...Physics... =(

[Sarah]

ok, I can't figure this out, since half of ya'll are in college you can help this high schooler here. College physics is a teensy bit hard but these questions are more common sense.. and I have none.

1. Nellie Newton hangs at rest from the ends of the rope. How does her reading on the scale compare to her weight? (She's holding both end of the ropes equally.

2. What is the upper limit of the weight the strongman can lift? (pic shows a guy trying to lift a block by a rope...

3. When the brakes are applied in a speeding car, how does the direction of the fristion between the tires and the road compare with the direction of motion of the car.

4. Place a heavy book on the table and the table pushes up on the book. Why doesn't this upward push cause the book to rise from the table?

5. An empty jug of weight W rests on a table. What is the support force exerted on the jug by the table? What is the support force when water of weight w is poured into the jug?

6. Two forces act on a parachutist falling in air; weight and air drag. If the fall is steady, with no gain or loss of speed, then the parachutist is in dynamic equilibruim. How do the magnitudes of weight and air drag compare?

7. Because the Earth rotates once per 24 hrs, the west wall in your room moves in a direction toward you at a linear speed that is probably more than 1000 km/h per hour (exact speed depends on your latitude). When you stand facing the wall you are carried along at the same speed, so you don't notice it. But when you jump upward, with your feet no longer in contact with the floor, why doesn't the high speed wall slam into you?

8. If youtoss a coin straight upward while riding in a train, where does the coin land when the motion of the train uniform along a straight-line track? When the train slows while the coin is in the air? When the train is turning?

9. Consider an airplane that flies due east on a trip, then returns flying due west. Flying in one direction, the plane flies with the rotation of the Earth, and in the opposite direction, against the Earth's rotation. But if there are no wides, the times of flight are equal either way. What is this so?

Gracias, I think i know a few, but i just need to confirm them.

[suaver]

1. where's the scale? on the floor while she is hanging? no idea =T

2. however much force his huge muscles can generate? lol

3. force of friction is opposite to that of motion of the car

4. force of the table on the book = force of the book on the table; therefore, no net force because no acceleration is observed (F=ma, and since no net force, no acceleration)

5. weight of empty jub = W therefore, support force of table = -W
similarly, weight of water = w therefore, support force of table in this new condition = -(W+w)

6. parachutist is in dynamic equilibrium = no net acceleration, no net force acting on him. i'm not really sure but i'm guessing that his weight and the air drag should be the same.

7. inertia? you're going at the same speed as the earth, so even though u jump up for a second, you're still traveling at that speed along with the wall. thus, you dont slam into it (unless you decide to jump towards it)

8. hmmm, lands opposite to the motion of the train? e.g., if you toss it and the train slows down while the coin is in the air, the coin should land ahead of you.

9. i always thought it was faster if you traveled opposite the rotation of the earth....just makes your destination come to you faster lol

[kyrien]

I can't answer all of them... but I'll do the ones I know.

1. I'm not sure what the set up looks like.

2. xD;; What I'm thinking.. *points to suaver's reply*

3. Friction opposite motion of car, this is why the cars slows down.

4. Think of it this way. Imagine that your palm when facing up is a table, or a resting surface. Disregarding the way our hands shake if we just hold them out, let's say your hand is stationary in the air. When you put an object of noticeable weight in your hand, let's say, a grape fruit, your hand hand has a tendency to descend a little bit from the weight, but you won't necessarily drop the grape fruit because of its weight. The weight of the grape fruit = the downward force. You holding up the grape fruit = the upward force. Both these forces are equal, so they cancel each other out, and that is why you don't drop the fruit, and why the fruit isn't flying up (unless you purposely throw it). Now imagine a table. It exerts an upward force against what is on top of it as it tries to retain its natural state/shape. If you put a lot of stuff on the table, it will start to sag due to table elasticity to keep itself from breaking as it continues to exert the upward force. If you break the table, that means the downward force was greater than the upward force the table could maintain at the instant that it breaks.

5. Same as suaver.

6. no idea

7. same (lmao at suaver's comment xD)

8. Same logic as in 7. When sitting in a stationary train, everything is at rest relative to the earth, yes? When the train starts up, you should feel yourself lean back from the initial jerk, and as the train speeds up (assuming it does so fairly quickly), you should feel yourself pressed against the back of your seat. However, once the train stops accelerating and maintains a relatively stable speed, you will no longer feel as if pressed into your seat. What is happening, is that when the train first starts, it's as if the surface you're sitting is getting pulled forwards, so your butt moves forward while the rest of your body is stationary, at least until the forward motion propogates from your butt to your head and feet. You are actually gaining speed as the train speeds up--sort of like catching up if you will. You don't go pummelling under the train tho, because you are inside the train so the train is carrying you. Your body just has to catch up in relation to the earth.

That said... when you hold a coin in your hand in a moving train, the coin is also moving with you, and both you and the coin are stationary with respect to the train, but it is moving with respect to the earth. I'd like to add, that the moment the coin leaves your hand, the train no longer affects it until it lands back on your hand or on the floor. So when the train turns and the coin is in the air, the coin will not turn with the train because there is nothing solid to force it to turn. The coin will continue traveling in the same horizontal direction.

9. Hmm I think it's because there is always the 1000m/s (or whatever) force on the plane, the speed it was "traveling" while stationary on earth. So when the plane travels with the earth, you need to add the earth's rotation speed to the speed of the plane, otherwise the plane will lag behind. Likewise, even if the plane is traveling in the opposite direction of the earth's rotation, there is still that 1000m/s force acting on the plane in the direction of the earth's motion. This relates to you jumping up and the wall not smashing you, or you throwing up a coin in a moving train and the coin not flying backwards. The difference is that the situation with the plane is at a much (much much much) larger scale.

[suaver]

damn dude, i didnt know u were such a physics nerd =P

[Winter]

1. Her weight will be 0.
2. Uh, it seems like there should be more to that question.
3. They got it *points up*
4. They got it again *points up*
5. Same
6. I'm pretty sure Suaver got it right
7. Suaver is right. It's all about inertia. When you jump, you're still moving at the same speed you were moving when you were standing still, therefore, even though you are airborne, you are still traveling at that great speed and won't slam into the wall.
8. When the train is moving, the coin will land back in your hand (assiming you are a good shot). If the train slows, then the coin will land either on your head or at your feet (depending on which way you are facing on the train), if the train turns, the coin will land a little to the left or to the right of where you tossed it.
9. Yeah, Kyrien got it

[phlyRy]

okay...my turn
1. "She's holding both end of the ropes equally." That sentence confused me, but if I ignore that sentence, then if the scale is just below her feet, her weight is 0. Like winter said.
2. how is the guy relative to the rope and weight? also the definition of lifting it. question is a little confusing.
3. 4. 5. kyrien and suaver got it.
6. suaver got it right...equal in magnitude, opposite in direction.
7 and 8 are basically inertia questions. 7. you are still moving at the speed of the ground / wall when your feet leaves the ground. 8. since the coin is moving at that speed of the train, if the train doesn't change speed the coin will land straight down. if the train slows, the coin will land ahead in the direction the train is moving. if the train turned, then it will land to the left or right.
9. something about kyrien's answer is a little off, because describing a force in a velocity measurement is just wrong. the question itself phrases something that is trying to confuse you. basically, it is also inertia. when the earth is spinning, the plane is also moving at ground velocity. if you are looking at it from the point of view of the earth being a round object spinning, then really, the plane is moving at a circular velocity, caused by the centripital force called gravity. this velocity is also equal to the circular velocity of the ground. so these two cancel each other, and all you have is the velocity of the plane. this is a question i pondered about when i was flying overseas and wondering how come it didn't take me less hours on the plane (not hours relative to time zones). so basically, kyrien was on the right track, there is a force acting on the plane, it's call gravity, creating circular speed. but the speed and the force should not be confused as the same thing.

[Sarah]

wow you guys are freakin' geniouses (sp?) I got a few wrong after reading your guy's reply.

Does any of you guys know how to graph vector additions? I have F1 and F2, I sorta forgot how to find R...maybe if you guys know?

[phlyRy]

without changing the orientation of the arrows, shift one of the two arrows (say F2) so that the "blunt end" of it is connected to the "tip end" of the other (F1). the answer would be an arrow drawn from the blunt end of F1 to the tip of F2.

[Sarah]

ohh I get it now, What I did was that F1 and f2, I drew them separately, like the blunt end would be at (0,0), I should first either draw F2 or F1 and then draw the second one on top of that one? Right?

Edit: Okay, I am utterly confused. :confused: (I wish I had a scanner at this moment) My book states that I should use the parallelogram method of vector addition, How would I graph that accurately, or graph it for all that matters? I have my first test today and its all on this so I need to master this before 6 pm tonight.

[suaver]

F1 = --->
F2 = ^
|


so F1 + F2 = ^ B
|
A --->


so the resultant vector R would be a vector from A to B (arrowhead on B) such that it is pointing NE (because F1 is E and F2 is N, thus, R would be NE )

hmmm, the lines arent coming out like i wanted them to =T

just listen to what phlyry said....

[Sarah]

oh ok i get it now, the parallelogram method just confused me...

[suaver]

cool :thumbup:

[Sarah]

I took the test last night, I mastered the vectors portion but there was a Q where it asking "An Aluminum rod has a diameter of 5 and the mass is 1.27" Find the length." (we got to use the density table) So I know I got one question wrong already, but I'm pretty sure the other ones may be right.

[kyrien]

D = M/V
V = M/D

V = pi*r*r*length
.... so
pi*r*r*length = M/D
length = M/(D*pi*r*r)

right? D:

[Sarah]

yay, I need more help...! lol

2. What is the tangential speed of a passenger on a Ferris wheel that has a radius of 10 m and rotates once in 30 s?

3. Neglect the weight of the meterstick and consider only the two weigts hanging from its ends. One is a 1-kg weight, as shown. Where is the center of mass of this system (the point of balance)? How does your answer relate to torque?

4. A 10,000-N vehicle is stalled one-quarter the way across the bridge. Calculate the additional reaction forces that are supplied at the supports on both ends of the bridge.

6. To tighten a bolt, you push a force of 80N at the end of a wrench handle that is 0.25m from the axis of the bolt. (a) What torque are you exerting?. (b) If you move your hand inward to be only 0.10 m from the bolt, what force do you have to exert to achieve the same torque? (c) Do your answers depend on the direction of your push relative to the direction of the wrench handle?

8. If the variation in g between one's head and feet is to be less than 1/100 g, then compared to one's height, what should be the minimum radius of the space habitat?

10. How much greater is the angular momentum of the Earth orbiting about the sun than the moon orbiting about the Earth? (Set up the ratio of angular momenta using data on the inside back cover) (ME: if you need any info, tell me and I'll look in my book!)

Thanks!

[Raiden_HC]

Sorry I was away on AIM. I'll answer any questions when we're both online.

[phlyRy]

it's been forever since i did torque stuff... so here goes

2. 10m * 2pi = C, take C / 30 s = tangential speed
3. the center of mass of the system is such that at that point, the system is at an equilibrium, thus the net torque at that point should be 0. if each weight is hanging at the end and has the same weight, the center of mass is the midpoint of the meterstick.
4. let x be the length of the bridge. we know that the bridge and the truck are both in torque equilibrium. the center of mass for the bridge is the midpoint. both ends of the bridge are exterting a force, let them be A and B. the vehicle is exerting force. the torque exerted by the vehicle is 10,000*x/4 (let this direction torque be positive). one of the torque exerted by one end (the one the vehicle is further from) actually is in the same direction as the torque of the vehicle (let the force be B). the other one counters both (let this force be A). Using the fact that the net torque is 0. This gives you the formula Ax/2 +Bx/2 + 10000x/4 = 0. If we do the same using the point at the center of the vehicle, the vehicle itself would not be exerting any torque and the formula we get is Ax/4+3Bx/4 = 0. Using algebra, we can derive B and A. B = 2500N. A = -7500N.
6. a) 80N*.25m = 20Nm b) 20Nm / .10m = 200N c) yes, if you push in the opposite direction, the torque is also in the opposite direction
8. I have no idea what space habitat you are talking about
10. angular momentum is r x p. where p = mass * v. then the angular momentum of the earth is M_earth * v_earth * D, where D is the distance between the sun and the earth. the angular momentum of the moon is M_moon *v_moon * d, where d is the distance between earth and the moon. the ratio would be M_earth * v_earth * D / M_moon * v_moon * d. you can probably find the mass of the earth and the moon and their distances from the sun / earth. the only thing you got to calculate is the velocity of earth (which you convert to seconds from 365 days) and the velocity of the moon which is 28 days. then you can do the math.

[Sarah]

thanks phlyry! :)

now who tends to like geometry? lol