course Phy 231 6/17 3:15 009. Forces exerted by gravity; nature of force; units of force Question: `q001. Note that there are 10 questions in this set.
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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 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 equation 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. Self-critique (if necessary): My slope was not exactly the same, but it was pretty close. Self-critique rating #$&*: 3 ********************************************* Question: `q002. Do the points seem to be randomly scattered around the straight line or does there seem to be some nonlinearity in your results? Your solution: The points on my graph, from what I can see, seem to be very close around the straight line. The does not seem to be curving. confidence rating #$&*:: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK Self-critique rating #$&*: OK ********************************************* Question: `q003. If the acceleration of the system is indeed proportional to the net force on the system, then your straight line should come close to the origin of your coordinate system. Is this the case? Your solution: It does seem like the line comes close to the origin. I found my y-intercept to be 15 cm/s^2, and the best fit line had an intercept of 12.8cm/s^2. These are both fairly near the origin, though not exactly. confidence rating #$&*:: 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): OK Self-critique rating #$&*: OK ********************************************* Question: `q003. What is it that causes the system to accelerate more when a greater proportion of the mass is suspended? Your solution: When more of the mass is suspended, then the force of gravity is greater. This means that there is more gravity force on the weight, which causes the acceleration to be faster. confidence rating #$&*:: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The gravitational forces exerted on the system are exerted two objects: • gravitational force is exerted on the suspended mass, i.e., the part of the original mass that has been removed from the cart and is now suspended • gravitational force is exerted on the cart (including the masses that remain in it). The frictional force and the force exerted by the ramp together counter the force of gravity on the cart and the masses remaining in it. The gravitational force on the cart and the masses in it therefore does not affect the acceleration of the system. However there is no force to counter the pull of gravity on the suspended masses. The net force on the system is therefore just the gravitational force acting which acts on the suspended mass. 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. Gravity is the source of the unbalanced force, so in a sense it is the cause. However the net force is different when different proportions of the mass are suspended, and to say that gravity is the cause does not explain the differing accelerations. The cause of the different accelerations is the configuration of the system. Self-critique (if necessary): The forces acting on the car are cancelled because of friction, so the only force in the system that matters is working on the suspended mass. The more mass that is suspended, the greater the gravitation force acting on it which causes more acceleration. Self-critique rating #$&*: 3 ********************************************* Question: `q004. The 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 non-gravitational 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? Your solution: The results indicate that to keep the same acceleration, the force is proportional to the mass being accelerated. You have the baseline acceleration with 1 car and 1 mass. Each time you add a car and 1 mass and the acceleration stays about the same. confidence rating #$&*:: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK Self-critique rating #$&*: OK ********************************************* Question: `q005. Now we note again that the force of gravity acts on the entire mass of the system when an entire system is simply released into free fall, and that this force results in an acceleration of 9.8 m/s^2. If we want our force unit to have the property that 1 force unit acting on 1 mass unit results in an acceleration of 1 m/s^2, then how many force units does gravity exert on one mass unit? The force unit has nothing to do with the acceleration of gravity. Gravity does exert forces. First we define our unit of force, then we use the observed acceleration of gravity to figure out how much force gravity exerts. 1 force unit accelerates 1 mass unit at 1 m/s^2. 9.8 force units would accelerate 1 mass unit at 9.8 m/s^2 The action of the gravitational force of Earth on object near its surface acts as follows: • The gravitational force on any freely falling mass accelerates it at 9.8 m/s^2. • In particular 1 mass unit would, like anything else, accelerate at 9.8 m/s^2 if in free fall. Thus the gravitational force accelerates 1 mass unit at 9.8 m/s^2. Our force unit is defined as follows: • 1 force unit acting on 1 mass unit results in an acceleration of 1 m/s^2. So, for example, • 2 force units acting on 1 mass unit would result in an acceleration of 2 m/s^2. • 3 force units acting on 1 mass unit would result in an acceleration of 3 m/s^2. • etc. We can say the same thing in slightly different words: • To accelerate 1 mass unit at 2 m/s^2 would require 2 force units. • To accelerate 1 mass unit at 3 m/s^2 would require 3 force units. • etc. Following this pattern, we come to the conclusion that • To accelerate 1 mass unit at 9.8 m/s^2 would require 9.8 force units. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Gravity exerts 9.8 force units on 1 mass unit, to get the correct acceleration. confidence rating #$&*:: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK Self-critique rating #$&*: OK ********************************************* Question: `q006. If we call the force unit that accelerates 1 mass unit at 1 m/s^2 the Newton, then how many Newtons of force does gravity exert on one mass unit? Your solution: Gravity exerts 9.8 Newtons on 1 mass unit. confidence rating #$&*:: 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): OK Self-critique rating #$&*: OK ********************************************* Question: `q007. The mass unit used here is the kilogram. How many Newtons of force does gravity exert on a 1 kg mass? Your solution: Gravity exerts 9.8 Newtons on 1 kg. confidence rating #$&*:: 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): OK Self-critique rating #$&*: OK ********************************************* Question: `q008. How much force would gravity exert on a mass of 8 kg? Your solution: 9.8 N for 1 kg 8kg*9.8 N = 78.4 Newtons confidence rating #$&*:: 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): OK Self-critique rating #$&*: OK ********************************************* Question: `q009. How much force would be required to accelerate a mass of 5 kg at 4 m/s^2? Your solution: 4m/s^2 would need 4 Newtons for 1 kg 4 N *5kg = 20 Newtons confidence rating #$&*:: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK Self-critique rating #$&*: OK ********************************************* Question: `q010. How much force would be required to accelerate the 1200 kg automobile at a rate of 2 m/s^2? Your solution: 2 N *1200kg = 2400 Newtons confidence rating #$&*:: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK Self-critique rating #$&*: OK "