I have 2 more physics questions. May someone help me? I would really appreciate! Please Register or Log in to view the hidden image! 1) Artificial gravity can be created by rotation of a spacecraft. An astronaut is in a rotating space station. The station has an inside diameter of 3.0 km. Draw a free body diagram of the astronaut and find the speed required to have his apparent weight equal in magnitude to his Earth-bound weight. [This is an example question in my textbook, the free body diagram only shows one force acting on the astonaut: normal force. But how about the force of gravity? Why is it 0? And if it is not 0, how can it solve this problem?] 2) Please Register or Log in to view the hidden image! First question, why is the direction of the tension given 27 degrees below the horizontal, not 27 degrees above the horizontal? (The tension force should point AWAY from the climber, right? i.e. 27 degrees above the horizontal) Does that matter? Second question, from the diagram I can see that there seems to have 2 ropes and so 2 tension forces? Is the tension on each rope 729N or do they "sum up" to 729N? Or is there just 1 rope? Thirdly, the answer provided in my textbook is 729N [27.0 degrees above the horizontal]. I can't make the slimmest sense out of it. Shouldn't the force exerted by the cliff on the climber's feet be horizontal?
Relax, picture's there. 1) There's no real gravity. The astronaut's 'apparent weight' is the magnitude of the normal force acting on him. 2) The tension acts both ways. It pulls on both the climber and the cliff. It doesn't really change the problem whether this tension force is in one rope or distributed between 2 ropes. And the book seems to have restated the tension force by accident instead of giving the answer.
I can see it. I don't know if the force on the climber's feet has to be horizontal or not, but it's easy to see that the textbook answer can't be right. The tension in the rope points both ways. It pulls down on the cliff, and up on the climber. I'd ignore the two ropes, and assume that either (1) there's just one rope, or (2) the tensions in the ropes add up to 729N (either way amounts to the same thing). I'm not sure, but I think that to fully solve this problem, you have to do two things: 1) Do a freebody diagram for the climber, including gravity, the tension in the rope, and the force of the cliff on her feet, and set them to sum to zero. 2) Consider the torque around the climber's feet and set it to sum to zero ensure that there is no rotation force on the climber. The first is definitely necessary, but I'm not sure about the second.
1) Why would there be no gravity? Although the space station is high above the earth, there would still be force of gravity acting on the astronaut, right? Then why did the textbook ignore it?
Or is the 729N acting on each rope, i.e. total tension = 729N x2? Because I see 2 ropes in the diagram. I haven't learnt torque in my physics course so I guess I can ignore it.
Because the astronaut and spacestation are in free-fall, so it doesn't matter. They might as well be in deep, empty space. This is the same reason that we don't worry about the force of the Sun's gravity foir situations on Earth - Since we're in freefall round the Sun, so it just doesn't matter (unless we're concerned about tidal effects). Well, that could be the case, but I'd have thought they'd explicitly say so in the question.
Ha! The textbook is right! I was thinking about it the wrong way before when I said it couldn't be right. Yes, it is definitely treated as a single rope, with tension of 729N. Just the free-body diagram is enough to solve it, and the answer is the the cliff is pushing on the climber with very close to the same force magnitude and at the same angle but up and away from the cliff. Interestingly, we can tell that the coefficient of static friction between the climber's boots and the cliff must be at least 0.51, otherwise she will slip and bump face-first into the cliff! We can also tell (from torques about the rope-climber attachment point) that the rope is acting on the climber at about twice the distance from the cliff as her centre of gravity - I think this would be at about head height normally. So the picture can't be accurate - it indicates that the climber is sitting in a sling of some kind, but this can't be right. Even if she were wearing heavy boots, I don't think her center of gravity could be below about shoulder height.
The force on the climber's feet can't possibly be the same as the tension on the rope. Look at it this way: If the climber was *much* further down, the rope would be almost vertical, and then the force on the climber's feet would be almost zero while the tension on the rope would be the same (or slightly more, with a longer and therefore heavier rope). Kingwinner: Yes, the force on the climber's feet should be horizontal. Tension is along the length of the rope, and is equal in both directions.
I was surprised too. Work it out! Yes, but the climber isn't further down. Nope, doesn't balance. If the force on the climber's feet is horizontal, then there is a net downward force on the climber.
Why does Fg not matter when the space station is free falling? I mean, the gravitational force is still there...I don't get it...
3) Decribe how an airplane can make a turn when the air resistance is low, starting with Bernoulli's principle. The airplane wings are flat at bottom and curved at top, so the air is moving faster than bottom relative to the plane. Fast-moving fluid means high pressure and since force goes from high to low pressure, the lift force is created. What I observe from past experience is that when an airplane turns, one side of the wing is higher than the other. This provide a component of the lift force acting sideways, so the plane can turn because it has a component of acceleration in the sideway direction. Am I right? I would like to ask the question of why and how can an airplane wing suddenly becomes higher than the wing on the other side?
But the normal force must be horizontal because the contact surfaces between the climber's shoes and the cliff is vertical. I have tried another approach. I used 729N x 2 as the total tension acting on the climber. the vertical compoent of the total tension balances out nicely with the weight of the climber. So does that mean the normal force is indeed acting horizontally on the climber?
Friction. There is friction between the climber's boots and the cliff. If there wasn't, the climber would slip down the cliff and bang their face on the rock. Yes, if there are two ropes each with 729N tension, then the force of the cliff on the climber is horizontal. But the textbook answer indicates that there is only one rope with 729N tension or two with half that.
The pilot alters the control surfaces (flaps, rudder, elevators, ailerons) to maneuver the plane. The ailerons are the ones that make the plane roll: Please Register or Log in to view the hidden image!
Oh! I think I get it now. If the climber climbed up even *higher* she'd have to press harder against the cliff; since she's above the 45° angle she's working the wrong side of the "lever", so to speak. So the 27° angle just happens to be the one where the forces balance out. As for doing the math-- no thanks, trig isn't fun. If the sling the climber is hanging from is at her center of mass, and she's spending the energy herself to hold her legs straight, then there's no vertical force-- either up or down-- on her feet; only a horizontal one, pressing against the cliff. In reality, yeah there should be some downward force on her feet, but then in reality she shouldn't be exactly horizontal-- a climber would have her body angled with head up and feet down. I think this is being ignored to make the problem easier for the student.
The airplane gets lift from both Bernoulli's principle *and* the "angle of attack" of the wings. Your textbook will almost certainly not bring up angle of attack, but look at any airplane or bird in flight and you'll see the wings are angled with the leading edge up and the trailing edge down. Might make an interesting classroom discussion, altho it's almost certainly not the answer the textbook or your teacher is looking for. Why does banking the plane make it turn? Because the plane is lifted in the direction perpendicular to its wings. If the plane is tilted, then the force of lift is tilted in the same direction. So if the plane tilts to the right, its lift will pull it to the right. But that's not the entire answer. That would cause the plane to move right, but that would just move it sideways-- not actually *turn* it. The rudder is what actually turns the tail of the plane in one direction or the other.
