#$&* course Phy 201 10/10 5 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: We know the initial velocity of the system and, for the period until it comes to rest, we know that its final velocity will be 0 m/s. If we can find the acceleration, we will have three of the five necessary variables with which to analyze the motion and we can therefore determine the time interval `dt and displacement `ds corresponding to this period. We begin by analyzing the forces acting on the system. Our first step is to choose the positive direction. The system can move in one of two directions, the direction in which the greater mass descends while the lesser ascends, or the direction in which the greater mass ascends while the lesser descends. Either choice is fine, but once we make the choice all directional quantities will have positive or negative signs relative to this direction. We arbitrarily choose the positive direction for the present system to be the direction in which the system moves as the greater of the two hanging masses descends. Thus the positive direction is the one in which the 4 kg mass is descending. Gravity exerts forces of 4 kg * 9.8 m/s^2 = 39.2 Newtons on the 4 kg mass and 3 kg * 9.8 m/s^2 = 29.4 Newtons on the 3 kg mass. Taking the positive direction to be the direction in which the system moves when the 4 kg mass descends, as stated earlier, then these forces would be +39.2 Newtons and -29.4 Newtons. The total mass of the system is 7 kg, so its total weight is 7 kg * 9.8 m/s^2 = 68.4 Newtons and the frictional force is therefore frictional force = .02 * 68.4 Newtons = 1.4 Newtons, approx.. The system is given an initial velocity in the direction which makes the lesser mass descend. This direction is opposite the direction we have chosen here to be positive, so the initial velocity is negative. The frictional force will tend to oppose the motion of the system. So if the system is moving in the negative direction, then the direction of the frictional force is positive. The frictional force is thus +1.4 Newtons. The net force on the system is therefore F_net = +39.2 Newtons - 29.4 Newtons + 1.4 Newtons = +11.2 Newtons. This results in an acceleration of +11.2 N / (7 kg) = 1.6 m/s^2. At this point we know that v0 = -5 m/s, vf = 0 and a = 1.6 m/s^2. From this we can easily reason out the desired conclusions: The change in velocity is +5 m/s and the average velocity is -2.5 m/s. At the rate of 1.6 m/s^2, the time interval to change the velocity by +5 m/s is `dt = +5 m/s / (1.6 m/s^2) = 3.1 sec, approx.. At an average velocity of -2.5 m/s, in 3.1 sec the system will be displaced `ds = -2.5 m/s * 3.1 s = -7.8 meters. These conclusions could also have been reached using equations: since vf = v0 + a `dt, `dt - (vf - v0) / a = (0 m/s - (-5 m/s) ) / (1.6 m/s^2) = 3.1 sec (appxox). Since `ds = .5 (v0 + vf) * `dt, `ds = .5 (-5 m/s + 0 m/s) * 3.1 s = -7.8 meters. PARTIAL STUDENT SOLUTION and QUESTION: Fnet = 11.2 N; Fnet / m = 1.6 m/s^2. a = 1.6 m/s^2, vf = 0, v0 = 5 m/s. Using the fourth equation `ds = (vf^2 - v0^2) / (2 a) = ... = -15.6 m `dt = `dv / a = -5 m/s / (1.6 m/s^2) = -3.1 sec. I'm still not clear however on my the change in velocity is a positive integer - shouldn't this be negative if 'dv = vf - v0 ?? INSTRUCTOR RESPONSE: You did well to notice and question the inconsistency. You made your net force positive. When you did so you implicitly declared that the positive direction is toward the greater mass. The 5 m/s initial velocity is in the direction of the lesser mass. So having made your net force positive, you should have made the initial velocity negative. My recommendation is to always declare your positive direction explicitly, and be careful that the signs of all your quantities agree with this declaration. Interpretation of your solution: The velocity at the initial instant (which could, for example, have been the instant at which the clock was started--the initial instant is not necessarily the instant at which the motion started) was 5 m/s, the direction of acceleration was the same as that of the initial velocity, and the system was at rest .521 seconds before the initial instant. STUDENT QUESTION: So Fnet=39.2-29.4-1.372= 8.428N (* this is what I thought would be the net force,but when I saw the solution,the frictional force is taken positive,CAN YOU PLEASE EXPLAIN THAT) INSTRUCTOR RESPONSE: Briefly we chose the positive direction, found that relative to this choice the velocity is positive, so the friction (being directed opposite to the velocity) is positive. You need to self-critique the given solution so I can tell what you do and do not understand: Truncating the corresponding section of the given solution a little bit, that solution tells you the following: • Our first step is to choose the positive direction. The system can move in one of two directions, the direction in which the greater mass descends while the lesser ascends, or the direction in which the greater mass ascends while the lesser descends. • Either choice is fine, but once we make the choice all directional quantities will have positive or negative signs relative to this direction. • We arbitrarily choose the positive direction for the present system to be the direction in which the system moves as the greater of the two hanging masses descends. Thus the positive direction is the one in which the 4 kg mass is descending. The system is given an initial velocity in the direction which makes the lesser mass descend. This direction is opposite the direction we have chosen here to be positive, so the initial velocity is negative. The frictional force will tend to oppose the motion of the system. So if the system is moving in the negative direction, then the direction of the frictional force is positive. The frictional force is thus +1.4 Newtons. Can you self-critique this part of the solution? That is, tell me, phrase by phrase, what you do and do not understand about this part? STUDENT QUESTION I thought I did everything correctly, however it looked like I got my signs backwards on when I went to find the ‘dt and the ‘ds, because both my time interval and displacement had the wrong signs. I don’t really get why my signs where backwards though. INSTRUCTOR RESPONSE (further explanation of positive and negative directions and choice of signs) If you have two 3-kg masses on opposite sides of the pulley, gravity exerts a downward 29.4 N force on each one. That would make a total force of 58.4 N downward. However the system is perfectly counterbalanced. Equal weights on the two sides of the pulley would balance each other perfectly, even without the help of friction. If the net force on this system was 58.4 N, however, the 6 kg mass of the system would accelerated at a = 58.4 N / (6 kg) = 9.8 m/s^2. So clearly the 58.4 N downward force is not the net force on the system. In fact since the system has zero acceleration, we know that the net force on it is zero. How can we explain this? If we regard the two 3-kg masses and the string connecting them as our system, then the two suspended masses tend to pull the system in opposite directions. That is, one can be said to pull in the positive direction, the other in the negative direction. The choice of which direction is which is arbitrary--that is, we can pick either direction to be positive, with the other being negative. When we have just the two 3 kg masses hanging there, before we put anything into motion or unbalance the two sides, we might not need not need to worry about which is which. We can just accept that the gravitational forces are in balance, one pulling the system in the positive direction, one in the negative. When we add the 1 kg object to one side, then set the system into motion in the direction of the 'lighter' side, it's time to choose directions. We could choose to let the positive direction be the direction in which the 'lighter' side descends, or we could choose the positive direciton to be the direction in which the 'heavier' side descends. In the given solution we make the latter choice. Since the initial velocity is toward the 'lighter' side, then the initial velocity must be negative. The 39.2 N gravitational force pulling the greater mass in our chosen positive direction is a positive force, and the 29.4 N gravitational force pulling the lesser mass is in the negative direction, the net force is +9.8 N. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q002. If the system in the previous example was again given an initial velocity of 5 m/s in the direction of the 3 kg mass, and was allowed to move for 10 seconds, then what would be its final displacement relative to its initial position? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 39.2N-29.4N-1.4N=8.4N 10s-3.2s=6.8s 'dv=(1.2m/s^2*6.8s 'dv=8.2m/s 8.2m/s=vf-v0 vf=13.2m/s (13.2m/s+-5m/s)/2=4.1m/s 4.1m/s='ds/6.8s 'ds=27.9m confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The acceleration of the system is different if it is moving in the positive direction than when it is moving in the negative direction. We only know how to analyze a uniform-acceleration situation. Since we have both positive and negative velocities this problem must the separated into two parts. As seen in the previous example, in the first 3.2 seconds the system displaces -7.8 meters. This leaves 6.8 seconds after that instant during which the system may accelerate from rest in the positive direction. We therefore analyze the motion from the instant the system comes to rest until the remaining 6.8 seconds has elapsed. Recall our choice of positive direction, made in the preceding problem. To avoid having to reorient ourselves, we will continue to use this choice: We arbitrarily chose the positive direction for the present system to be the direction in which the system moves as the greater of the two hanging masses descends. Thus the positive direction is the one in which the 4 kg mass is descending. During the final 6.8 seconds the system will be moving in the positive direction, with the 4 kg mass descending. The frictional force during this time is directed opposite the motion so the frictional force will be negative. The net force on the system will therefore be F_net = + 39.2 Newtons -29.4 Newtons - 1.4 Newtons = 8.4 Newtons, and the acceleration will be a = F_net / m = 8.4 Newtons / (7 kg) = 1.2 m/s^2, approx.. The initial velocity during this phase is 0, the time interval is 6.8 sec and the acceleration is 1.2 m/s^2. We therefore conclude that the velocity will change by `dv = 1.2 m/s^2 * 6.8 sec = 8.2 m/s, approx, ending up at 8.2 m/s since this phase started at 0 m/s. This gives an average velocity of 4.1 m/s; during 6.8 sec the object therefore displaces `ds = v_Ave * `dt = 4.1 m/s * 6.8 sec = 28 meters approx.. These results are easily reasoned out from the definitions of velocity and acceleration. They could have also been obtained equally easily using the equations. For the entire 10 seconds, the displacements were -7.8 meters and +28 meters, for a net displacement of approximately +20 meters. That is, the system is at this instant about 20 meters in the direction of the 4 kg mass from its initial position. STUDENT COMMENTS: Both the calculation of Fnet and acceleration are apparent and appear easy, now that I've seen the correct #'s to plug in. I took note that v0 for this 2nd phase is 0 - that makes sense because the vf from the 1st stage was 0 m/s after 3.2 s. Having read through your solution, I seem to understand how you structured it, though prior to seeing it worked out, I was obviously way off in my calculation. I guess my real problem in solving these questions is in determining how best to set the problem up. The calculations themselves seem reasonably easy, but setting up the problem in the correct manner is throwing me off every time! I've read and taken notes in all of the """"ideas"""" and """"problem sets"""" and I try to draw diagrams and the such. Any suggestions for improvement here? INSTRUCTOR RESPONSE You're doing the right things and asking the right questions. I suspect everything will become clear as you work subsequent problems. If not, I'll make additional suggestions at the end. STUDENT QUESTION: now I am again stuck with the question about frictional force, if that is negative here or positive. Please explain on that. INSTRUCTOR RESPONSE: The velocity is now positive, relative to our selected positive direction. So the frictional force is negative. Please see my notes on the previous problem. Having understood (or self-critiqued) that, you'll be in a position to understand it for the new direction of the velocity ADDITIONAL INSTRUCTOR NOTE: The direction of the frictional force depends on the direction in which the system is moving. The system here moves in one direction, reverses, and moves in another. So the frictional force changes when the direction of motion reverses. STUDENT QUESTION When I first read the question i thought that the system was continuing in the same direction as before but for a longer duration. Once I re-read and reworked I noticed the from the initial position of -7.8 meters at v0=0. I guess it would have to go in the opposing direction at that point, but it does seem confusing that the question says that the direction has continued on the 3kg side being the lighter side????????? It the systme still favoring the lighter side and going in the opposite direction??????????????????????????? INSTRUCTOR RESPONSE The system came to rest in 3.2 seconds at position -7.8 meters, relative to its original position. So at this point we start a new acceleration interval, on which v0 is zero, as you say. If I'm reading correctly the rest of your question is related to which direction is positive. The choice of positive direction is arbitrary, and was made in the preceding problem; in the given solution the direction in which the 4 kg object descends was chosen as positive. You might well have chosen the opposite direction as positive, since the initial velocity was in the direction of descent of the 3 kg object. Neither choice of positive direction has any particular advantage over the other, and whichever is chosen, the final interpretation of the solution will be the same. The choice of positive direction has nothing to do with which side the system is 'favoring'. The choice is purely one of convenience. For this situation: the total of the gravitational forces is in the direction in which the 4 kg object descends the initial velocity in the original statement is in the direction of descent of the 3 kg object, and in whichever direction the system is moving the frictional force is in the opposite direction (since in the current problem the system reverses its direction of motion, the frictional force must also reverse its direction of motion) Once the positive direction is chosen, the signs of the forces, accelerations, velocities, etc. are determined accordingly. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q003. An automobile has a mass of 1400 kg, and as it rolls the force exerted by friction is .01 times the 'normal' force between its tires and the road. The automobile starts down a 5% incline (i.e., an incline with slope .05) at 5 m/s. How fast will it be moving when it reaches the bottom of the incline, which is 100 meters away (neglect air friction and other forces which are not part of the problem statement)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1400kg*9.8m/s^2=13720N 13720N*.05=686N 13720*.01=137.2N 686N-137N=549N 549N=1400kg*a a=.39m/s^2 vf^2=5m/s+2(.39m/s^2)(100m) vf=10.4m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We are given the initial velocity and displacement of the automobile, so we need only find the acceleration and we can analyzes problem as a standard uniform acceleration problem. The automobile experiences a net force equal to the component of its weight which is parallel to the incline plus the force of friction. If we regard the direction down the incline as positive, the parallel component of the weight will be positive and the frictional force, which must be in the direction opposite that of the velocity, will be negative. The weight of the automobile is 1400 kg * 9.8 m/s^2 = 13720 Newtons, so the component of the weight parallel to the incline is parallel weight component = 13720 Newtons * .05 = 686 Newtons. The normal force between the road and the tires is very nearly equal to the weight of the car because of the small slope, so the magnitude of the frictional force is approximately frictional force = 13720 Newtons * .01 = 137 Newtons, approx.. The frictional force is therefore -137 Newtons and the net force on the automobile is Fnet = 686 Newtons - 137 Newtons = 550 Newtons (approx.). It follows that the acceleration of the automobile must be a = Fnet / m = 550 Newtons / 1400 kg = .4 m/s^2 (approx.). We now have a uniform acceleration problem with initial velocity v0 = 5 meters/second, displacement `ds = 100 meters and acceleration a = .4 m/s^2. We can easily find the final velocity using the equation vf^2 = v0^2 + 2 a `ds, which gives us vf = +- `sqrt(v0^2 + 2 a `ds) = +- `sqrt( (5 m/s)^2 + 2 * .4 m/s^2 * 100 m) = +- `sqrt( 25 m^2 / s^2 + 80 m^2 / s^2) = +-`sqrt(105 m^2 / s^2) = +- 10.2 m/s. It is obvious that the final velocity in this problem is the positive solution +10.2 m/s, since the initial velocity and acceleration are both positive. STUDENT COMMENTS I believe I understood everything, with the exception of how the slope of the incline played into it. """"Parallel weight component"""" - OK, I have committed this to my notes. So in addition to the frictional force, we have this parallel weight component force, which is essentially the WEIGHT of the auto multiplied by the slope of the incline. Also interesting to note that the frictional force is unaffected the slope of the incline. However, the frictional force in the solution is a negative integer, which is actually contrary to what I would have originally thought. Parallel component force less frictional force (or plus a negative frictional force) = our Fnet for this problem. I understand how the accel was calculated and how the additional calculations for vf are calculated. INSTRUCTOR RESPONSE 1400 kg * 9.8 m/s^2 is the gravitational force. On a level road this force would be balanced by the force of the road pushing up on the car, and the combination of these two forces would make no net contribution to the net force. You included the 1400 kg * 9.8 m/s^2 as part of your net force, which threw off the rest of your results. Other than this, your solution was correct. A little more about inclines: On an incline the road doesn't push straight up--it pushes in a direction perpendicular to the road surface, so it doesn't completely balance the gravitational force. On a small incline it still balances most of the gravitational force, but not all. A component of the gravitational force remains. (These force components are analyzed in upcoming assignments using vectors; for right now we simply make the statement that the force component parallel to the (small) incline is approximately equal to the slope multiplied by the weight). Imagine that you are trying to hold back a small car on an incline. The steeper the incline the greater the force you have to exert. What you are doing is holding the car back against the parallel component of the gravitational force; you have to exert a force which is equal to this component (less the rolling friction of the tires on the road). The frictional force is in fact affected by the slope of the incline, but for small inclines the difference is negligible. (We will see soon, when we analyze the vectors, that the frictional force depends on the cosine of the angle of the incline, which changes very little for small inclines). STUDENT COMMENT: 9.8 m/s^2 * 1400 kg = 13720 Newtons After this, Im pretty confused on how to factor in the 5% slope. I believe it would have to do with the incline being parallel to the mass of the automobile but Im confused on exactly how to put that into an equation. INSTRUCTOR RESPONSE: The mass of the automobile is not a vector quantity; it doesn't have a direction. So it can't be parallel or perpendicular to anything. You might be thinking of the weight of the automobile. The weight of the automobile is directed in the downward vertical direction. If the automobile was on a horizontal surface, its weight would therefore be at a right angle to the surface. Assuming the surface is strong enough to support the automobile, it pushes up with a force that is equal and opposite to the weight of the car. Now if you tilt the surface to form an incline, the automobile will tend to roll down the incline. If the tilt is slight, it won't be hard to hold the automobile back. As the angle of the tilt increases, it gets harder and harder to hold the automobile back. If you are trying to hold the automobile back, you will be pushing or pulling parallel to the incline, and you will perceive the automobile as being subject to a force which is directed along the incline. The force you perceive is actually gravitational in nature. However you aren't trying to pick up the whole car, so you don't need to exert a force equal to the gravitational force. On a 'small' incline, not too far from horizontal, the incline is still pretty much supporting the car. However, the tilted incline can't push straight upward in the vertical direction, so it can't completely balance the force of gravity. It can only balance the part of the weight that's perpendicular to the incline. The part that isn't balanced is the 'parallel component' of the weight, the component which is parallel to the incline. That's the force that tends to accelerate the auto down the incline, and it's this force you have to push back against if you want to hold the automobile still. Soon enough we will use vectors and trigonometry to figure out these components. This is not particularly difficult, but for right now you are just told that the component of weight parallel to an incline is approximately equal to the slope multiplied by the weight, provided the slope is small. If you were on an incline with slope .01, the parallel component of the gravitational force would be 13720 N * .01 = 137 N. If you were on an incline with slope .02, the parallel component of the gravitational force would be 13720 N * .02 = 274 N. The steeper the incline the greater the parallel component and the harder you have to push or pull to hold the auto stationary. The present slope is .05, which is still a small slope, so to find the parallel component you just multiply the 13720 N by .05. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q004. If the automobile in the previous example started at the bottom of the incline with velocity up the incline of 11.2 m/s, how far up the hill would it be able to coast? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -686N-137N=-823N -823N=1400kg*a a=.59m/s^2 'dt='dv/a 'dt=11.2m/s/.59m/s^2 'dt=19s vAvew=(11.2m/s+0)/2=5.6m/s*19s=106m confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We again have a uniform acceleration situation where the initial velocity is known and we have the information to determine the acceleration. The other quantity we can deduce is the final velocity, which at the furthest point up the hill will be zero. Since the automobile is coasting up the incline, we will now take the upward direction as positive. The frictional force will still be 137 Newtons and will again be directed opposite the velocity, and will therefore be negative. The parallel component of the weight will be the same as before, 686 Newtons, but being directed down the incline will be in the direction opposite to our chosen positive direction and will therefore also be negative. The net force on the automobile therefore be net force = -686 Newtons - 137 Newtons = -820 Newtons (approx.). Its acceleration will be a = Fnet / m = -820 Newtons / 1400 kg = -.6 m/s^2 (approx.). We see now that v0 = 11.2 m/s, vf = 0 and a = -.6 m/s^2. Either by direct reasoning or by using an equation we now easily find that `dt = (-11.2 m/s) / (-.6 m/s^2) = 19 sec (approx) and `ds = (11.2 m/s + 0 m/s) / 2 * 19 sec = 5.6 m/s * 19 sec = 106 meters (approx). STUDENT COMMENTS: OK, so both frictional force AND parallel component force are negative? It is this key mistake that once again led me astray. Had I done that part correctly, I would have gotten the answer correct!! It seems there is always something about the set up of the problems that I've left out or done slightly incorrectly - perfect example is the Fnet of this problem. I would have suspected that the Ffrict was now in the opposite direction and would therefore be positive, but I guess it makes imperical sense that frictional force is always negative? I feel like I'm getting closer to a point where my understanding for the material is improved, but I can't quite seem to make it over the hill, where I'm understanding the questions/problems to the extent that I don't make key mistakes like in this problem. Maybe it's as simple as going to the text and doing additional problems until my comprehension is better. Do you have any suggestions? INSTRUCTOR RESPONSE: You are following all the correct steps, and you have mastered the mathematics of the situation. You're asking the right questions and thinking about your answers. As you correctly indicate, you are right on the verge of mastering this. For an automobile rolling up or down an incline, the rolling friction will be in the direction opposite motion. Having declared a positive direction and given the velocity the appropriate sign (+ or -), the force of rolling friction will have the sign opposite the velocity. Once you get into the habit of being very explicit and careful with the signs, everything should come together nicely. STUDENT QUESTION Why did the acceleration change? I thought this was the same system? INSTRUCTOR RESPONSE It's the same system, but moving in a different direction, which changes the direction of the frictional force. STUDENT QUESTION I can solve the rest of the problem as I now understand that the frictional force and the gravitational componenet would both be negative now as the direction of motion is uphill.But why is velocity not taken as negative.I don’t know where but in one of the questions we even did change the sign for velocity when we reversed the direction of motion. Please suggest on this. INSTRUCTOR RESPONSE The solution to this problem specifies that the upward direction is now chosen as positive. Had you chosen the downward direction as positive, which would be a valid choice, your signs should have all been opposite those in the given solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating:3 -------------------------------------------------------------------------------- Summary of Key Relationships: The key relationships here are F_net = m a `dW = F * `ds `dW_net_ON = `dKE (where `dW_net_ON designated the work done by the net force acting ON the mass). It is important to be very detailed in your use of units and in the calculation of units. Most errors will result in incorrect units, and if units don't work out it indicates an error either in the calculation of the units or in your solution. The units of these quantities: Unit of force = unit of mass * unit of acceleration = kg * m/s^2, also called Newtons. Unit of work = unit of force * unit of displacement = N * m, or kg m/s^2 * m = kg m^2 / s^2, also called Joules. Unit of kinetic energy = unit of mass * (unit of velocity)^2 = kg * (m/s)^2 = kg m^2 / s^2. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: -------------------------------------------------------------------------------- Summary of Key Relationships: The key relationships here are F_net = m a `dW = F * `ds `dW_net_ON = `dKE (where `dW_net_ON designated the work done by the net force acting ON the mass). It is important to be very detailed in your use of units and in the calculation of units. Most errors will result in incorrect units, and if units don't work out it indicates an error either in the calculation of the units or in your solution. The units of these quantities: Unit of force = unit of mass * unit of acceleration = kg * m/s^2, also called Newtons. Unit of work = unit of force * unit of displacement = N * m, or kg m/s^2 * m = kg m^2 / s^2, also called Joules. Unit of kinetic energy = unit of mass * (unit of velocity)^2 = kg * (m/s)^2 = kg m^2 / s^2. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!