course phy201 Query 9.1009. 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 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. 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): ok, I had to fill in all the data points first and then find the slope.
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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): ??????????????????Is this because the best fit line does not completly go through all the data points, even though it looks linear, or am I plotting wrong??????????????????????????
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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): I think I understand the concepts ------------------------------------------------ Self-critique rating #$&*:2 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q003. What is it that causes the system to accelerate more when a greater proportion of the mass is suspended? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: because when the mass is suspended it is putting more force on the downward motion of the cart, when the mass is in the cart it is more evenly distributed in the cart, but when it is suspended in front it causes the cart to accelerate forward. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, so the gravitational force on the suspended masses is more than that within the cart and causes greater acceleration. ------------------------------------------------ Self-critique rating #$&*:2 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* 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: to acheive a given acceleration the amount of force neccessary is in fact proportional to the amount of mass being acceleration because even though more carts are being added, more suspended mass is also being added which equals out the weightin the cart to the weight suspended from the cart so acceleration stays the same. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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, so the mass and the suspended mass and the net force are equal. ------------------------------------------------ Self-critique rating #$&*:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q005. Now we note again that the force of gravity acts on the entire mass of the system when an entire system is simply released into free fall, and that this force results in an acceleration of 9.8 m/s^2. If we want our force unit to have the property that 1 force unit acting on 1 mass unit results in an acceleration of 1 m/s^2, then how many force units does gravity exert on one mass unit? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am not sure confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):
so, normally 1 force unit = 9.8m/s^@ acceleration from gravity?
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.
so force must be equal to 9.8 force units to be 1m/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.
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Self-critique rating #$&*:1
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Question: `q006. If we call the force unit that accelerates 1 mass unit at 1 m/s^2 the Newton, then how many Newtons of force does gravity exert on one mass unit?
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Your solution:
9.8 newtons of force
confidence rating #$&*:2
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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, I was trying to write it as a whole unit like newtons per kilogram
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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. STUDENT COMMENT I think I did it right but am I not supposed to multiply out 8 * 9.8 Newtons to get 78.6 Newtons? I understand the units wouldn’t cancel, but am I supposed to leave it without “completing” it? INSTRUCTOR RESPONSE It's fine to complete it, and that's what you are generally expected to do. The given solution didn't do that because I wanted to emphasize the logic of the solution without the distraction of the (obvious) arithmetic. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, I understood this. ------------------------------------------------ Self-critique rating #$&*:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q009. How much force would be required to accelerate a mass of 5 kg at 4 m/s^2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 9.8 = 5kg * 4m/s^2=9.8newtons = 20 kg confidence rating #$&*: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): ok, so force is equal to mass time acceleration, ???????????????But I am a little confused as to the 9.8 newtons in question 8...it seems you would incorporate this into the calculation somehow???????????????????
9.8 Newtons came out of a previous situation in which the acceleration was 9.8 m/s^2. That isn't the acceleration here, so 9.8 Newtons isn't relevant to this question.
9.8 Newtons is the force exerted by gravity on 1 kg.
This is because 1 Newton of force accelerates 1 kg at 1 m/s^2, and gravity accelerates a mass at 9.8 m/s^2.
1 Newton accelerates 1 kg at 1 m/s^2, so 9.8 N are required to accelerate 1 kg at 9.8 m/s^2.
However the number 9.8 doesn't enter
into the current situation at all.
We are asked in this question how much force is required to accelerate 5 kg at 4
m/s^2. We could use the following chain of reasoning:
1 Newton accelerates 1 kg at 1 m/s^2, so to accelerate 1 kg at 4 m/s^2 requires 4 Newtons.
4 Newtons accelerate 1 kg at 4 m/s^2, so to accelerate 5 ig at 4 m/s^2 takes 5 times as much, or 20 Newtons.
Our calculation
5 kg * 4 m/s^2 = 20 Newtons
summarizes this reasoning process.
The result of our reasoning can be generalized:
To get the force required to give a given mass a given acceleration, we multiply the mass by the acceleration. ------------------------------------------------ Self-critique rating #$&*:1 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q010. How much force would be required to accelerate the 1200 kg automobile at a rate of 2 m/s^2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F= 1200 * 2 = 2400newtons confidence rating #$&*: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): I got it using f=m*a But I am still a little confused on the 9.8 newtons in question 8 ------------------------------------------------ Self-critique rating #$&*:3 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? mass=8kg 3/50=.06= slope 8* .06= .48 I am not sure about this...I think I have this wrong? The direction of force is down the incline, I think that the slope is the acceleration, but I am not sure. I know that the Force= the mass * the accleration and the mass is 8 kg. I am unsure of the accleration.???????????????????????????????????????
You're on the right track.
The weight of the 8 kg object is the force exerted on it by gravity, which is 8 kg * 9.8 m/s^2 = 79 N, approx.. The component of the gravitational force down the incline, for small slopes, is very nearly equal to slope * weight. So for the present slope, which is small, we have component of gravitational force down incline = slope * weight = .06 * 79 N = 4.7 N, approx..Excellent questions. See my notes, and let me know if you have additional questions.