Reading quiz: In the SI (metric) system, the unit of force is the.. A) galileo B) joule C) newton D) eiffel E) none of these

Transcription

1 Reading quiz: In the SI (metric) system, the unit of force is the.. A) galileo B) joule C) newton D) eiffel E) none of these

2 5-1: An astronaut floating weightlessly in orbit shakes a large iron anvil rapidly back and forth. She reports back to Earth that A: the shaking costs her no effort because the anvil has no inertial mass in space. B: the shaking costs her some effort but considerably less than on Earth. C: although weightless, the inertial mass of the anvil is the same as on Earth. Answer: C. Just as it says! Mass is the same wherever you are. It's a measure of inertia

3 5-2: A constant force is exerted on a cart (initially at rest) on an air track. Neglect friction. The force acts for a short time and gives the cart a certain final speed. To reach the same final speed with a force that is only half as big, the force must be exerted on the cart for a time interval A: four times as long as B: twice as long as C: equal to D: half as long as... that for the stronger force. OR E: (Not enough information given) Answer: B. Force causes acceleration = delta v/delta t. If you want the SAME delta v, with HALF the force, it'll take TWICE the delta t.

4 5-3: A constant force is exerted for a short time on a cart (initially at rest) on an air track. This force gives the cart a certain final speed. The same force is exerted for the same length of time on another cart, also initially at rest, that has twice the mass of the first one. The final speed of the heavier cart is A: one-fourth B: four times C: half D: double E: the same as...that of the lighter cart. Answer: F = m delta v / delta t... If you have the same force, and the same time, but TWICE the mass, you'll get HALF the change in velocity. Since we started from rest, it means v(final) is half.

5 5-4: A constant force is exerted for a short time on a cart that is initially at rest on an air track. This force gives the cart a certain final speed. We repeat the exp't but, instead of starting from rest, the cart is already moving with constant speed in the direction of the force at the moment we applied the force. After we exert the same constant force for the same short time interval, the increase in the cart s speed is A: two times its initial speed. B: the square of its initial speed. C: four times its initial speed. D: the same as when it started from rest. E: cannot be determined from the info provided. Answer: Newton's law says F = m Delta v / Delta t. Same force, same time, same mass means delta v is always the same. The CHANGE, the INCREASE is always the same. Answer is D.

6 5-5: A constant rightward force (F) is exerted for a short time interval (t) on a cart that is initially at rest on an air track. This gives the cart a certain final velocity, vf. Suppose we begin with the cart moving with the velocity vf above, and then we apply a leftward force (-F) for the same short time interval (t). What will the cart be doing after time t? A: Moves forward with 2* initial speed. B: Moves forward with initial speed. C: Final speed is 0. D: Cannot be determined from the info provided. Answer: Negative force means delta v is negative. We went from 0 to vf before, so we'll go from vf to 0 now.

7 5.5x1 Three people are pulling on a ring in a "2-D" tug of war. Shown is a "top view". No one is winning. 100 N 100 N If the pulls are configured as shown, (teams 1 and 2 are each pulling with a force of 100 N) How hard is team 3 pulling? 45 0 T3? 45 0 A) 100 N B) 200 N C) 141 N D) 71 N E) 0 N Hint: sin(45) = cos(45) =.707 If they are pulling harder, why aren't they winning? Answer: If no one is winning, that must mean Fnet on the ring = 0. (It's not accelerating) We have to add the three vectors, and they add to zero. T1+T2+T3 = 0, Or, in other words, T3 = -(T1+T2). That's a vector equation. So, the X components must match, AND the y components. The x components are kind of easy: T3 has zero, T1 and T2 each have x components, but by inspection they are equal and opposite, so cancel to zero. Nothing learned there. T1 and T2 each have y components, both are 100 N * sin(45) = 70.7 N. These ADD. So T3(y) = - 2(70.7) = -141 N. Thus, T3(y) = (0, -141 N), which has magnitude 141.

8 5-6: An object is lowered by a rope at a constant speed. The net force on the object is A: upward B downward C: zero D: not enough information given. (v is down, and constant) b) What is the direction of the acceleration of the object? A: up B: down C: a=0 c) How does the tension T in the string compare with the weight "mg" of the object? A: T=mg B: T > mg C: T < mg D: Not enough information given. Answer: If v is constant, the acceleration is ZERO. That means, from N-II, ZERO net force. Which means T = mg. Remember, Net force doesn't cause velocity. It causes CHANGE in velocity!

9 5-7: An object is being lowered on a cord at a speed which is decreasing. There are two forces on the object, the weight (magnitude mg) and the tension, magnitude T, in the cord. Which equation is true: A: T = mg B: T > mg C: T < mg (v is down, and decreasing) T mg HINT: What is the direction of the acceleration? A: up B: down C: a=0 Answer: If it's moving down, but SLOWING, that means acceleration is UP. (Convince yourself, don't just take my word for it!) If the acceleration is up, the net force must be upwards, since F=ma. If the net force is upwards, T > mg (look at the picture)

10 5-8: A coffee filter floats gently downward, at some constant ("terminal") velocity v. v The net force on the coffee filter is A: Upwards B: Downwards C: Zero. Answer:: N-II says F=ma. If you have a constant v, your acceleration is zero. So F(net)=0. There are TWO forces: mg downward, and friction upwards. Apparently, they must BALANCE when the coffee filter gently floats!

11 5-9: You are a passenger in a car and not wearing your seat belt. Without increasing or decreasing its speed, the car makes a sharp left turn, and you find yourself colliding with the right-hand door. Which is the correct analysis of the situation? A: During the turn, there is a rightward force pushing you into the door. B: During the turn, the door exerts a leftward force on you. C: both of the above D: neither of the above Answer: B is correct. There is no rightward force on you. You tend to follow a straight line. As the car swerves left, it must push you left in order that you stay with the car, moving leftwards, instead of going straight.

12 5-10: Consider a horse pulling a buggy. Is the following statement true? "The weight of the horse and the normal force exerted by the ground on the horse constitute an interaction pair that are always equal and opposite according to Newton s third law." A: yes B: no Answer: B, no. The forces ARE equal and opposite, in this case, but it has nothing to do with N-III. The textbook explains this in some detail. If you label forces as F ON a BY b, the weight of the horse is F ON horse BY planet earth, but the normal force is the F ON horse BY ground. The labels tell the story: N-III ONLY tells you that F ON a BY b = - F ON b BY a (nothing else!)

13 5-11: A locomotive engine pulls a series of train cars. Which is the correct analysis of the situation? A: The train moves forward because the locomotive pulls forward slightly harder on the cars than the cars pull backward on the locomotive. B: The locomotive's force on the cars equals the force of the cars on the locomotive, but the frictional force on the locomotive is forward and large, while the backward frictional force on the cars is smaller. Answer: A is false, it violates N-III. B is correct. It is friction on the wheels that drives an engine forward. The NET force on the locomotive is not zero (the NET force on the cars is also not zero). My lecture notes explain this in detail in the "horse and cart" story.

14 5-12: A giant, extremely massive moon is headed straight for an inhabited planet. The Enterprise has decided to "tractor" the moon out of the planet's path. The Capt. orders the tractor beam on at maximum strength, (but forgets to ALSO order their impulse engines to be turned on) What happens? A: The moon is tractored out of the planet's path, slowly but surely, and the Enterprise saves the day once again. B: The Enterprise accelerates rapidly towards the moon and crashes into it, killing all aboard and ending the series. Answer: Alas, poor Enterprise, even in the 23rd century, Newton's laws will still hold, despite all the cool technology they develope. The answer is B.

15 5-13: Consider a car at rest. We can conclude that the downward gravitational pull of Earth on the car and the upward contact force of Earth on it are equal and opposite because A: the two forces form an "action reaction" pair. B: the net force on the car is zero. C: neither of the above Answer: B. Think about it! (Look at the problem earlier about the up and down forces on a horse)

16 5-14: Steve and a Sumo wrestler are having a tug-of-war. So far, no one is winning. What is the direction of the force of friction on Steve's feet (Fs)? A: left B: right Answer: A. Net force on me is zero (no one is winning, I'm not moving, F(net)=0. There's clearly a rightward force from the rope. So friction must be to the left. If there WAS no friction, picture it - I'd move rightwards, wouldn't I?

17 5-15: Steve and a Sumo wrestler are having a tug-of-war. So far, no one is winning. How does Fs (the magnitude of the force of friction on Steve's feet) compare with Fw (the force of friction on the feet of the sumo wrestler) A: Fs > Fw B: Fs=Fw C: Fs < Fw. The forces are equal in magnitude. The free-body diagrams for Skinny and Fatty: F S F Rope on S F Rope on F F F Skinney Fatty FS = FR on S, FF = FR on F, FR on S = FR on F F S = F F

18 5-16: In the 17th century, Otto von Güricke, a physicist in Magdeburg, fitted two hollow bronze hemispheres together and removed the air from the resulting sphere with a pump. Two eight-horse teams could not pull the halves apart even though the hemispheres fell apart when air was readmitted. Suppose von Güricke had tied both teams of horses to one side and bolted the other side to a heavy tree trunk. In this case, the tension on the hemispheres would be A: twice B: exactly the same as C: half what it was before. Answer: Draw a force diagram! Let's call the force of an 8-horse team "F". Before, you had "F" to the left, and "F" to the right. They balanced, but there was a tension in the system. Now, you put both teams on one side, which means you have "2F" on that side. The other side is connected to a tree. Assuming the tree doesn't budge, there must be "2F" acting the opposite direction. Once again, these forces balance (the sphere doesn't accelerate), but the TENSION is twice as much as before.

19 5-17: Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the person is A: larger than B: identical to C: smaller than the downward weight W of the person. Answer: Larger. There must be a net force on the person, because they accelerate upwards. The only forces acting on the person are "N" up, and "W" down. If the vector sum is "up", then N must be larger.

20 5-18: Consider a person standing in an elevator that is moving upward with constant velocity. The upward normal force N exerted by the elevator floor on the person is A: larger than B: identical to C: smaller than the downward weight W of the person. Answer: the same as.. Since v = const, a = 0. Since a = 0, F net = 0. Since F net = 0, the upward force N must cancel the downward force mg. Newton rules! Aristotle is bogus! Just because you are moving upward does not mean there is an upward net force.

21 5-19: How does the force exerted on the cart by the string (T) compare with the weight of body B? (Assume all surfaces are frictionless) ma mb A: T = mb g. B: T < mb g. C: T > mb g. Answer: I hope it's reasonable to you that the system WILL accelerate, Ma will accelerate to the right, Mb will accelerate down (at the same rate, they're tied together!) So look at Mb: it accelerates down, so the net force on it is DOWN (by N-II). There are only 2 forces acting on it: mb*g down, and T up. If the NET force on Mb is down, then mb*g must "win". B is the correct answer. If you think mb*g = T, then that means you think there is ZERO net force on Mb. Which means it would just sit there... but there's no friction, so it really will be pulled down, and start to accelerate.

22 5-20: An Atwood's machine is a pulley with two masses connected by a string as shown. HERE, The mass of A is twice the mass of B. How does the force exerted on the mass B by the string (T) compare with the weight of body B? (Assume a frictionless, massless pulley) A: T = mb g. B: T < mb g. A B C: T > mb g. D: not enough information Answer: Since the heavier mass A accelerates downward there must be a net force acting on it. In order for there to be a net force downward on A, we must have m A g > T. By similar reasoning with B, we must have T > m B g. The tensions are the same on both sides of the pulley by Newton's third law (assuming that the string is massless). T T a A B a m A g m B g Suppose I now tell you mb is larger than ma. What is your new answer? Answer:: I'll let you work it out!

23 5-21: A piano mover raises a 100- kg piano at a constant rate using the frictionless pulley system shown here. With how much force is she pulling on the rope? Ignore friction, let g = 10 m/s^2. A: 2,000 N B: 1,000 N C: 500 N D: none of the above E: impossible to determine Answer: Draw force diagram for the piano. There are TWO ropes pulling up. If there's a tension "T" in the rope, it'll be everywhere. SO we have 2T up, and 100kg*g = 1000 N down. These balance, so 2T = 1000, or T = 500. That's the same as the force the woman applies.

24 5-22: Two blocks, with masses M2>M1, are connected by ropes. You pull to the right on a second rope, with external force "T1". The blocks sit on a frictionless surface. M2 T2 M1 T1 How do the magnitudes of the tensions in the rope compare? A: T2 > T1 B: T2 < T1 C: T2 = T1 Answer: Lots of ways to think about this. Here's my favorite: M1 and M2 each must accelerate the SAME ('cause they're tied together). The net force on M2 is M2*a. That's supplied by T2. So, T2 = M2*a. Now consider "M1+M2" as a system. That SYSTEM has mass M1+M2, and acceleration a. The only external force on it is T1. So T1 = (M1+M2)*a This is clearly bigger than M2*a. SO T1 > T2. (T1 has to accelerate everything, T2 only has to accelerate M2) This is a subtle one. You need to think it through for yourself! There will be a tutorial on this, it's not trivial at first!)

25 5-23: A glider is pulled along an air track with a string at an angle (theta) from the horizontal, with a constant force. F q The direction of the net force on the glider is A: Horizontal. B: At an angle theta above the horizontal. C: At some angle above the horizontal, but not necessarily theta. Answer: A. Net force = ma, that's Newton's law. "Air track" implies no friction, so this glider is clearly going to be accelerating. Assuming it stays on the track (implied by "pulled along an air track"), it must be accelerating to the right. So by N-II, Net force is horizontal! Apparently, the vertical component of F, plus N (normal force), minus mg, must all cancel - the net result is OBSERVED, the glider is moving to the right, so the net force must be to the right!

26 5-24: A mass m is pulled along a frictionless table by constant force external force F EXT at some angle above the horizontal. (The magnitudes of the forces on the freebody diagram have not been drawn carefully, but the directions are correct.) Fext N mg Which statement below must be true? Fext A: mg > N B: N > mg C: N=mg Answer: See the last problem. If N was equal to (or worse still, greater than) mg, then the "up" forces could not possibly cancel out. (Because Fext ALSO adds in a little upwards force) Yet we know they must... because the mass moves ALONG the table, horizontally, which means there is no NET force in the up or down direction. So it must be the case that N < mg.

27 5-25: A mass m is accelerated down along a frictionless inclined plane. The magnitudes of the forces on the freebody diagram have not been drawn carefully, but the directions of the forces are correct. a N mg Which statement below must be true? A: mg > N B: N > mg C: N=mg Answer: Pick your "axes" to be tilted, so that "x" points down the ramp, and "y" is perpendicular to the ramp. We *observe* an acceleration in the pure x direction. So the NET force in the y direction better be zero - there's no acceleration perpendicular to the ramp! We have +N in the "y" direction, and opposing that is not mg, but a PART of mg. So N must balance a PART of mg, which means mg itself must be LARGER than N. Tricky! Think about it yourself!