course Phy 231 7/9 around 8 30 009. Forces exerted by gravity; nature of force; units of force
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Given Solution: Since there are 25 equal masses and the mass of the cart is equivalent to another 25 of these masses, each mass is 1/50 = .02 of the total mass of the system. Thus the first 10 data points should might have been something like (.02, 21 cm/s^2), (.04, 48 cm/s^2), (.06, 55 cm/s^2), (.08, 85 cm/s^2), (.10, 101 cm/s^2), (.12, 125 cm/s^2), (.14, 141 cm/s^2), (.16, 171 cm/s^2), (.18, 183 cm/s^2), (.20, 190 cm/s^2). The data given in the problem would correspond to alternate data points. The slope of the best-fit line is 925 x + 12.8, indicating a slope of 925 and a y intercept of 12.8. The 967.5 is in units of rise / run, or for this graph cm/s^2. If you calculated the slope based on the points (.04, 48 cm/s^2) and (.20, 190 cm/s^2) you would have obtained 142 cm/s^2 / (.16) = 890 cm/s^2 (very approximately). Whether this is close to the best-fit value or not, this is not an appropriate calculation because it uses only the first and last data points, ignoring all data points between. The idea here is that you should sketch a line that fits the data as well as possible, then use the slope of this line, not the slope between data points. STUDENT COMMENT If slope is rise / run (190 - 21) / (.20 - .02) = 939, then how is 925 the slope? _INSTRUCTOR RESPONSE 925 is the ideal slope, which you are unlikely to achieve by eyeballing the position of the best-fit line. Your selected points will be unlikely to give you the ideal slope, and the same is so for the point selected in the given solution.__The last paragraph says 'If you calculated the slope based on the points (.04, 48 cm/s^2) and (.20, 190 cm/s^2) ...'__That paragraph doesn't say this selection of point will give the ideal slope. You are unlikely to get the ideal slope based on a graphical selection of points; however you can come reasonably close. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: ********************************************* Question: `q002. Do the points seem to be randomly scattered around the straight line or does there seem to be some nonlinearity in your results? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope of the line should be someplace between 900 cm/s^2 and 950 cm/s^2. The best possible straight line should be drawn from the randomly scattered points. This type of judiously exacting experiments show that the acceleration of this system by nature is to a great degree of precisely proportionate to the weight with which it is suspended Confidence rating:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The slope of your line should probably be somewhere between 900 cm/s^2 and 950 cm/s^2. The points should be pretty much randomly scattered about the best possible straight line. Careful experiments of this nature have shown that the acceleration of a system of this nature is to a very high degree of precision directly proportional to the proportion of the weight which is suspended. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: ********************************************* Question: `q003. If the acceleration of the system is indeed proportional to the net force on the system, then your straight line should come close to the origin of your coordinate system. Is this the case? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If all information given of this system is represented correctly then the reprentation of the y coordinate of the straight line will be a constant multiple of the x coordinate, this means you can always find the y coordinate by multiplying the x coordinate by a specific number and this specific number is the same for all x coordinates. For example if the x coordinate is zero then the y coordinate will be 0 times 0. Because of unavoidable errors your graph may not precisely pass through the origin. But if you have not made too many errors your line should pass very close to the origin. In this experiment the y-intercept was 12.8. This is realistically small given the scale of the data used here plus the randomness of the data points above and below the straight-line. The variation is consistent with the situation where precise measurements would actually expose a straight line through the origin. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If the acceleration of the system is proportional to the net force, then the y coordinate of the straight line representing the system will be a constant multiple of the x coordinate--that is, you can always find the y coordinate by multiplying the x coordinate by a certain number, and this 'certain number' is the same for all x coordinates. The since the x coordinate is zero, the y coordinate will be 0 times this number, or 0. Your graph might not actually pass through the origin, because data inevitably contains experimental errors. However, if experimental errors are not too great the line should pass very close to the origin. In the case of this experiment the y-intercept was 12.8. On the scale of the data used here this is reasonably small, and given the random fluctuations of the data points above and below the straight-line fit the amount of deviation is consistent with a situation in which precise measurements would reveal a straight line through the origin. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: ********************************************* Question: `q003. What is it that causes the system to accelerate more when a greater proportion of the mass is suspended? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The gravitational forces exerted on the system are exerted on the suspended masses and the cart and the masses that remain in the cart. The force exerted by the ramp counters the gravitational force on the cart and the masses inside the cart, therefore the gravity in this system is not affected by the acceleration. But then there is no opposition to the pull of gravity on the suspended masses and as a result the net force is acting on the entire mass of the cart-and mass system. The gravitational force on the suspended mass is proportional to how many masses are suspended – for example two times as many masses are twice the force. Consequently the greater gravitational force on the suspended masses results in greater acceleration. Confidence rating:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The gravitational forces exerted on the system are exerted on the suspended masses and on the cart and the masses remaining in it. The supporting force exerted by the ramp counters the force of gravity on the cart and the masses remaining in it, and this part of the gravitational force therefore does not affect the acceleration of the system. However there is no force to counter the pull of gravity on the suspended masses, and this part of the gravitational force is therefore the net force acting on the mass of the entire cart-and-mass system. The force exerted by gravity on the suspended masses is proportional to the number of suspended masses--e.g, if there are twice as many masses there is twice the force. Thus it is the greater gravitational force on the suspended masses that causes the greater acceleration. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: ********************************************* Question: `q004. This results of this sort of experiment, done with good precision, support the contention that for a given mass the acceleration of a system is indeed proportional to the net force exerted on the system. Other experiments can be done using rubber bands, springs, fans and other nongravitational sources of force, to further confirm this result. In another sort of experiment, we can see how much force is required on different masses to obtain a certain fixed acceleration. In such experiments we find for example that if the mass is doubled, it requires twice the force to achieve the same acceleration, and that in general the force required to achieve a given acceleration is proportional to the amount of mass being accelerated. In a certain experiment using the same cart and masses as before, plus several additional identical carts, a single cart is accelerated by a single suspended mass and found to accelerate at 18 cm/s^2. Then a second cart is placed on top of the first and the two carts are accelerated by two suspended masses, achieving an acceleration of 20 cm / s^2. Then a third cart is placed on top of the first to and the three carts are accelerated by three suspended masses, achieving and acceleration of 19 cm/s^2. A fourth cart and a fourth suspended mass are added and an acceleration of 18 cm/s^2 is obtained. Adding a fifth cart in the fifth suspended mass an acceleration of 19 cm/s^2 is obtained. All these accelerations are rounded to the nearest cm/s^2, and all measurements are subject to small but significant errors in measurement. How well do these results indicate that to achieve a given acceleration the amount of force necessary is in fact proportional to the amount of mass being accelerated? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since there is very little variation in the accelerations then it could be a human error but without proper instructions on how they found out the accelerations then we cannot completely say this. Now since the second through the fifth systems had 2, 3, 4, and 5 times the mass of the first with 2, 3, 4, and 5 times the suspended mass so we know that the acceleration is proportional to the mass of the system. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The accelerations obtained are all about the same, with only about a 10% variation between the lowest and the highest. Given some errors in the observation process, it is certainly plausible that these variations are the result of such observation errors; however we would have to have more information about the nature of the observation process and the degree of error to be expected before drawing firm conclusions. If we do accept the conclusion that, within experimental error, these accelerations are the same then the fact that the second through the fifth systems had 2, 3, 4, and 5 times the mass of the first with 2, 3, 4, and 5 times the suspended mass and therefore with 2, 3, 4, and 5 times the net force does indeed indicate that the force needed to achieve this given acceleration is proportional to the mass of the system. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: ********************************************* Question: `q005. Now we note again that the force of gravity acts on the entire mass of the system when an entire system is simply released into free fall, and that this force results in an acceleration of 9.8 m/s^2. If we want our force unit to have the property that 1 force unit acting on 1 mass unit results in an acceleration of 1 m/s^2, then how many force units does gravity exert on one mass unit? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since we know that the force of gravity is 9.8 this means that the one mass unit must exert the same amount of force on the object. Confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since gravity gives 1 mass unit an acceleration of 9.8 m/s^2, which is 9.8 times the 1 m/s^2 acceleration that would be experienced from 1 force unit, gravity must exerted force equal to 9.8 force units on one mass unit. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: ********************************************* Question: `q006. If we call the force unit that accelerates 1 mass unit at 1 m/s^2 the Newton, then how many Newtons of force does gravity exert on one mass unit? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since the one mass unit is equal to 9.8 then the gravity must exert 9.8 newtons on the object. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since gravity accelerates 1 mass unit at 9.8 m/s^2, which is 9.8 times the acceleration produced by a 1 Newton force, gravity must exert a force of 9.8 Newtons on a mass unit. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: ********************************************* Question: `q007. The mass unit used here is the kilogram. How many Newtons of force does gravity exert on a 1 kg mass? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Gravity must exert 9.8 newtons of force on the 1 kg Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Gravity exerts a force of 9.8 Newtons on a mass unit and the kg is the mass unit, so gravity must exert a force of 9.8 Newtons on a mass of 1 kg. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: ********************************************* Question: `q008. How much force would gravity exert on a mass of 8 kg? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Gravity exerts a force 8 times the force on 1 kg then 8 kg so it will exert 8*9.8 newtons Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Gravity exerts 8 times the force on 8 kg as on 1 kg. The force exerted by gravity on a 1 kg mass is 9.8 Newtons. So gravity exerts a force of 8 * 9.8 Newtons on a mass of 8 kg. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: ********************************************* Question: `q009. How much force would be required to accelerate a mass of 5 kg at 4 m/s^2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since we have 5 times the mass and 4 times the accleration then we need 20 times the force Confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Compared to the 1 Newton force which accelerates 1 kg at 1 m/s^2, 2e have here 5 times the mass and 4 times the acceleration so we have 5 * 4 = 20 times the force, or 20 Newtons. We can formalize this by saying that in order to give a mass m an acceleration a we must exert a force F = m * a, with the understanding that when m is in kg and a in m/s^2, F must be in Newtons. In this case the calculation would be F = m * a = 5 kg * 4 m/s^2 = 20 kg m/s^2 = 20 Newtons. The unit calculation shows us that the unit kg * m/s^2 is identified with the force unit Newtons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: ********************************************* Question: `q010. How much force would be required to accelerate the 1200 kg automobile at a rate of 2 m/s^2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This will require 1200 kg times 2 m/s2 so 2400 newtons Confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This force would be F = m * a = 1200 kg * 2 m/s^2 = 2400 kg * m/s^2 = 2400 Newtons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: Questions related to Class Notes 1. An ideal cart of mass 8 kg rolls frictionlessly down an incline. The incline has a length of 50 cm and its 'rise', from low end to high end, is 3 cm. According to the relationship between the accelerating force, the slope and the weight of an object on an incline, as described in the Class Notes: . What are the magnitude and direction of the force that accelerates the cart down the incline? What therefore is the acceleration of the cart? The direction is downward and the magnitude of it is 9.8 m/s so the acceleration is 9.8 m/s2 "