open qa-assingnment 9

course Phy 231

6/21 11:30

009. Forces exerted by gravity; nature of force; units of force

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Question: `q001. Note that there are 10 questions in this set.

You have done the Introductory Force Experiment in which you used rubber bands and bags of water, and you understand that, at

least in the vicinity of the Earth's surface, gravity exerts downward forces. You have also seen that forces can be measured

in units called Newtons. However you were not given the meaning and definition of the Newton as a unit of force. You also

know that the standard unit in which mass is measured is the kilogram.

Here we are going to develop, in terms of an experiment, the meaning of the Newton as a force unit.

We begin with a cart containing a number of masses.

• We suppose the cart contains 25 equal masses. We will call these the 'small masses'.

• The cart itself is equal in mass to the combined total of the 25 small masses (this can be verified by balancing them

on a beam at equal distances from a fulcrum, the 25 small masses on one end and the cart on the other). So the entire system

has a mass which is 50 times that of the small masses.

• The cart is placed on a slight downward incline and a weight hanger of negligible mass is attached to the cart by a

light string and suspended over a low-friction pulley at the lower end of the ramp.

The incline is adjusted until the cart, when given a slight push in the direction of the pulley, is observed to move with

unchanging, or constant, velocity (and therefore zero acceleration).

• At this slope the weight of the cart (which acts vertically downward), the force exerted by the incline in response

to the weight (which acts perpendicular, or normal to the incline and is called the normal force) and the frictional force

(which acts in the direction opposite motion) are all in balance. That is the sum of all these forces is zero.

• If any additional force is exerted in the direction of motion, that force will therefore be the net force.

The small masses are then moved one at a time from the cart to the hanger.

• In each case we regard the system as the cart and the suspended masses.

• Transfer of a mass from cart to hanger doesn't change the mass of the system, since the transferred mass is just

moved from one part of the system to the other.

•

The gravitational, normal and frictional forces on cart and the masses that remain in it still add up to zero, for the same

reasons as before.

However the suspended mass is no longer in the cart, and the force exerted on it by gravity is no longer balanced by the

normal and frictional forces.

• The net force on the system is therefore now equal to the gravitational force on the suspended mass.

As masses are transferred one at a time, the system is therefore accelerated first by the force of gravity on one of the

masses, then by the the gravitational force on two of the masses, etc..

• With the transfer of each mass we observer the time required for the system to accelerate from rest through a chosen

displacement.

The acceleration of each system is then calculated, using the data for that system.

• The acceleration is graphed vs. the proportion of the total mass of the system which is suspended over the pulley.

Note that if the entire mass of the system, including the cart, is placed on the weight hanger, there will be no mass left on

the incline and the entire weight will fall freely under the acceleration of gravity.

The proportion of the total mass of the system which is now on the hanger is calculated for each system.

• recall that the entire system is equivalent in mass to 50 of the small masses

• so for example if a single small mass is on the hanger, that corresponds to 1/50 the mass of the system and the

proportion of the mass which is suspended is 1/50 = .02

• if two small masses are on the hanger the proportion is 2/50 = .04, etc.

Suppose the data points obtained for the second, fourth, sixth, eighth and tenth systems were

• (.04, 48 cm/s^2), (.08, 85 cm/s^2), (.12, 125 cm/s^2), (.16, 171 cm/s^2), (.20, 190 cm/s^2)

Sketch these points on an accurate graph of acceleration vs. proportion of weight suspended and determine the slope and

y-intercept of the line.

• What is your slope and what is the y intercept?

• What is the equation of the line?

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Your solution:

(.04, 48 cm/s^2), (.20, 190 cm/s^2)

142/.16 slope is 887.5

(.08, 85 cm/s^2), (.12, 125 cm/s^2)

Slope is 1000

So average slope is 943

According to the solution the best fit is 925

Thus the best equation is 925x+12.8

Confidence rating #$&* 3

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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.

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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?

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Your solution:

Seem to be randomly scattered around the line

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.

STUDENT QUESTION

Is this because the best fit line does not completely go through all the data points, even though it looks linear, or am I

plotting wrong??????????????????????????

INSTRUCTOR RESPONSE

The slope you gave in the first problem is the average slope between two actual data points. The line through these two data

points is unlikely to be the best fit to the data. You should have sketched the line you think fits the data best, coming as

close as possible on the average to the data points (and likely not actually going through any of the data points), then used

two points on this line to calculate the slope.

However, whatever straight line you used, it won't go through all of the data points, because they don't all lie on a

straight line.

So there is some variation in the straight lines people will sketch in their attempt to approximate the best-fit line, and

there is variation in the slopes of these estimated lines.

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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?

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Your solution:

This is not the case with my graph my (0,12.8) 12.8 being my y intercept

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.

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Question: `q003. What is it that causes the system to accelerate more when a greater proportion of the mass is suspended?

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Your solution:

When twice as many masses there is twice the force. Meaning a greater gravitational force on the suspended masses that

causes the greater acceleration.

Confidence rating #$&* 3

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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.

STUDENT QUESTION

I am pretty sure I understand this question. Basically the only thing acting in this equation

to cause a different net force or any type of acceleration is gravity?

INSTRUCTOR RESPONSE

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.

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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?

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Your solution:

The forces required to achieve the accelerate amount by the above does seem to be seen.

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.

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Self-critique (if necessary):1

I found this hard to follow due to the ups and down between 1-5 it seemed to me that the added masses caused accelerations

that weren’t steady increases that I expected to see.

Self-critique rating #$&*2

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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?

STUDENT QUESTION

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

INSTRUCTOR COMMENT

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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Your solution:

Gravity must exert 9.8 force unit to accelerate a 1 mass unit 9.8m/s^2

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.

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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:

9.8N of force/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.

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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:

9.8N/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.

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Question: `q008. How much force would gravity exert on a mass of 8 kg?

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Your solution:

78.4N

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.

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.

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Question: `q009. How much force would be required to accelerate a mass of 5 kg at 4 m/s^2?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

1N/kg=1m/s

5kg and 4m/s would require 20N

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.

STUDENT QUESTION

9.8 = 5kg * 4m/s^2=9.8newtons = 20 kg

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?

INSTRUCTOR RESPONSE

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.

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Question: `q010. How much force would be required to accelerate the 1200 kg automobile at a rate of 2 m/s^2?

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Your solution:

1N/1kg=1m/s

2*1200=2400N

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.

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"

Good.

Those 'ups and downs' you mentioned appear to have been the expected statistical fluctuations in actual data. These would be smoothed out by fitting a straight line to the data. I believe you understand this, but let me know if not.

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