Ch5 Page 1 Chapter 5: Applications of Newton's laws Tuesday, September 17, 2013 10:00 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:
Ch5 Page 2 Example: Static equilibrium a. b. c. Calculate the tension in the strings if (assume the strings have no mass) the labelled angles are both 30 the labelled angles are both 10 one labelled angle is 40 and the other labelled angle is 60
Ch5 Page 3
Ch5 Page 4 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
Ch5 Page 5 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 exerted on you by the surface you're standing on) depends on your acceleration. Example: Apparent weight a. b. c. Calculate the apparent weight of a 70 kg person in an elevator that is accelerating upward at 4 m/s 2. moving upward at a constant speed of 5 m/s. accelerating downward at 2 m/s 2.
Ch5 Page 6 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. 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.
Ch5 Page 7 µ r << µ k < µ s Sample coefficients of friction are found on Page 144 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 a. b. 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 = 0.73. Determine the force needed to get the refrigerator moving. the force needed to keep the refrigerator moving at a constant speed once it has already started moving.
Ch5 Page 8 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 980-715 = 265 N. Thus, the refrigerator accelerates to the right. Air drag
Ch5 Page 9 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 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.
Ch5 Page 10 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 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.
Ch5 Page 11 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 various heights. Or, better yet, let's use the kinematics equations to determine the
Ch5 Page 12 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. a. b. Blocks and pulleys Example: Calculate the acceleration of the system and the magnitudes of the inter-block forces if a 10 N force acts on the left-most block towards the right. 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.
Ch5 Page 13
Ch5 Page 14
Ch5 Page 15
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). Ch5 Page 16
Ch5 Page 17
Ch5 Page 18
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. Ch5 Page 19
Ch5 Page 20 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:
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. Ch5 Page 21
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. Ch5 Page 22
Ch5 Page 23 (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: 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
Ch5 Page 24 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: 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.
Ch5 Page 25 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. 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: 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
Ch5 Page 26 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. http://en.wikipedia.org/wiki/hyatt_regency_walkway_collapse 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.
Ch5 Page 27 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.