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SPECIFICS OF FORCE
In review, we've looked at forces and put them into two categories: contact and non contact. In addition, we have examined what happens to objects when combinations of forces are applied. We have seen that even though a number of forces may act on an object, if the forces cancel each other, that object will keep moving at the same speed in a straight line (keep its state of motion). Along the same line, if the forces don't cancel, a net force will result and that object will change its state of motion (i.e. accelerate). In addition, we have come to realize that net forces not only cause changes in translation, but they can cause changes in rotation as well. When a force is applied at some distance away from an axis of rotation, it can create a torque, which alters the rotational state of motion of the object (angular acceleration). We've developed techniques that make it easier for us to setup and solve problems. Free body diagrams and setting up Newton's 2nd law equations will remain with you for the remainder of the year. What we haven't done with force is look inside what creates the forces themselves. Non contact forces are created by field interactions which can be evaluated in terms of the physical characteristics of the objects. Contact forces occur as a result of the direct interaction of two objects. Consequently, our analysis of force can get a little more in depth. Net Force and Units Although I have referred to these concepts already , it never hurts to re-examine them in more detail to gain mastery of the subject. Net forces result from combinations of forces that don't cancel each other out. When this occurs, an object's state of motion is changing over time, what we have come to call acceleration. The net force is characterized by the product of the mass of the object experiencing the net force and it's acceleration. Or, more simply put, ![]() To analyze the units of force, we need a little background. This year we will use the SI (international standard--metric system) of measurement. Within this system there are two common sets of measurements: cgs (centimeter-gram-second) for small puny things--like Rattet, and mks (meter-kilogram-second) for large scale things--like Quiocho. For our purposes, it will be most convenient to use the mks system of nomenclature (names). In the mks system, the unit of mass is the kilogram (kg) , length is the meter (m), and time is the second (s) . In mechanics, all of the other units result from combinations of these. For example, as we study kinematics, you find that velocity has units of meters per second (m/s) and acceleration has units of meters per second squared (m/s2) . With this knowledge, we can figure out what the units of force ought to be. Since F = ma, and m has units of kg, while a has units of m/s2, then force must have units of kg•m/s2. Well , any real physicist is far too lazy to write down kg•m/s2 every time they need to express the units of force, so we invented a name for the unit of force--the newton (N). Thus, whenever you compute any force, net or otherwise, the units that follow a numerical answer ought to be newtons (N), so long as you are working with the mks system. For argument's sake: Earlier we showed that the net force acting on an object traveling in a circular path has a special version of "ma", which is ![]() Show that this expression yields the same units as "ma". Non Contact Forces All non contact forces occur as a result of the interaction of fields. As previously discussed, three non contact forces exist: gravity, electric, and magnetic. We will examine each of these "fields" of study in depth. Gravitation: Any thing that has mass has a gravitational field surrounding it. This field spreads out radially, forever, in all directions. As far as we know the gravitational field surrounding any object extends into the farthest reaches of the universe. Anytime two fields interact, a gravitational force is exerted on the objects generating those fields. Consequently, nothing is ever truly "weightless," because its field interacts with all of the other fields in the universe, no matter its location. Since mass is what generates a field, it mak es sense that the more mass an object has, the larger the field is that surrounds that object. As far as we know, all gravitational fields are attractive. The result of which implies that gravitational fields do not have a net effect of canceling each other out at great distances. That is, the combined field from two masses, when observed from a very large distance from the masses, appears like the field from one, larger mass. However, if we want to evaluate gravitational fields and how they interact with each other, then we need to know how big they are at some point. So, the other factor that needs consideration is distance. How does distance affect the size of the gravitational field? If you were to take an educated guess, you'd probably say that the field would decrease in strength as you move farther away from an object. Whether it would decrease as 1/r (where r represents the distance away from the object), 1/r2, o r 1/r3, is a little harder to guess. In order to understand what the relationship actually is we need to look at the geometry surrounding an object in 3 dimensions. To assist, I'm going to attempt to come up with a fitting analogy . Imagine, if you will , a ball of chalk dust with a tiny explosive device placed in the center of the ball. You then take the ball into space such that it is far from any other object that might influence how it scatters once it explodes. Around the ball you setup 5 observation shells that will record all of the dust particles that pass through them. Each observation shell forms a spherical shell around the ball at one meter intervals. When the ball is exploded, it spreads out evenly in all directions and eventually drifts beyond all of the observation points. All of the shells count the same number of dust particles that passed by. However, how close the dust particles were to each other at each shell varied tremendously. In fact, since the area of each observation shell can be computed as A = 4πr2, the density of the particles at the 1 meter shell were 4 times greater than the 2 meter shell , 9 times greater than the 3 meter shell, 16 times greater than the 4 meter shell , and 25 times greater than the 5 meter shell . Therefore, the particle density decreased with the square of the distance. The number of dust particles passing an observation point are analogous to something called gravitational flux. The density of the dust at a particular shell equates to gravitational field strength. Consequently , as you get farther away from a mass, the strength of its field decreases with the square of the distance. Equation wise, the field strength appears as: ![]() where m is the mass and r is the distance from the center of the mass. Notice that the field strength is only proportional to and not equal to m/r2. The reason for this has to do with the units we chose to measure gravitational field. It would be possible to manipulate the values of the units such that an equality resulted, but it is simpler to keep the same units of measurement and introduce something called a proportionality constant. Essentially, what a proportionality constant does, is change a proportional equation to an equality . So, if you graphed field strength versus m/r2, you'd get a direct variation graph with a small slope. The small slope is your proportionality constant. For gravitation, the value of the proportionality constant is 6.67 x 10-11 N•m2/kg2. The number has a special name--the universal gravitation constant. The stuff behind the number are merely the units of the constant. The significance of this number is that it applies to any gravitational field or force interaction in the universe. In reality , the constant was not determined by a field strength vs. m/r2 graph, but a gravitational force versus m1m2/r2 graph. Even so, the concepts behind how the constant was found are the same. In 1798 a guy named Henry Cavendish used a torsional balance (this is a scale that measures force by an axle being twisted) to determine the value of the universal gravitation constant. By bringing two large known masses close to two small known masses suspended from the torsional axle, the gravitational force between them would cause the axle to twist until it resisted being twisted further. Thus, he could record the force and then carefully measure the distance that separated the centers of the masses. After taking gazillions of measurements he was able to graph the gravitational force versus m1m2/r2 data, establish a line of best fit, and arrive at a value for the universal gravitational constant. Not much different than what you were supposed to do in your spring lab (except for the gazillions of measurements part). ![]() Cavendish was working from the model of gravitation proposed by Isaac Newton in his theory of gravitation. Wherein, Newton proposed that all objects are gravitationally attracted to each other according to the product of the masses divided by the square of the distance between their centers of mass. Or, in equation form: ![]() Cavendish's contribution allows us to write a more exact version of this equation. ![]() Once again physicists are pretty lazy , so we called the co nstant "G", thus making the equation: ![]() The only thing missing here is a direction. Since gravity always pulls one object toward the center of the other (and vice versa), regardless of where that object is located, then the force must be radially directed. EXERCISES: | |
| 1. |
Determine the initial acceleration of a cat dropped from a high building
near the surface of the earth. Use 6.37 X 106 m for the radius between the
masses, 10 kg for the mass of the cat, and 5.98 X 1024 kg for the mass of the earth. |
| a) |
Repeat this for a dog(m = 20 kg), a high school principal (m= 80 kg),
and an elephant (m= 4000 kg). | |
| b) |
Explain how you can justify doing no further calculations to arrive
at an expression for the acceleration of any object near the surface
of the earth. | |
| c) |
What physical quantity that we discussed looked a lot like the
expression for acceleration (before you plugged in the numbers)? |
| 2. |
Given the distance between the earthand moon is 3.84 x 108 m and the mass
of the moon is 7.36 x 1022 kg, determine the speed of the moon as it orbits
the earth. |
| 3. |
Using the information from #2, determine the time it takes the moon to
orbit the earth once. |
| 4. |
Show that the acceleration due to gravity near the earth's surface is
independent of the mass being accelerated. |
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A special case for gravitation is what occurs near the surface of the earth. As shown in the
previous exercise, acceleration due to gravity is independent of mass and maintains a value of
approximately 9.8 m/s2. Although the force due to gravity anywhere is frequently referred to
as weight, it is most often near the surface of the earth where the "weight" has the equation: Applications for gravitational force equations: |
| I. |
W = mg | |
| a. |
Used in a FBD for most anything near the surface of the earth. | |
| b. |
Used to determine mass when the weight of an object is casually mentioned. | |
| c. |
Used to determine weight when mass is casually mentioned. | |
| d. |
Typically involved in linear translation and torque problems. | |
| II. |
![]() | |
| a. |
Used to determined the force on a mass from any other mass or group of masses. | |
| b. |
Used in planetary problems. | |
| c. |
Used in proportion problems (the next section). | |
| d. |
Can be used in linear or circular problems, but not torque problems. | |
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PROPORTION PROBLEMS: Proportion problems use known information in conjunction with certain equations. Using the knowledge of the variables in the equation, you then have to determine the size of the force or acceleration without performing a lot of number crunching. The AP test makes like to throw in proportion problems, because they believe that you need to really understand the physics behind the problem in order to do these problems. No comment. A typical problem reads: What is the acceleration due to gravity near the surface of a planet whose mass is twice earth's and whose radius is half earth's? You need to realize three things. First, the acceleration due to gravity is about 10 m/s2 on earth. Second, acceleration due to gravity is always directly proportional to the mass of the planet. Third, acceleration due to gravity is inversely proportional to the square of the distance from the center of the earth. Once you realize these three things, you can reason out what the acceleration must be at the surface of the other planet. Here's the reasoning for the previously stated problem: The planet has twice earth's mass, twice the mass would double the acceleration making it 20 m/s2, since a ∝ m. In addition, the radius is cut in half. Given that a ∝ 1/r2, then halving the radius would also increase the acceleration, this time by a multiple of 4, thus making the acceleration 80 m/s2 at the at the surface of the unknown planet (10 times 2 times 4) . If pure math works better than reasoning for you, then the previous problem could be solved as follows: ![]() Try a couple for yourself: Acceleration Proportion Problems They AP test makers don't restrict themselves to acceleration, they also love to work with gravitational forces in similar capacities. In one typical problem, they take a person or object of known weight on earth and place them on another planet. This type of problem work very similar to the acceleration problem, but it can have a twist. Sample 1: Cheez (m = 100 kg) is sent to investigate another planet, that is made of wiz. Upon his arrival he finds that the mass of the planet is 2/3 earth's and its radius is 3/4 earth's. What is Cheez' weight on that planet? First, find Cheez' weight on earth: W = mg = 100(10) = 1000N Then, look at the relation: ![]() If Cheez' mass is 100 kg and remains the same then the approach is exactly the same as acceleration. Let's take the mathematical approach to solve the problem this time. ![]() However, the problems can be made a tiny bit tric kier, as seen in the following example: Sample 2: Cheez , who weighs 1000N on planet X, is sent to planet Y whose radius and mass is twice that of X. If he weighs 800N on planet Y, by how much did his all donut diet increase or decrease his mass in his flight over to planet Y? You are still going to use the relation: ![]() , only this time Cheez' mass (m1) will be in the picture. In this case the setup looks like this: ![]() The mass ratio of Cheez' new mass to old is 1.6, or Cheez increased his mass by a factor of 1.6 by consuming donuts on his journey. Try a few yourself: Weight Proportion Problems PROBLEMS: |
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| 1. |
Which exerts a greater force on the other, the earth or the moon? Need a hint? Check your answer? |
| 2. |
Which direction do all gravitational forces act? Need a hint? Check your answer? |
| 3. |
What sets of coordinates have we studied thus far and how do they differ? Need a hint? Check your answer? |
| 4. |
Is it possible for anything to be truly weightless? Why ? Need a hint? Check your answer? |
| 5. |
How many forces act on the moon to keep it orbiting the earth? Need a hint? Check your answer? |
| 6. |
What type of motion is the moo n's orbit? Need a hint? Check your answer? |
| 7. |
If the moon's speed is constant what is the angle between the net force vector and the
velocity vector? Need a hint? Check your answer? |
| 8. |
Briefly describe, in your own words, what a field is and why a field is inversely
proportional to the distance squared. Need a hint? Check your answer? |
| 9. |
The acceleration due to gravity on the moon is 1/6th that of the earth. What is the mass
of a 10 kg object on the moon? Need a hint? Check your answer? |
| 10. |
What is the weight of a 10 kg object on the moon? Need a hint? Check your answer? |
| 11. |
How can you determine the distance around one complete orbit? What would you need
to know to calculate the average speed of the orbit? Need a hint? Check your answer? |
| 12. |
The mass of the moon was only guessed at until a satellite orbited it. Once a satellite
completed an orbit, the mass was computed as 7.36 x 1022 kg. If the period (time to go
around once) was 2.5 hours, what was the radius of the satellite's orbit? Need a hint? Check your answer? |
| 13. |
Two objects hang from a wheel with an axle as shown below. If
the wheel/axle combination have a moment of inertia, I, of 25
kg•m2, the smaller radius is 0.5 m, the larger radius is 2.0 m,
and the masses of the blocks are 3 kg and 1 kg respectively,
what is the direction of the rotation and size of the angular
acceleration?![]() Need a hint? Check your answer? |
| 14. |
A 5 kg mass is separated by 3 meters from a 2 kg mass. a) What is the net force on the 5 kg mass? b) What is the net force on the 2 kg mass? Need a hint? Check your answer? |
| 15. |
Three masses are placed on an axis as shown below. Mass
A has a mass of 2 kg, mass B has a mass of 5 kg, and mass
C has a mass of 3 kg. Determine the size and direction of
the force and the acceleration experienced by mass C.![]() Need a hint? Check your answer? |
| 16. |
Two masses are separated by 6 m. The left mass is 4 kg
and the right mass is 6 kg, where could a third mass be
placed such that the net force on that mass is zero?![]() Need a hint? Check your answer? |
| 17. |
The radius of the moon's orbit is 3.84 x 108 m and it takes the moon 27.3 days to orbit
the earth. Using this information, find the mass of the earth. Need a hint? Check your answer? |
| 18. |
A geosynchronous orbit is one where a satellite maintains its position over the same spot
on the earth continuously. In other words the satellite orbits once every 24 hours. At
what distance from the center of the earth must the satellite be placed such that it is in a
geosynchronous orbit? Need a hint? Check your answer? |
| 19. |
Ralph (mass = 120 kg) sojourns to another planet whose mass is three earth's and radius
is 4 times earth's. What is the acceleration due to gravity on that planet? What is Ralph's
weight on that planet? Need a hint? Check your answer? |
| 20. |
Describe a general procedure for finding a proportionality constant , like G. Need a hint? Check your answer? |