Open query 9

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course Phy 121

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

NOTE: Parts of this qa involve a cart on a ramp. The situation is on video in Class Notes #9. You don't need to worry about the analysis, but if you are having trouble visualizing the situation, a quick look at the videos of the cart setup might be helpful.

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Question: `q001. Note that there are 13 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:

The mass of the cart is equivalent to another 25 of there 25 masses. Each mass is 1/50 = 0.02 total mass. The first 10 points are (.02, 21cm/s^2), (.04, 48 cm/s^2), (.06, 55cm/s^2), (.08, 85 cm/s^2), (.10, 101cm/s^2), (.12, 125cm/s^2), (.14, 141cm/s^2), (.16, 171cm/s^2), (.18, 183cm/s^2), and (.20, 190 cm/s^2).

The best fit line is 925x + 12.8 which is 925 for slope and 12.8 for the y - intercept.

confidence rating #$&*:

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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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Self-critique (if necessary):

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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:

Slope of the line is around 925cm/s^2. The points are randomly scattered around the line.

confidence rating #$&*:

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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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Self-critique (if necessary):

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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:

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.

confidence rating #$&*:

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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:

The gravitational forces exerted on the system are exerted two objects: gravitational force is exerted on the suspended mass and gravitational force is exerted on the cart.

The gravitational force on the cart and the masses in it therefore does not affect the acceleration of the system. 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.

confidence rating #$&*:

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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.

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 accelerations obtained are all about the same, with only about a 10% variation between the lowest and the highest.

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.

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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:

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.

1 force unit acting on 1 mass unit results in an acceleration of 1 m/s^2.

confidence rating #$&*:

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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:

It must exert a force of 9.8 Newtons on a mass unit

confidence rating #$&*:

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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?

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Your solution:

9.8 N on a mass of 1 kg

confidence rating #$&*:

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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:

8 * 9.8 = 78.6 Newtons

confidence rating #$&*:

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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?

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Your solution:

5 * 4 = 20 times the force = 20 Newtons

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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:

F = m * a = 1200 kg * 2m/s^2 = 2400kg * m/s^2 = 2400 Newtons.

confidence rating #$&*:

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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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Question: `q011. A net force F_net accelerates a certain mass m at 30 cm/s^2.

What would be the acceleration of the same mass if subjected to a net force twice as great?

What would be the acceleration if the original net force acted on a mass half as great?

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Your solution:

60 Force Units for the net force twice as great and 15 Force Units for the net force half as great.

confidence rating #$&*:

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Question: `q012. What net force would be required to accelerate a 50 kg mass at 4 m/s^2?

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Your solution:

50 kg * 4m/s^2 = 200kg * m/s^2 = 200 Newtons

confidence rating #$&*:

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Question: `q013. What would be the acceleration of a 40 kg mass subjected to a net force of 20 Newtons?

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Your solution:

20N = 40 kg * m/s^2 = 0.5m/s^2

confidence rating #$&*:

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Question: `q015. What is the mass of an object which when subjected to a net force of 100 Newtons accelerates at 4 m/s^2?

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Your solution:

100 N = kg * 4m/s^2 = 25kg

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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? "

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Question: `q011. A net force F_net accelerates a certain mass m at 30 cm/s^2.

What would be the acceleration of the same mass if subjected to a net force twice as great?

What would be the acceleration if the original net force acted on a mass half as great?

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Your solution:

60 Force Units for the net force twice as great and 15 Force Units for the net force half as great.

confidence rating #$&*:

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Question: `q012. What net force would be required to accelerate a 50 kg mass at 4 m/s^2?

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Your solution:

50 kg * 4m/s^2 = 200kg * m/s^2 = 200 Newtons

confidence rating #$&*:

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Question: `q013. What would be the acceleration of a 40 kg mass subjected to a net force of 20 Newtons?

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Your solution:

20N = 40 kg * m/s^2 = 0.5m/s^2

confidence rating #$&*:

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Question: `q015. What is the mass of an object which when subjected to a net force of 100 Newtons accelerates at 4 m/s^2?

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Your solution:

100 N = kg * 4m/s^2 = 25kg

confidence rating #$&*:

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Questions related to 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? "

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Open query 9

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