assume a solar intensity of 1,340 W / m^2.
Originally posted by RamnedWithout gravitational effects, there's not much of an orbit 😉
#10-20 Challenger
[b]One possible means of achieving space flight is to place a perfectly reflecting aluminized sheet into Earth's orbit and to use the light from the Sun to push this solar sail. Suppose such a sail, of area 6.00 X 10^4 m^2 and mass 6,000 kg is placed in orbit facing the sun. (A) What force is exerted on the sail? (B) What is the sa ...[text shortened]... 8 m away? Ignore all gravitational effects, and assume a solar intensity of 1,340 W / m^2.[/b]
Assume the sail starts from rest and gets no orbital boost or detriment (no gravity effects) and travels in a straight line with no other forces acting on it. Assume it reflects perfectly and remains perpendicular to the direction to the sun.
Without deriving everything:
Let S be power per unit area
c = speed of light = 2.99792 x 10^8 m/s
Pressure on a solar sail = S/c (that’s just for absorbing the photons)
If all energy reflected, then total pressure = 2S/c (absorbing the photons gives the sail energy, then perfectly reflecting them at the same speed gives the same amount of energy again)
Pressure = Force[F]/Area[A]
So F/A = 2S/c
F = 2SA/c
Part A: F = 2*(1340 W/m^2) * (6.00 x 10^4 m^2) / (2.99792 x 10^8 m/s)
F = 0.53637 N, rounding to 0.536 N with the correct number of significant digits.
Part B: F = ma (Newton’s 2nd law); a = F/m
a = 0.53637 N / 6000 kg = 8.93953 x 10^-5 m/s/s, or 0.0000894 m/s/s with the correct number of significant digits. (m/s/s is meters per second per second, which is the same as meters per second squared.)
Part C:
Let time = t
Integrate acceleration twice to get the distance equation, with a starting velocity of 0:
Distance traveled = 0.5 * a * t^2
t = square root of (2s/a)
t = sqrt [(2 * 3.84 x 10^8 m)/(0.0000893953 m/s/s)]
t = 2,930,000 seconds with the correct number of significant digits, or about 33 days, 22 hours, and 11 minutes.