General strategy for using Newton's second law to solve problems:

Transcription

1 Chapter 4B: Applications of Newton's Laws Tuesday, September 17, :00 PM General strategy for using Newton's second law to solve problems: 1. Draw a diagram; select a coördinate system 2. Identify relevant objects/agents 3. Sketch an interaction scheme 4. Draw a free-body diagram, using the coördinate system chosen in Step 1 5. Apply Newton's second law to each relevant object, in each direction specified by the coördinate system chosen in Step 1 Example: Static equilibrium Determine the tension in the string. Solution: Ch4B Page 1

2 Example: Static equilibrium Calculate the tension in the strings if (assume the strings have no mass) a. the labelled angles are both 30 b. the labelled angles are both 10 c. one labelled angle is 40 and the other labelled angle is 60 Ch4B Page 2

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5 We mentioned in class that you can test the previous example for yourself by hanging a heavy picture frame that has a wire attached to the back. If you pull the wire extremely taut, the tension in the wire when it is hanging will be extremely high. Leaving some slack in the wire means that once the picture hangs from the hook, the angles of the wire relative to the horizontal will be a little larger, and so the tension will decrease. We also mentioned that the same idea can be used to pull a car out of a ditch; you just have to attach one end of a rope to the car, attach the other end to an object that will not move (very sturdy tree, for example), and then tighten the rope as much as you can. Pressing on the middle of the rope sideways will exert a very large tension in it, which will pull the car out a little bit. Then tighten the rope and repeat. (Caution: I've never actually pulled a car out of a ditch like this, so one should try this out with less massive objects before trying it with a real car. I can see the tree being uprooted, as the tension will be very high!) Check out the following link: Mass and Weight Apparently we don't "feel" gravitational forces directly; what we do feel is contact forces. This explains why we feel weightless when falling, and why astronauts feel weightless when they are in orbit around the Earth (in "free fall"). When you stand on your bathroom scale, you "feel" the normal force exerted by the scale on you; this is what the scale also measures. You feel a similar normal force when you sit on a chair. If you are just standing on your bathroom scale, the normal force from the scale balances your weight (i.e., the gravitational force that the Earth exerts on you), which is why the scale reading (which measures this normal force) equals your weight. Make sure you understand the difference between mass and weight; your mass is a measure of the substances that you consist of, and is independent of your location. Your weight is a measure of the gravitational force that the Earth exerts on you; although it's proportional to your mass, it is not the same as your mass (W = mg). If you were on the Moon, your weight would be different, because what you feel as your weight would be the normal force that the surface of the Moon exerts on you, which is equal to the gravitational force that the Moon exerts on you. When you are accelerating, your apparent weight (i.e., the normal force Ch4B Page 5

6 exerted on you by the surface you're standing on) depends on your acceleration. Example: Apparent weight Calculate the apparent weight of a 70 kg person in an elevator that is a. accelerating upward at 4 m/s 2. b. moving upward at a constant speed of 5 m/s. c. accelerating downward at 2 m/s 2. Ch4B Page 6

7 Thus, if the cable snaps, then the normal force becomes zero, which means the apparent weight of the passenger becomes zero; the passenger feels weightless. Friction You may have noticed that it is difficult to push a refrigerator. The reason for this is that the bottom of the refrigerator is rough, and so is the floor. You may also have noticed that it's more difficult to move the refrigerator over a carpet than over a smooth floor, because the carpet is rougher than the floor. Thus, the frictional force between two surfaces depends on the surfaces. This dependence is so complicated that we don't have any good theory for predicting the friction between two surfaces; the best we can do is just measure the frictional forces in experiments. Further reflection will reveal other facts about friction. You may have noticed that when you begin to push on the refrigerator, it doesn't move. However, as you gradually increase your pushing force, eventually the refrigerator will move. Once the refrigerator is moving, it takes less pushing force to keep it moving than the force needed to get it moving in the first place. Conclusion: the coefficient of static friction is greater than the coefficient of kinetic friction. Check out the following link: Furthermore, if you stack another heavy object on the refrigerator, it will be harder to move it. Similarly, if someone presses down on the refrigerator, that also makes it harder to move. Thus, frictional forces depend on the normal force acting between the two surfaces in contact. Conclusions: f s µ s n f k = µ k n Typically, the coefficient of kinetic friction is less than the coefficient of static friction for the same pair of surfaces; the coefficient of rolling friction is much less than the coefficient of static friction for the same surfaces. Ch4B Page 7

8 µ r << µ k < µ s Sample coefficients of friction are found on Page 101 of your textbook. For example, for rubber on concrete, approximate typical values are µ s = 1.00, µ k = 0.80, µ r = 0.02 Example: A refrigerator of mass 100 kg is resting on a horizontal floor. The coefficients of friction between the refrigerator and the floor are µ s = 1.00 and µ k = Determine a. the force needed to get the refrigerator moving. b. the force needed to keep the refrigerator moving at a constant speed once it has already started moving. Ch4B Page 8

9 In the previous problem, what happens if the original 980 N applied force is maintained once the refrigerator is moving? That is, the kinetic friction force opposing the motion is 715 N, but the applied force is greater than the frictional force, so the net horizontal force on the refrigerator is = 265 N. Thus, the refrigerator accelerates to the right. Ch4B Page 9

10 Motion on a Ramp Example: Consider a block of mass 5 kg that is sitting at rest on a ramp inclined at an angle of 30 degrees relative to the horizontal. a. Determine the acceleration of the block down the ramp if there is no friction between the block and the ramp. b. Determine the minimum coefficient of static friction needed to keep the block at rest. Solution: (a) Draw a diagram, and then draw a free-body diagram of the block. Now choose a coördinate system wisely, and separate any forces that are not on the axes into components: Ch4B Page 10

11 There are various ways to separate the weight vector mg into components; the following diagram hints at this (for example, you could use the black pair of components, or the blue pair, or one of each): Now write down Newton's second law of motion in the y-direction: Now write down Newton's second law of motion in the x-direction: Equation 2 is enough to solve part (a) of this problem, as the only unknown is the acceleration. Equation 1 is not relevant for this problem; it gives us the strength of the normal force, which may be interesting, but we weren't specifically asked about this. Solving equation 2, we get: Note that the result is independent of the mass of the block. Is this reasonable? (b) Now draw a new free-body diagram, including the frictional force. Ch4B Page 11

12 Once again, a wise choice of coördinate system is helpful: Once again, write down Newton's second law for each component; first in the y-direction: Newton's second law in the x-direction is: We have two equations for two unknowns, which is fine, but we were asked about the coefficient of static friction, which doesn't appear in our equations so far. OK, so we need to include the connection between the normal force and the friction force, which will bring in the coefficient of static friction: Now solve equation 1 for the normal force, solve equation 2 for the frictional force, and substitute the values into equation Ch4B Page 12

13 3 to determine the coefficient of static friction: Thus, the minimum value of the coefficient of static friction between the block and the ramp to keep the block from moving is It's interesting that the value is independent of the mass of the block, and also independent of the acceleration due to gravity. This means that the same values would solve the problem on the Moon or some other planet. A related problem, that we'll examine later involves banking of curves on highways. In the absence of friction (wet or icy conditions), banking the road helps to provide the turning force that friction would otherwise provide. It turns out that the best banking angle is also independent of the mass, which is great, because once you get the banking angle ideal, the same angle is ideal for transport trucks, motorcycles, and everything in between. The following material on air drag and terminal speed is OPTIONAL, and will NOT be included in tests and the final exam for this course. Air drag very complicated, much like friction; we don't have good theories of air drag, so we can't predict very well the amount of air drag that will be present in some situation; the best we can do is to experiment, sometimes with scale models (wind tunnels, for example, are used for experimentation) As an approximation, that is reasonably good in certain circumstances, the force of air drag (i.e., air resistance) in newtons is about: D = 0.3Av 2 Ch4B Page 13

14 where A is the cross-sectional area of the object that is moving through the air (in square metres), and the speed is measured in m/s. Notice the quadratic dependence on speed; this means if the speed doubles, the air drag quadruples. This makes it very difficult indeed to get things moving through the air at high speeds. A lot of engineering design work goes into reducing air drag by designing shapes and using materials for objects that will reduce the coefficient "0.3" to a smaller number. A more detailed formula is D = 0.25ρAv 2 where ρ represents the density of air; a typical value for the density of air at sea level is ρ = 1.22 kg/m 3. Remember that these formulas are very rough approximations, and experts in various fields will have their own better approximations that deal specifically with their own situations. A consequence of air drag is terminal speed. If you drop an object from rest, the gravitational force on the object is approximately constant for the entire fall, but the air drag starts off very small (because the speed is small) and then increases rapidly with increasing speed. At some point, if the object falls for long enough, the air drag will grow so large that it balances the gravitational force. After this point, the net force acting on the falling object will be zero, so the object's speed will be constant; this is called the object's terminal speed. Example: Terminal speed 1. Calculate the terminal speed of a human being of mass 70 kg. 2. Calculate the area of a parachute needed to reduce the terminal speed of a human being of mass 70 kg to (i) 10 km/h, and (ii) 5 km/h. Assume that the person falls feet-first, so that we can estimate his or her cross section as approximately 0.1 m 2. (We can probably stand on a 1-foot square kitchen tile without "sticking out", right?) Thus, the drag force on the person is about Ch4B Page 14

15 When the drag force is equal to the force of gravity on the person, then the net force on the person is zero, and the person's speed is constant. Thus, the terminal speed satisfies This is very fast. You can minimize the terminal speed by increasing the crosssectional area, which increases the drag force. One way to do this is to fall "belly-flop" style instead of feet-first. Another way is to use a parachute, which is explored in Part (b). Remember that such calculations are only approximate. The actual drag force depends on many factors that are not included in the formula, such as the shape of the object, the material on the surface of the object, and so on. Ch4B Page 15

16 Do the results seem reasonable? Not really, because typical sky-diving parachutes have cross-sectional areas in the range of 10 m 2 to 50 m 2. Evidently, sky divers hit the ground at speeds significantly greater than what we assumed. Suppose we take a typical parachute cross-sectional area of 50 m 2. Here is the resulting terminal speed: This seems awfully fast. Would you want to be hitting the ground at this speed? We can compare this terminal speed with speeds attained by jumping from Ch4B Page 16

17 various heights. Or, better yet, let's use the kinematics equations to determine the height from which you would have to drop a ball so that it hits the ground at a speed of 6.8 m/s (assuming no air resistance): This seems reasonable, as the height is less than the height of a typical one-storey roof. I've seen athletic teenagers jump off one-storey roofs with no ill effects (I'd probably break my legs), so jumping from a lesser height is probably safe. It's reasonable to expect a parachute to slow a novice sky-diver to similar speeds. (End of optional material.) Sliding blocks Imagine pushing a heavy refrigerator across the floor. If there is a fragile sculpture, made of thin glass, on the far side of the fridge, it's conceivable that the sculpture might slide along, being pushed by the fridge, without being damaged. However, if you pushed from the other direction, the sculpture would surely crumble as you pushed it against the fridge. We can understand this situation in the context of Newton's laws as follows. Example: Calculate the acceleration of the system and the magnitudes of the inter-block forces if a. a 50 N force acts on the left-most block towards the right. b. a 50 N force acts on the right-most block towards the left. Assume there is no friction between the blocks and the surface they rest on. Ch4B Page 17

18 Solution: (a) Sketch a free-body diagram for each block. Notice that we have five unknowns, so we need five independent equations to be able to solve this problem. Applying Newton's second law to each component direction for each block produces 4 equations; where are we going to get one more equation? After thinking for a while, you may realize that the two interblock forces have equal magnitudes, according to Newton's third law: OK, let's write down Newton's second law for each component direction of each block. (Note that because the blocks move together, they have the same acceleration, so we can use one single symbol to represent the acceleration of each block.) Ch4B Page 18

19 Combining equations 1 and 5 gives us and substituting the result into equation 3, and solving for a, gives us: Using equation 5, we can calculate the inter-block force as (b) Now I invite you to complete part (b) for yourself. If you compare your solution for part (b) to the solution for part (a), you will see that the solution processes are exactly the same; the only difference is that the roles of the two blocks are interchanged. Thus, you will find that the acceleration in part (b) is the same as in part (a) (does this make sense?), but the inter-block forces in part (b) are 100 times greater than in part (a). Does this make sense? Does this explain to you what we intuitively understand, that the delicate glass sculpture gets smashed in part (b) but not in part (a)? The next example is much like the previous one, but with three blocks instead Ch4B Page 19

20 of two. Example: Calculate the acceleration of the system and the magnitudes of the inter-block forces if a. a 10 N force acts on the left-most block towards the right. b. a 10 N force acts on the right-most block towards the left. Assume there is no friction between the blocks and the surface they rest on. Ch4B Page 20

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23 Blocks and pulleys Ch4B Page 23

24 Example: Calculate the tension in each string and the acceleration of each block. Assume that the strings are massless and don't stretch, and that the pulley is massless and there is no friction at the bearing. Further assume that the string slides on the pulley without friction (or, more accurately, turns the pulley without slipping, so that friction need not be considered). Ch4B Page 24

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27 Example: Calculate the tension in the string and the acceleration of each block. Assume that the strings are massless and don't stretch, and that the pulley is massless and there is no friction at the bearing. Further assume that the string slides on the pulley without friction (or turns the pulley without friction needing to be considered). Suppose that the coefficient of static friction between the 2 kg block and the horizontal surface is 0.8, and that the coefficient of kinetic friction between the same surfaces is 0.5. Ch4B Page 27

28 Now for an important bit of strategy: Note that if the tension force is less than the maximum value of the static friction force, then the blocks will not move. The static friction force will adjust itself to match the applied force, up to a point; if the applied force exceeds the maximum static friction force, then the blocks move, and then kinetic friction will apply. Thus, let's first assume that the blocks do not move, and we'll calculate the tension force. Then we'll compare the tension force to the maximum force of static friction. If the tension force is greater than the maximum static friction force, then we know the blocks will move, and we'll re-solve a part of the problem using the kinetic friction force Ch4B Page 28

29 to determine the actual accelerations of the blocks. Here goes the first step, using static friction: The tension force is greater than the maximum static friction force, so the blocks actually move. Thus, we can go back to equation (*) and replace f by the force of kinetic friction to determine the actual tension in the string. Ch4B Page 29

30 Example: Calculate the tension in the string and the acceleration of each block. Assume that the strings are massless and don't stretch, and that the pulley is massless and there is no friction at the bearing. Further assume that the string slides on the pulley without friction (or turns the pulley without friction needing to be considered). (a) Suppose that the surface is frictionless. (b) Suppose that the coefficient of static friction between the 2 kg block and the slanted surface is 0.8, and that the coefficient of kinetic friction between the same surfaces is 0.5. Ch4B Page 30

31 (a) If the surface is frictionless, then f = 0, and the equations above simplify to (also noting that the sine of 30 degrees is 1/2): Because we are only interested in calculating the tension in the string and the acceleration, we can ignore equation (1). Solve equation (2) for T and substitute the result in equation (3), to obtain: Ch4B Page 31

32 Substituting the value of the acceleration into equation (2), we can solve for the tension in the string: (b) Now for an important bit of strategy: Note that if the tension force is less than the maximum value of the static friction force, then the blocks will not move. The static friction force will adjust itself to match the applied force, up to a point; if the applied force exceeds the maximum static friction force, then the blocks move, and then kinetic friction will apply. Thus, let's first assume that the blocks do not move, and we'll calculate the tension force. Then we'll compare the tension force to the maximum force of static friction. If the tension force is greater than the maximum static friction force, then we know the blocks will move, and we'll re-solve a part of the problem using the kinetic friction force to determine the actual accelerations of the blocks. Here goes the first step, using static friction: Ch4B Page 32

33 The tension force is greater than the maximum static friction force, so the blocks actually move. Thus, we can go back to equation (*) and replace f by the force of kinetic friction to determine the actual tension in the string. Note that the acceleration is less in this problem than in the previous problem, which is not surprising, since (because of the slope of the slanted surface) gravity works against the acceleration of the 2-kg block. Ch4B Page 33

34 Next consider a pulley problem that is a little more sophisticated. Example: Determine the accelerations of each block and the tension in the rope. (Because we assume that the rope is massless and does not stretch, and the pulleys are massless and frictionless, the tension is the same in each part of the rope.) Solution: Draw three free-body diagrams; one each for the two outside masses, and another one for the central pulley. As usual, we'll choose positive directions for each free-body diagram so that the accelerations will all be compatible. We have three equations for four unknowns, so we need another equation. Because the rope does not stretch, the accelerations are all connected, but in a way that is a little more complex than the previous problems in which there were only two blocks. Note that if the middle block is displaced downward so that its displacement Ch4B Page 34

35 is x 2, the sum of the displacements of the other two blocks is twice as large as x 2, because two lengths of ropes are attached to the middle pulley. Because if the pulleys are released from rest, then Thus, (Calculus lovers could derive equation 4 in an alternative way by simply starting with the displacement relation above and differentiating each term twice.) So now we have four equations for four unknowns, and the problem is solvable. One way to complete the solution is to solve each of the first three equations for the accelerations and substitute the resulting expressions into equation 4: Ch4B Page 35

36 We can now back-substitute the tension into the first three equations to determine the accelerations: Are the results reasonable? Well, the total mass of the outside blocks is 9 kg, and the mass of the central block is 6 kg, so it seems reasonable that all three blocks accelerate in the negative directions. To get a better feel for what can happen in general, it's worth solving the problem in general. This is done in the following example. Example: Determine expressions for the accelerations of each block and the tension in the rope. Ch4B Page 36

37 Solution: As in the previous example, we'll draw a free-body diagram for each of the two outer blocks and one for the middle pulley. The relations among the accelerations is the same as in the previous example, by the same reasoning used there. We can solve the problem by using the same solution process as in the previous example; solve each of the first three equations for the accelerations, then substitute the resulting expressions into equation 4 to obtain an equation involving just the tension. After solving this equation for the tension, we can then backsubstitute into each of the first three equations to obtain expressions for the accelerations. Ch4B Page 37

38 Thus, Similarly (do the algebra!), and Ch4B Page 38

39 Now it would be interesting to play with these expressions to see some examples of possible motions. For example: It's possible for the first block to have zero acceleration. One way to do this is to arrange for the first block and the third block to have the same mass and the middle block to have twice the mass of one of the others. But this solution is a bit boring (although it makes sense, doesn't it?), because then all three blocks have zero acceleration. A more interesting solution is to arrange for the first block to have twice the mass of the third block, and middle block to have four times the mass of the first block. Then the first block has zero acceleration, but the third block flies up with an acceleration of g, and the middle block moves down with an acceleration of g/2. You can arrange for the third block to have an acceleration of zero by just following the same choices as in the previous item, but interchanging the roles of the first and third blocks. You can arrange for the middle block to have an acceleration of zero by choosing the masses of the middle block and the third block to be equal, and the mass of the first block to be a third of the mass of the middle block. Then the middle block does not accelerate, but the third block accelerates downward with magnitude g/2, and the first block accelerates upward with magnitude g/2. There are many other interesting possibilities, and the right sort of person would have fun playing with this on a rainy day. If you're good at programming, you might like to program a simulation, so that you could see the results of changing the values of the masses in various ways. That would also be fun, would it not? Finally, notice that in all of the problems in this chapter, time has been ignored. This is the usual tactic in first-year physics: We simplify as much as possible at the beginning, sometimes at the cost of oversimplification. As an example of a situation where time must be included in the analysis, consider the following massive block suspended by a thread, and having another thread attached below it: Ch4B Page 39

40 Suppose that you pull down on the bottom thread; what happens? Well, it depends on how gradually you increase the force as you pull down. If you give the bottom thread a sudden jerk, then the bottom thread will snap. If you give the bottom thread a slow, gradual increase in force, then the top thread will snap. You can explain this for yourself if you draw free-body diagrams for each thread (that's right, the threads, not the block), and notice that the tension in the top thread is greater, because it also has to support the weight of the block. However, if you give the lower thread a sudden jerk, there is not enough time for the force to be transmitted to the upper thread, and so the lower thread will break. It's a bit like a train engine pulling on a long train of cars; not every car begins moving at once. Instead, the first car begins moving, and there is a slight time delay until the coupling between the first car and the second car engages, which gets the second car moving. Then, after another short time delay, the third car begins moving, and so on. You can see the same phenomenon (for a different reason) when a line of cars gets moving when the traffic light changes from red to green; there is a time delay for each car in the line to get moving. You can experience this phenomenon (with the block and threads) in every day life when you break a paper towel from its roll, or break toilet paper from its roll. If you "snap" the paper, then it will break along one of the perforations, but if you pull steadily the paper will unroll without breaking off. The same thing happens at the grocery store when you break off a plastic bag from its roll in the produce department. A sudden snap breaks a bag from the roll, but a gradual pull just unrolls the bags. There are other practical consequences to the tension experienced by different members of hanging structures. Consider the following hanging walkway. Ch4B Page 40

41 Moral: Critical thinking involves actively searching for errors. It's unreasonable to expect that no errors will be made; however, one needs systematic searches for errors. This is one of the great strengths of science: Nothing is taken for granted, and nothing is taken by authority. Each idea is always tested, over and over again. The upside is that errors are found and corrected; the system of science is "self-correcting." The downside is that new ideas are often subject to vicious attacks, and sensitive people are often turned away from scientific research because of its rough-and-tumble nature. Even the very greats suffer the slings and arrows of attack (to be fair, this is common in every field of creative endeavour, not just science), and sometimes they succumb; witness Boltzmann's suicide, which may have been driven by the vicious attacks on his revolutionary ideas in statistical mechanics. Resilience is helpful in all fields of endeavour, so perhaps it's not surprising that it is helpful for survival in scientific research too. Ch4B Page 41