NOTES: Newton's Laws ( a story of a boy meeting an apple)

Isaac Newton (a 17th century scientist) put forth a variety of laws that explain why objects move (or don't move) as they do. These three laws have become known as Newton's three laws of motion. 

There is a popular story that Newton was sitting under an apple tree, an apple fell on his head, and he suddenly thought of the Universal Law of Gravitation. As in all such legends, this is almost certainly not true in its details, but the story contains elements of what actually happened.

The most famous anecdote of discovery in science is how Newton’s apple inspired the notion of gravity as a universal force. In 1752, William Stukeley set down the story in his manuscript “Memoirs of Sir Isaac Newton’s life”, as it was told to him by Newton himself. “Why should that apple always descend perpendicularly to the ground, thought he to himself, occasioned by the fall of an apple, as he sat in contemplative mood.”  http://blog.physicsworld.com/2010/01/18/newtons-apple-the-birth-of-a/ 
http://royalsociety.org/library/moments/newton-apple/

THE LAW OF INERTIA

Up to the time of Galileo, it was thought that one must exert force in order to cause and preserve motion, as claimed by the physics of Aristotle. Indeed, when we look at the world surrounding us, we see that in order to continue movement we must exert force. Thus, for example, in order to conserve the speed of a car, the engine must work. Objects on which no force is exerted to preserve their movement eventually come to a stop. Galileo understood that one can explain the stopping of bodies by the common experience that we always encounter a force of friction which resists the motion of bodies. However, without such resistance force, the bodies would continue to move at their previous speed.

BUT:

An object at rest stays at rest and an object in motion stays in motion with the same

 speed and in the same direction unless acted upon by an unbalanced force. 

There are two parts to this statement - one that predicts the behavior of stationary objects and the other that predicts the behavior of moving objects. The two parts are summarized in the following diagram.

There are many more applications of Newton's first law of motion. Several applications are listed below. Perhaps you could think about the law of inertia and provide explanations for each application.

• Blood rushes from your head to your feet while quickly stopping when riding on a descending elevator.
• The head of a hammer can be tightened onto the wooden handle by banging the bottom of the handle against a hard surface
• To dislodge ketchup from the bottom of a ketchup bottle, it is often turned upside down and thrusted downward at high speeds and then abruptly halted.
• Headrests are placed in cars to prevent whiplash injuries during rear-end collisions.
• While riding a skateboard (or wagon or bicycle), you fly forward off the board when hitting a curb or rock or other object that abruptly halts the motion of the skateboard.

First Experiment:

                        Conclusions:

The object /child/ will roll down with increasing speed. It begins at rest, i.e., its speed is equal to zero, and then gradually gathers speed. The longer the inclined plane, the greater its speed.
We call this increase in speed "acceleration." The opposite situation, in which a body gradually slows down, is known as "deceleration." Thus, we have seen that a body moving down an inclined plane accelerates downward.
What will happen to the child after we give it an upward push? The speed of the child rolling up an inclined plane will gradually decrease, while that of a child rolling down will gradually increase.

Second Experiment:

Is there a difference between a bicycle ride along a moderate incline and a steep incline?
The more moderate a plane, the slower the acceleration of a body along it: i.e., the body's speed will increase at a slower rate up the plane. The more moderate the incline, the lower the deceleration of the body and the greater the distance it traverses.
We know that the force of gravity pulls heavy objects downward, toward the center of the earth. For this reason, bodies on an inclined plane are drawn downward. The smaller the incline, the lower the acceleration of the bodies moving down the plane and the lower the deceleration of objects moving up the plane.

(http://muse.tau.ac.il/museum/galileo/the_law_of_inertia.html)

nertia is one of the greatest enemies of the human race. You can explain about 90% or so of people’s actions if you apply the principle of inertia to psychology. Call it “maintaining the status quo” or “going along to get along” or “just floating” if you want to, but it’s all inertia.

Why do people spend hours aimlessly surfing the Internet? Again, it’s inertia. To stop and do something else would require their will to be a stronger force than the draw of the Web. If you think about it, StumbleUpon and sites like it are counting on the principle of human inertia.

Why will a man work at a job he utterly despises with people he wouldn’t give the air in a jug to if they were dying? It’s a safe, known quantity. Finding another job requires exerting a force to overcome the stable — if soulkilling — station he’s in. It’s often passed off by saying “the devil you know is better than the one you don’t,” but is it really? Most people never find out unless a greater force — say massive corporate cutbacks — acts to move them.

Inertia explains why people stay in houses they’d like to move away from, relationships with people they’d like to get away from, and activities they don’t really enjoy. Changing — overcoming personal inertia — takes more force of will than most people today can muster. It’s tons easier to ride along in the ruts than it is to pull off onto a side street. Funny thing about ruts, especially deep ones, they look just like a grave with both ends knocked out.

People are fat because it takes too much willpower to overcome the inertia. Junkies stay on drugs because it’s easy to keep moving and difficult to stop. People don’t go back to school because of inertia. In most anything it’s just easier to “let nature take its course” than it is to exert any real control over our lives and I admit to being the chiefest of sinners.

(http://grocerystorefeet.wordpress.com/2011/10/12/newtons-first-law/)

Newton's conception of inertia stood in direct opposition to more popular conceptions about motion. The dominant thought prior to Newton's day was that it was the natural tendency of objects to come to a rest position. Moving objects, so it was believed, would eventually stop moving; a force was necessary to keep an object moving. But if left to itself, a moving object would eventually come to rest and an object at rest would stay at rest; thus, the idea that dominated people's thinking for nearly 2000 years prior to Newton was that it was the natural tendency of all objects to assume a rest position.

WILL IT STOP!?!?!?!

Galileo and the Concept of Inertia

Galileo, a premier scientist in the seventeenth century, developed the concept of inertia. Galileo reasoned that moving objects eventually stop because of a force called friction. In experiments using a pair of inclined planes facing each other, Galileo observed that a ball would roll down one plane and up the opposite plane to approximately the same height. If smoother planes were used, the ball would roll up the opposite plane even closer to the original height. Galileo reasoned that any difference between initial and final heights was due to the presence of friction. Galileo postulated that if friction could be entirely eliminated, then the ball would reach exactly the same height.

Galileo further observed that regardless of the angle at which the planes were oriented, the final height was almost always equal to the initial height. If the slope of the opposite incline were reduced, then the ball would roll a further distance in order to reach that original height.

Galileo's reasoning continued - if the opposite incline were elevated at nearly a 0-degree angle, then the ball would roll almost forever in an effort to reach the original height. And if the opposing incline was not even inclined at all (that is, if it were oriented along the horizontal), then ... an object in motion would continue in motion... .

Forces Don't Keep Objects Moving

Isaac Newton built on Galileo's thoughts about motion. Newton's first law of motion declares that a force is not needed to keep an object in motion. Slide a book across a table and watch it slide to a rest position. The book in motion on the table top does not come to a rest position because of the absence of a force; rather it is the presence of a force - that force being the force of friction - that brings the book to a rest position. In the absence of a force of friction, the book would continue in motion with the same speed and direction - forever! (Or at least to the end of the table top.) A force is not required to keep a moving book in motion. In actuality, it is a force that brings the book to rest.

Mass as a Measure of the Amount of Inertia

All objects resist changes in their state of motion. All objects have this tendency - they have inertia. But do some objects have more of a tendency to resist changes than others? Absolutely yes! The tendency of an object to resist changes in its state of motion varies with mass. Mass is that quantity that is solely dependent upon the inertia of an object. The more inertia that an object has, the more mass that it has. A more massive object has a greater tendency to resist changes in its state of motion.

A common physics demonstration relies on this principle that the more massive the object, the more that object resist changes in its state of motion. The demonstration goes as follows: several massive books are placed upon a teacher's head. A wooden board is placed on top of the books and a hammer is used to drive a nail into the board. Due to the large mass of the books, the force of the hammer is sufficiently resisted (inertia). This is demonstrated by the fact that the teacher does not feel the hammer blow. (Of course, this story may explain many of the observations that you previously have made concerning your "weird physics teacher.") A common variation of this demonstration involves breaking a brick over the teacher's hand using the swift blow of a hammer. The massive bricks resist the force and the hand is not hurt. (CAUTION: do not try these demonstrations at home.)

http://www.youtube.com/watch?v=21itZZ_DvwM

QUESTIONS:

1. Imagine a place in the cosmos far from all gravitational and frictional influences. Suppose that you visit that place (just suppose) and throw a rock. The rock will

a. gradually stop.

b. continue in motion in the same direction at constant speed.

A: According to Newton's first law, the rock will continue in motion in the same direction at constant speed.

2. A 2-kg object is moving horizontally with a speed of 4 m/s. How much net force is required to keep the object moving at this speed and in this direction?

A: Answer: 0 N

An object in motion will maintain its state of motion. The presence of an unbalanced force changes the velocity of the object.

3. Mac and Tosh are arguing in the cafeteria. Mac says that if he flings the Jell-O with a greater speed it will have a greater inertia. Tosh argues that inertia does not depend upon speed, but rather upon mass. Who do you agree with? Explain why.

Tosh is correct. Inertia is that quantity which depends solely upon mass. The more mass, the more inertia. Momentum is another quantity in Physics which depends on both mass and speed. Momentum will be discussed in a later unit.

4. Supposing you were in space in a weightless environment, would it require a force to set an object in motion?

Even in space objects have mass. And if they have mass, they have inertia. That is, an object in space resists changes in its state of motion. A force must be applied to set a stationary object in motion. Newton's laws rule - everywhere!

5. Fred spends most Sunday afternoons at rest on the sofa, watching pro football games and consuming large quantities of food. What effect (if any) does this practice have upon his inertia? Explain.

A:Fred will increase his mass if he makes a habit of this. And if his mass increases, then his inertia increases.

6. Ben Tooclose is being chased through the woods by a bull moose that he was attempting to photograph. The enormous mass of the bull moose is extremely intimidating. Yet, if Ben makes a zigzag pattern through the woods, he will be able to use the large mass of the moose to his own advantage. Explain this in terms of inertia and Newton's first law of motion.

A:The large mass of the bull moose means that the bull moose has a large inertia. Thus, Ben can more easily change his own state of motion (make quick changes in direction) while the moose has extreme difficulty changing its state of motion. Physics for better living!

 7. Two bricks are resting on edge of the lab table. Shirley Sheshort stands on her toes and spots the two bricks. She acquires an intense desire to know which of the two bricks are most massive. Since Shirley is vertically challenged, she is unable to reach high enough and lift the bricks; she can however reach high enough to give the bricks a push. Discuss how the process of pushing the bricks will allow Shirley to determine which of the two bricks is most massive. What difference will Shirley observe and how can this observation lead to the necessary conclusion?

 A: The bricks, like any object, possess inertia. That is, the bricks will resist changes in their state of motion. If Shirley gives them a push, then the bricks will offer resistance to this push. The one with the most mass will be the one with the most inertia. This will be the brick which offers the most resistance. This very method of detecting the mass of an object can be used on Earth as well as in locations where gravitational forces are negligible for bricks.

Inertia: tendency of an object to resist changes in its velocity.

 An object in motion with a velocity of 2 m/s, East will (in the absence of an unbalanced force) remain in motion with a velocity of 2 m/s, East. Such an object will not change its state of motion (i.e., velocity) unless acted upon by an unbalanced force. Objects resist changes in their velocity.

1. A group of physics teachers is taking some time off for a little putt-putt golf. The 15th hole at the Hole-In-One Putt-Putt Golf Course has a large metal rim that putters must use to guide their ball towards the hole. Mr. S guides a golf ball around the metal rim When the ball leaves the rim, which path (1, 2, or 3) will the golf ball follow?

The answer is 2. Once leaving the rim, the ball will follow an "inertial path" (i.e., a straight line). At the instant shown in the diagram, the ball is moving to the right; once leaving the rim, there is no more unbalanced forces to change its state of motion. Paths 1 and 3 both show the ball continually changing its direction once leaving the rim.

2. A 4.0-kg object is moving across a friction-free surface with a constant velocity of 2 m/s. Which one of the following horizontal forces is necessary to maintain this state of motion?

a. 0 N

b. 0.5 N

c. 2.0 N

d. 8.0 N

If an object is in motion, then it will stay in motion with those very same motion characteristics. It doesn't take any force to maintain that same state of motion. In fact, the presence of a force would "ruin" such a state of motion and cause an acceleration.

Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

Newton's second law of motion can be formally stated as follows:

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

This verbal statement can be expressed in equation form as follows:

a = F_net / m

The above equation is often rearranged to a more familiar form as shown below. The net force is equated to the product of the mass times the acceleration.

::::NET FORCE::::

Determining the Net Force

If you have been reading through Lessons 1 and 2, then Newton's first law of motion ought to be thoroughly understood.

An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

In the statement of Newton's first law, the unbalanced force refers to that force that does not become completely balanced (or canceled) by the other individual forces. If either all the vertical forces (up and down) do not cancel each other and/or all horizontal forces do not cancel each other, then an unbalanced force exists. The existence of an unbalanced force for a given situation can be quickly realized by looking at the free-body diagram for that situation. Free-body diagrams for three situations are shown below. Note that the actual magnitudes of the individual forces are indicated on the diagram.

In each of the above situations, there is an unbalanced force. It is commonly said that in each situation there is a net force acting upon the object. The net force is the vector sum of all the forces that act upon an object. That is to say, the net force is the sum of all the forces, taking into account the fact that a force is a vector and two forces of equal magnitude and opposite direction will cancel each other out. At this point, the rules for summing vectors (such as force vectors) will be kept relatively simple. Observe the following examples of summing two forces:

Observe in the diagram above that a downward vector will provide a partial or full cancellation of an upward vector. And a leftward vector will provide a partial or full cancellation of a rightward vector. The addition of force vectors can be done in the same manner in order to determine the net force (i.e., the vector sum of all the individual forces). Consider the three situations below in which the net force is determined by summing the individual force vectors that are acting upon the objects.

The Big Misconception

So what's the big deal? Many people have known Newton's first lawsince eighth grade (or earlier). And if prompted with the first few words, most people could probably recite the law word for word. And what is so terribly difficult about remembering that F = ma? It seems to be a simple algebraic statement for solving story problems. Thebig deal however is not the ability to recite the first law nor to use the second law to solve problems; but rather the ability to understand their meaning and to believe their implications. While most people know what Newton's laws say, many people do not know what they mean (or simply do not believe what they mean).

The most common misconception is one that dates back for ages; it is the idea that sustaining motion requires a continued force.

The misconception has already been discussed in a previous lesson, but will now be discussed in more detail. This misconception sticks out its ugly head in a number of different ways and at a number of different times. As your read through the following discussion, give careful attention to your own belief systems. View physics as a system of thinking about the world rather than information that can be dumped into your brain without evaluating its consistency with your own belief systems.

Newton's laws declare loudly that a net force (an unbalanced force) causes an acceleration; the acceleration is in the same direction as the net force. To test your own belief system, consider the following question and its answer as seen by clicking the button.

Are You Infected with the Misconception?

Two students are discussing their physics homework prior to class. They are discussing an object that is being acted upon by two individual forces (both in a vertical direction); the free-body diagram for the particular object is shown at the right. During the discussion, Anna Litical suggests to Noah Formula that the object under discussion could be moving. In fact, Anna suggests that if friction and air resistance could be ignored (because of their negligible size), the object could be moving in a horizontal direction. According to Anna, an object experiencing forces as described at the right could be experiencing a horizontal motion as described below.

Noah Formula objects, arguing that the object could not have any horizontal motion if there are only vertical forces acting upon it. Noah claims that the object must be at rest, perhaps on a table or floor. After all, says Noah, an object experiencing a balance of forces will be at rest. Who do you agree with?

Answer: Noah Formula may know his formulas but he does not know (or does not believe) Newton's laws. If the forces acting on an object are balanced and the object is in motion, then it will continue in motion with the same velocity. Remember: forces do not cause motion. Forces cause accelerations.

  

Remember last winter when you went sledding down the hill and across the level surface at the local park? (Apologies are extended to those who live in warmer winter climates.)

 

Imagine a the moment that there was no friction along the level surface from point B to point C and that there was no air resistance to impede your motion. How far would your sled travel? And what would its motion be like? Most students I've talked to quickly answer: the sled would travel forever at constant speed. Without friction or air resistance to slow it down, the sled would continue in motion with the same speed and in the same direction. The forces acting upon the sled from point B to point C would be the normal force (the snow pushes up on the sled) and thegravity force (see diagram at right). These forces are balanced and since the sled is already in motion at point B it will continue in motion with the same speed and direction. So, as in the case of the sled and as in the case of the object that Noah and Anna are discussing, an object can be moving to the right even if the only forces acting upon the object are vertical forces. Forces do not cause motion; forces cause accelerations.

Newton's first law of motion declares that a force is not needed to keep an object in motion. Slide a book across a table and watch it slide to a rest position. The book in motion on the table top does not come to a rest position because of the absence of a force; rather it is the presence of a force - that force being the force of friction - that brings the book to a rest position. In the absence of a force of friction, the book would continue in motion with the same speed and direction - forever (or at least to the end of the table top)! A force is not required to keep a moving book in motion; and a force is not required to keep a moving sled in motion; and a force is not required to keep any object horizontally moving object in motion.

Free Fall and Air Resistance

In a previous unit, it was stated that all objects (regardless of their mass) free fall with the same acceleration - 9.8 m/s/s. This particular acceleration value is so important in physics that it has its own peculiar name - the acceleration of gravity - and its own peculiar symbol - g. But why do all objects free fall at the same rate of acceleration regardless of their mass? Is it because they all weigh the same? ... because they all have the same gravity? ... because the air resistance is the same for each? Why? 

In addition to an exploration of free fall, the motion of objects that encounter air resistance will also be analyzed. In particular, two questions will be explored:

• Why do objects that encounter air resistance ultimately reach a terminal velocity?
• In situations in which there is air resistance, why do more massive objects fall faster than less massive objects?

To answer the above questions, Newton's second law of motion (F_net = m•a) will be applied to analyze the motion of objects that are falling under the sole influence of gravity (free fall) and under the dual influence of gravity and air resistance.

Free Fall Motion

As learned in an earlier unit, free fall is a special type of motion in which the only force acting upon an object is gravity. Objects that are said to be undergoing free fall, are not encountering a significant force of air resistance; they are falling under the sole influence of gravity. Under such conditions, all objects will fall with the same rate of acceleration, regardless of their mass. But why? Consider the free-falling motion of a 1000-kg baby elephant and a 1-kg overgrown mouse.

If Newton's second law were applied to their falling motion, and if a free-body diagram were constructed, then it would be seen that the 1000-kg baby elephant would experiences a greater force of gravity. This greater force of gravity would have a direct affect upon the elephant's acceleration; thus, based on force alone, it might be thought that the 1000-kg baby elephant would accelerate faster. But acceleration depends upon two factors: force and mass. The 1000-kg baby elephant obviously has more mass (or inertia). This increased mass has an inverse affect upon the elephant's acceleration. And thus, the direct affect of greater force on the 1000-kg elephant is offset by the inverse affect of the greater mass of the 1000-kg elephant; and so each object accelerates at the same rate - approximately 10 m/s/s. The ratio of force to mass (F_net/m) is the same for the elephant and the mouse under situations involving free fall.

This ratio (F_net/m) is sometimes called the gravitational field strength and is expressed as 9.8 N/kg (for a location upon Earth's surface). The gravitational field strength is a property of the location within Earth's gravitational field and not a property of the baby elephant nor the mouse. All objects placed upon Earth's surface will experience this amount of force (9.8 N) upon every 1 kilogram of mass within the object. Being a property of the location within Earth's gravitational field and not a property of the free falling object itself, all objects on Earth's surface will experience this amount of force per mass. As such, all objects free fall at the same rate regardless of their mass. Because the 9.8 N/kg gravitational field at Earth's surface causes a 9.8 m/s/s acceleration of any object placed there, we often call this ratio the acceleration of gravity.

Falling with Air Resistance

As an object falls through air, it usually encounters some degree of air resistance. Air resistance is the result of collisions of the object's leading surface with air molecules. The actual amount of air resistance encountered by the object is dependent upon a variety of factors. To keep the topic simple, it can be said that the two most common factors that have a direct affect upon the amount of air resistance are the speed of the object and the cross-sectional area of the object. Increased speeds result in an increased amount of air resistance. Increased cross-sectional areas result in an increased amount of air resistance.

Double Trouble (a.k.a., Two Body Problems)

Our study thus far has been restricted to the analysis of single objects moving under the influence of Newton's laws. But what happens if there are two objects connected together in one way or another? For instance, there could be a tow truck hauling a car down a highway. How is such an analysis conducted? How is the acceleration of the tow truck and the car determined? What about the force acting between the tow truck and the car? In this part of Lesson 3, we will make an attempt to analyze such situations. We will find that the analysis is conducted in the same general manner as when there is one object - through the use of free-body diagrams and Newton's laws.

Situations involving two objects are often referred to as two-body situations. When appearing as physics problems, two-body problems are characterized by a set of two unknown quantities. Most commonly (though not always the case), the two unknowns are the acceleration of the two objects and the force transmitted between the two objects. Two body-problems can typically be approached using one of two basic approaches. One approach involves a combination of a system analysis and an individual body analysis. In the system analysis, the two objects are considered to be a single object moving (or accelerating) together as a whole. The mass of the system is the sum of the mass of the two individual objects. If acceleration is involved, the acceleration of the system is the same as that of the individual objects. A system analysis is usually performed to determine the acceleration of the system. The system analysis is combined with an individual object analysis. In the individual object analysis, either one of the two objects is isolated and considered as a separate, independent object. A free-body diagram is constructed and the individual forces acting upon the object are identified and calculated. An individual object analysis is usually performed in order to determine the value of any force which acts between the two objects - for example, contact forces or tension forces.

The dual combination of a system analysis and an individual object analysis is one of two approaches that are typically used to analyze two-body problems. A second approach involves the use of two separate individual object analyses. In such an approach, free-body diagrams are constructed independently for each object and Newton's second law is used to relate the individual force values to the mass and acceleration. Each individual object analysis generates an equation with an unknown. The result is a system of two equations with two unknowns. The system of equations is solved in order to determine the unknown values.

As a first example of the two approaches to solving two-body problems, consider the following example problem.

Example Problem 1:

A 5.0-kg and a 10.0-kg box are touching each other. A 45.0-N horizontal force is applied to the 5.0-kg box in order to accelerate both boxes across the floor. Ignore friction forces and determine the acceleration of the boxes and the force acting between the boxes.

The first approach to this problem involves the dual combination of a system analysis and an individual object analysis. As mentioned, the system analysis is used to determine the acceleration and the individual object analysis is used to determine the forces acting between the objects. In the system analysis, the two objects are considered to be a single object. The dividing line that separates the objects is ignored. The mass of the system of two objects is 15.0 kg. The free-body diagram for the system is shown at the right. There are three forces acting upon the system - the gravity force (the Earth pulls down on the 15.0 kg of mass), the normal force (the floor pushes up on the system to support its weight), and the applied force (the hand is pushing on the back part of the system). The force acting between the 5.0-kg box and the 10.0-kg box is not considered in the system analysis since it is an internal force. Just as the forces holding atoms together within an object are not included in a free-body diagram, so the forces holding together the parts of a system are ignored. These are considered internal forces; only external forces are considered when drawing free-body diagrams. The magnitude of the force of gravity is m•g or 147 N. The magnitude of the normal force is also 147 N since it must support the weight (147 N) of the system. The applied force is stated to be 45.0 N. Newton's second law (a = F_net/m) can be used to determine the acceleration. Using 45.0 N for F_net and 15.0 kg for m, the acceleration is 3.0 m/s².

Now that the acceleration has been determined, an individual object analysis can be performed on either object in order to determine the force acting between them. It does not matter which object is chosen; the result will be the same in either case. Here the individual object analysis is conducted on the 10.0 kg object (only because there is one less force acting on it). The free-body diagram for the 10.0-kg object is shown at the right. There are only three forces acting upon it - the force of gravity on the 10.0-kg, the support force (from the floor pushing upward) and the rightward contact force (F_contact). As the 5.0-kg object accelerates to the right, it will be pushing rightward upon the 10.0-kg object; this is known as a contact force (or a normal force or an applied force or …). The vertical forces balance each other since there is no vertical acceleration. The only unbalanced force on the 10.0-kg object is the Fcontact. This force is the net force and is equal to m•a where m is equal to 10.0 kg (since this analysis is for the 10.0-kg object) and awas already determined to be 3.0 m/s². The net force is equal to 30.0 N. This net force is the force of the 5.0-kg object pushing the 10.0-kg object to the right; it has a magnitude of 30.0 N. So the answers to the two unknowns for this problem are 3.0 m/s² and 30.0 N.

Now we will consider the solution to this same problem using the second approach - the use of two individual object analyses. In the process of this second approach, we will ignore the fact that we know what the answers are and presume that we are solving the problem for the first time. In this approach, two separate free-body diagram analyses are performed. The diagrams below show the free-body diagrams for the two objects.

Note that there are four forces on the 5.0-kg object at the rear. The two vertical forces - F_grav and F_norm - are obvious forces. The 45.0-N applied force (F_app) is the result of the hand pushing on the rear object as described in the problem statement and depicted in the diagram. The leftward contact force on the 5.0-kg object is the force of the 10.0-kg object pushing leftward on the 5.0-kg object. As an attempt is made to push the rear object (5.0-kg object) forward, the front object (10.0-kg object) pushes back upon it. This force is equal to and opposite of the rear object pushing forward on the front object. This force is simply labeled as F_contact for both of the free-body diagrams. In the free-body diagram for the 10.0-kg object, there are only three forces. Once more, the two vertical forces - F_grav and F_norm - are obvious forces. The horizontal force is simply the 5.0-kg object pushing the 10.0-kg object forward. The 45.0 N applied force is not exerted upon this 10.0-kg object; it is exerted on the 5.0-kg object and has already been considered in the previous free-body diagram.

Now the goal of this approach is to generate system of two equations capable of solving for the two unknown values. Using F_net = m•a with the free-body diagram for the 5.0-kg object will yield the Equation 1 below:

45.0 - F_contact = 5.0•a

Using F_net = m•a with the free-body diagram for the 10.0-kg object will yield the Equation 2 below:

F_contact = 10.0•a

(Note that the units have been dropped from Equations 1 and 2 in order to clean the equations up.) If the expression 10.0•a is substituted into Equation 1 for F_contact, then Equation 1 becomes reduced to a single equation with a single unknown. The equation becomes

45.0 - 10.0•a = 5.0•a

A couple of steps of algebra lead to an acceleration value of 3.0 m/s². This value of a can be substituted back into Equation 2 in order to determine the contact force:

F_contact = 10.0•a = 10.0 •3.0
F_contact = 30.0 N

As can be seen, using the second approach to solve two body problems yields the same two answers for the two unknowns. Now we will try the same two approaches on a very similar problem that includes a friction force.

Example Problem 2:

A 5.0-kg and a 10.0-kg box are touching each other. A 45.0-N horizontal force is applied to the 5.0-kg box in order to accelerate both boxes across the floor. The coefficient of kinetic friction is 0.200. Determine the acceleration and the contact force.

Our first solution to this problem will involve the dual combination of a system analysis and an individual object analysis. As you likely noticed, Example Problem 2 is similar to Example Problem 1 with the exception that the surface is not frictionless in Example Problem 2. So when conducting the system analysis in this second example, the friction on the 15-kg system must be considered. So the free-body diagram for the system now includes four forces - the same three as in Example Problem 1 plus a leftward force of friction. The force of friction on the system can be calculated as μ•F_normwhere F_normis the normal force experienced by the system. The F_norm of the system is equal to the force of gravity acting upon the 15.0-kg system; this value is 147 N. So

F_frict = μ•F_norm = (0.200)•(147 N) = 29.4 N

The vertical forces balance each other - consistent with the fact that there is no vertical acceleration. The horizontal forces do not balance each other. The net force can be determined as the vector sum of F_app and F_frict. That is, F_net = 45.0 N, right + 29.4 N, left; these add to 15.6 N, right. The acceleration can now be calculated using Newton's second law.

a = F_net / m = (15.6 N/15.0 kg) = 1.04 m/s²

Now that the system analysis has been used to determine the acceleration, an individual object analysis can be performed on either object in order to determine the force acting between them. Once more, it does not matter which object is chosen; the result would be the same in either case. The 10.0-kg object is chosen for the individual object analysis because there is one less force acting upon it; this makes the solution easier. There are four forces acting upon the 10.0-kg object. The two vertical forces are obvious - the force of gravity (98.0 N) and the normal force (equal to the force of gravity). The horizontal forces are the friction force to the left and the force of the 5.0-kg object pushing the 10.0-kg object forward; this is labeled as F_contact on the free-body diagram. The net force - vector sum of all the forces - can always be found by adding the forces in the direction of the acceleration and subtracting those that are in the opposite direction. This F_net is equal to F_contact - F_frict. Applying Newton's second law to this object yields the equation:

F_contact - F_frict = (10.0 kg)•(1.04 m/s²)

The friction force on this 10.0-kg object is not the same as the friction force on the system (since the system wasweightier). The F_frict value can be computed as μ•F_norm where F_norm is the normal force experienced by the 10.0-kg object. The F_normof the 10.0-kg is equal to the force of gravity acting upon the 10.0-kg object; this value is 98.0 N. So

F_frict = μ•F_norm = (0.200)•(98.0 N) = 19.6 N

So now the value of 19.6 N can be substituted into the above equation and F_contact can be calculated:

F_contact - 19.6 N = (10.0 kg)•(1.04 m/s²)
F_contact = (10.0 kg)•(1.04 m/s²) + 19.6 N
F_contact = 30.0 N

So using the dual combination of the system analysis and individual body analysis allows us to determine the two unknown values - 1.04 m/s² for the acceleration and 30.0 N for the F_contact. Now we will see how two individual object analyses can be combined to generate a system of two equations capable of solving for the two unknowns. Once more we will start the analysis by presuming that we are solving the problem for the first time and do not know the acceleration nor the contact force. The free-body diagrams for the individual objects are shown below.

There are now five forces on the 5.0-kg object at the rear. The two vertical forces - F_grav and F_norm - are obvious forces. The 45.0-N applied force (F_app) is the result of the hand pushing on the rear object. The leftward contact force on the 5.0-kg object is the force of the 10.0-kg object pushing leftward on the 5.0-kg object. Its value is the same as the contact force that is exerted on the front 10.0-kg object by the rear 5.0-kg object. This force is simply labeled as F_contactfor both of the free-body diagrams. Finally, the leftward friction force is the result of friction with the floor over which the 5.0-kg object moves. In the free-body diagram for the 10.0-kg object, there are now four forces. The two vertical forces - F_grav and F_norm - are obvious. The rightward contact force (F_contact) is simply the 5.0-kg object pushing the 10.0-kg object forward. And the leftward friction force is the result of friction with the floor. Once more, the 45.0 N applied force is not exerted upon this 10.0-kg object; it is exerted on the 5.0-kg object and has already been considered in the previous free-body diagram. The friction force for each object can be determined as μ•Fnorm where F_norm is the normal force experienced by the individual objects. Each object experiences a normal force equal to its weight (since vertical forces must balance). So the friction forces for the 5.0-kg object (49.0 N weight) and 10.0-kg object (98.0 N weight) are 0.200•49.0 N and 0.200•98.0 N, respectively.

Using these F_frict values and Newton's second law, a system of two equations capable of solving for the two unknown values can be written. Using F_net = m•a with the free-body diagram for the 5.0-kg object will yield Equation 3 below:

45.0 - F_contact - 9.8 = 5.0•a

Using F_net = m•a with the free-body diagram for the 10.0-kg object will yield the Equation 4 below:

F_contact - 19.6 = 10.0•a

(Note that the units have been dropped from Equations 3 and 4 in order to clean the equations up.) From Equation 4, F_contact = 10.0•a + 19.6. Substituting this expression for F_contact into Equation 3 and performing proper algebraic manipulations yields the acceleration value:

45.0 - (10.0•a + 19.6) - 9.8 = 5.0•a
45.0 - 19.6 - 9.8 = 15.0•a
15.6 = 15.0•a
a = (15.6/15.0)= 1.04 m/s²

This acceleration value can be substituted back into the expression for F_contact in order to determine the contact force:

F_contact = 10.0•a + 19.6 = 10.0•(1.04) + 19.6
F_contact = 30.0 N

Again we find that the second approach of using two individual object analyses yields the same set of answers for the two unknowns. The final example problem will involve a vertical motion. The approaches will remain the same.

Example Problem 3:

A man enters an elevator holding two boxes - one on top of the other. The top box has a mass of 6.0 kg and the bottom box has a mass of 8.0 kg. The man sets the two boxes on a metric scale sitting on the floor. When accelerating upward from rest, the man observes that the scale reads a value of 166 N; this is the upward force upon the bottom box. Determine the acceleration of the elevator (and boxes) and determine the forces acting between the boxes.

Both approaches will be used to solve this problem. The first approach involves the dual combination of a system analysis and an individual object analysis. For the system analysis, the two boxes are considered to be a single system with a mass of 14.0 kg. There are two forces acting upon this system - the force of gravity and the normal force. The free-body diagram is shown at the right. The force of gravity is calculated in the usual manner using 14.0 kg as the mass.

F_grav = m•g = 14.0 kg • 9.8 N/kg = 137.2 N

Since there is a vertical acceleration, the vertical forces will not be balanced; the F_grav is not equal to the F_norm value. The normal force is provided in the problem statement. This 166-N normal force is the upward force exerted upon the bottom box; it serves as the force on the system since the bottom box is part of the system. The net force is the vector sum of these two forces. So

F_net = 166 N, up + 137.2 N, down = 28.8 N, up

The acceleration can be calculated using Newton's second law:

a = F_net /m = 28.8 N/14.0 kg = 2.0571 m/s² = ~2.1 m/s²

Now that the system analysis has been used to determine the acceleration, an individual object analysis can be performed on either box in order to determine the force acting between them. As in the previous problems, it does not matter which box is chosen; the result will be the same in either case. The top box is used in this analysis since it encounters one less force. The free-body diagram is shown at the right. The force of gravity on the top box is m•g where m = 6.0 kg. The force of gravity is 58.8 N. The upward force is not known but can be calculated if the F_net = m•a equation is applied to the free-body diagram. Since the acceleration is upward, the Fnet side of the equation would be equal to the force in the direction of the acceleration (F_contact) minus the force that opposes it (F_grav). So

F_contact - 58.8 N = (6.0 kg)•(2.0571 m/s2)

(Notice that the unrounded value of acceleration is used here; rounding will occur when the final answer is determined.) Solving for F_contact yields 71.14 N. This figure can be rounded to two significant digits - 71 N. So the dual combination of the system analysis and the individual body analysis leads to an acceleration of 2.1 m/s² and a contact force of 71 N.

Now the second problem-solving approach will be used to solve the same problem. In this solution, two individual object analyses will be combined to generate a system of two equations capable of solving for the two unknowns. We will start this analysis by presuming that we are solving the problem for the first time and do not know the acceleration nor the contact force. The free-body diagrams for the individual objects are shown below.

Note that the F_grav values for the two boxes have been included on the diagram. These were calculated using F_grav = m•g where m=6.0 kg for the top box and m=8.0 kg for the bottom box. The contact force (F_contact) on the top box is upward since the bottom box is pushing it upward as the system of two objects accelerates upward. The contact force (F_contact) on the bottom box is downward since the top box pushes downward on the bottom box as the acceleration occurs. These two contact forces are equal to one another since they result from a mutual interaction between the two boxes. The third force on the bottom box is the force of the scale pushing upward on it with 166 N of force; this value was given in the problem statement.

Applying Newton's second law to these two free-body diagrams leads to Equation 5 (for the 6.0-kg box) and Equation 6 (for the 8.0-kg box).

F_contact - 58.8 = 6.0 • a

166 - F_contact - 78.4 = 8.0 • a

Now that a system of two equations has been developed, algebra can be used to solve for the two unknowns. Equation 5 can be used to write an expression for the contact force (F_contact) in terms of the acceleration (a).

F_contact = 6.0 • a + 58.8

This expression for F_contact can then be substituted into equation 6. Equation 6 then becomes

166 - (6.0 • a + 58.8) - 78.4 = 8.0 • a

The following algebraic steps are performed on the above equation to solve for acceleration.

166 - 6.0 • a - 58.8 - 78.4 = 8.0 • a
166 - 58.8 - 78.4 = 8.0 • a + 6.0 • a
28.8 = 14.0 a
a = 2.0571 m/s² = ~2.1 m/s²

Now the value for acceleration (a) can be substituted back into the expression for F_contact (F_contact = 6.0 • a + 58.8) to solve for F_contact. The contact force is 71.14 N (~71 N).

It should be noted that the second approach to this problem yields the same numerical answers as the first approach. Students are encouraged to use the approach that they are most comfortable with.

For additional practice, consider the following two-body problems. A shortened version of the solution has been provided for each problem. The topic of two-body problems will be returned to in the next chapter when we consider situations involving pulleys and objects moving in different directions.

Check Your Understanding

1. A truck hauls a car cross-country. The truck's mass is 4.00x10³ kg and the car's mass is 1.60x10³ kg. If the force of propulsion resulting from the truck's turning wheels is 2.50x10⁴ N, then determine the acceleration of the car (or the truck) and the force at which the truck pulls upon the car. Assume negligible air resistance forces.

2. A 7.00-kg box is attached to a 3.00-kg box by rope 1. The 7.00-kg box is pulled by rope 2 with a force of 25.0 N. Determine the acceleration of the boxes and the tension in rope 1. The coefficient of friction between the ground and the boxes is 0.120.

3. A tractor is being used to pull two large logs across a field. A chain connects the logs to each other; the front log is connected to the tractor by a separate chain. The mass of the front log is 180 kg. The mass of the back log is 220 kg. The coefficient of friction between the logs and the field is approximately 0.45. The tension in the chain connecting the tractor to the front log is 1850 N. Determine the tension in the chain that connects the two logs.

4. Two boxes are held together by a strong wire and attached to the ceiling of an elevator by a second wire (see diagram). The mass of the top box is 14.2 kg; the mass of the bottom box is 10.4 kg. The elevator accelerates upwards at 2.84 m/s². (Assume the wire is relatively massless.)

(a) Find the tension in the top wire (connecting points A and B).

(b) Find the tension in the bottom wire (connecting points C and D).

T.B.C.