Motion in one dimension

Sections 2.1 - 2.5

Motion in 1 dimension

We live in a 3-dimensional world, so why bother analyzing 1-dimensional situations? Basically, because any translational (straight-line, as opposed to rotational) motion problem can be separated into one or more 1-dimensional problems. Problems are often analyzed this way in physics; a complex problem can often be reduced to a series of simpler problems.

The first step in solving a problem is to set up a coordinate system. This defines an origin (a starting point) as well as positive and negative directions. We'll also need to distinguish between scalars and vectors. A scalar is something that has only a magnitude, like area or temperature, while a vector has both a magnitude and a direction, like displacement or velocity.

In analyzing the motion of objects, there are four basic parameters to keep track of. These are time, displacement, velocity, and acceleration. Time is a scalar, while the other three are vectors. In 1 dimension, however, it's difficult to see the difference between a scalar and a vector! The difference will be more obvious in 2 dimensions.

Displacement

The displacement represents the distance traveled, but it is a vector, so it also gives the direction. If you start in a particular spot and then move north 5 meters from where you started, your displacement is 5 m north. If you then turn around and go back, with a displacement of 5 m south, you would have traveled a total distance of 10 m, but your net displacement is zero, because you're back where you started. Displacement is the difference between your final position (x) and your starting point (xo) :

Speed and velocity

Imagine that on your way to class one morning, you leave home on time, and you walk at 3 m/s east towards campus. After exactly one minute you realize that you've

left your physics assignment at home, so you turn around and run, at 6 m/s, back to get it. You're running twice as fast as you walked, so it takes half as long (30 seconds) to get home again.

There are several ways to analyze those 90 seconds between the time you left home and the time you arrived back again. One number to calculate is your average speed, which is defined as the total distance covered divided by the time. If you walked for 60 seconds at 3 m/s, you covered 180 m. You covered the same distance on the way back, so you went 360 m in 90 seconds.

Average speed = distance / elapsed time = 360 / 90 = 4 m/s.

The average velocity, on the other hand, is given by:

Average velocity = displacement / elapsed time.

In this case, your average velocity for the round trip is zero, because you're back where you started so the displacement is zero

We usually think about speed and velocity in terms of their instantaneous values, which tell us how fast, and, for velocity, in what direction an object is traveling at a particular instant. The instantaneous velocity is defined as the rate of change of position with time, for a very small time interval. In a particular time interval delta t, if the displacement is  , the velocity during that time interval is:

The instantaneous speed is simply the magnitude of the instantaneous velocity.

Acceleration

An object accelerates whenever its velocity changes. Going back to the example we used above, let's say instead of instantly breaking into a run the moment you turned around, you steadily increased your velocity from 3m/s west to 6 m/s west in a 10 second period. If your velocity increased at a constant rate, you experienced a constant acceleration of 0.3 m/s per second (or, 0.3 m/s2).

We can figure out the average velocity during this time. If the acceleration is constant, which it is in this case, then the average velocity is simply the average of the initial and final velocities. The average of 3 m/s west and 6 m/s west is 4.5 m/s west. This average velocity can then be used to calculate the distance you traveled during your

acceleration period, which was 10 seconds long. The distance is simply the average velocity multiplied by the time interval, so 45 m.

Similar to the way the average velocity is related to the displacement, the average acceleration is related to the change in velocity: the average acceleration is the change in velocity over the time interval (in this case a change in velocity of 3 m/s in a time interval of 10 seconds). The instantaneous acceleration is given by:

As with the instantaneous velocity, the time interval is very small (unless the acceleration is constant, and then the time interval can be as big as we feel like making it).

On the way out, you traveled at a constant velocity, so your acceleration was zero. On the trip back your instantaneous acceleration was 0.3 m/s2 for the first 10 seconds, and then zero after that as you maintained your top speed. Just as you arrived back at your front door, your instantaneous acceleration would be negative, because your velocity drops from 6 m/s west to zero in a small time interval. If you took 2 seconds to come to a stop, your acceleration is -6 / 2 = -3 m/s2.

Kinematics equations when the acceleration is constant

When the acceleration of an object is constant, calculations of the distance traveled by an object, the velocity it's traveling at a particular time, and/or the time it takes to reach a particular velocity or go a particular distance, are simplified. There are four equations that can be used to relate the different variables, so that knowing some of the variables allows the others to be determined.

Note that the equations apply under these conditions:

1. the acceleration is constant2. the motion is measured from t = 03. the equations are vector equations, but the variables are not normally written in

bold letters. The fact that they are vectors comes in, however, with positive and negative signs.

The equations are:

The equations above are all derived in section 2.5.

Constant Acceleration

9-10-99

Sections 2.6 - 2.7

Applying the equations

Doing a sample problem is probably the best way to see how you would use the kinematics equations. Let's say you're driving in your car, approaching a red light on Commonwealth Avenue. A black Porsche is stopped at the light in the right lane, but there's no-one in the left lane, so you pull into the left lane. You're traveling at 40 km/hr, and when you're 15 meters from the stop line the light turns green. You sail through the green light at a constant speed of 40 km/hr and pass the Porsche, which accelerated from rest at a constant rate of 3 m/s2 starting at the moment the light turned green.

(a) How far from the stop line do you pass the Porsche?

(b) When does the Porsche pass you?

(c) If a Boston police officer happens to get you and the Porsche on the radar gun at the instant the Porsche passes you, will either of you be pulled over for speeding? Assume the speed limit is 50 km/hr.

Step 1 - Write down everything you know. Define an origin - the stop line is a good choice in this problem. Then choose a positive direction. In this case, let's take the positive direction to be the direction you're traveling. Decide on a system of units...meters and seconds is a good choice here, so convert your speed to m/s from km/hr. Drawing a diagram is also a good idea.

Origin = stop line Positive direction = the direction you're traveling

Step 2 - Figure out what you need to solve for. At the instant you pass the Porsche, the x values (yours and the Porsche's) have to be equal. You're both the same distance from the stop line, in other words. Write out the expression for your x-value and the Porsche's. We'll use the equation:

For you : x = -15 + 11.11 t + 0

For the Porsche : x = 0 + 0 + 1/2 (3) t2 = 1.5 t2

At some time t, when you pass the Porsche, these x values will be the same. So, we can set the equations equal to one another and solve for time, and then plug the time back in to either x equation to get the distance from the stop line. Doing this gives:

-15 + 11.11 t = 1.5 t2

Bringing everything to one side gives:

1.5 t2 - 11.11 t + 15 = 0

This is a quadratic equation, which we can solve using the quadratic formula:

where a = 1.5, b = -11.11, and c = 15

This gives two values for t, t = 1.776 s and t = 5.631 s.

What do these two values mean? In many cases only one answer will be relevant, and you'll have to figure out which. In this case both are relevant. The smaller value is when you pass the Porsche, while the larger one is when the Porsche passes you back.

To get the answer to question (a), plug t = 1.776 into either of your x expressions. They should both give you the same value for x, so you can use one as a check.

For you, at t = 1.776, x = 4.73 m. For the Porsche, at t = 1.776 s, x = 4.73 m.

We've actually already calculated the answer to (b), when the Porsche passes you, which is at t = 5.6 s.

To get the answer to part (c), we already know that you're traveling at a constant speed of 40 km/hr, which is under the speed limit. To figure out how fast the Porsche is going at t = 5.631 seconds, use:

v = vo + a t = 0 + (3) (5.631) = 16.893 m/s.

Converting this to km/hr gives a speed of 60.8 km/hr, so the driver of the Porsche is in danger of getting a speeding ticket.

Free fall

Objects falling straight down under the influence of gravity are excellent examples of objects traveling at constant acceleration in one dimension. This also applies to anything you throw straight up in the air which, because of the constant acceleration downwards, will rise until the velocity drops to zero and then will fall back down again.

The acceleration experienced by a dropped or thrown object while it is in flight comes from the gravitational force exerted on the object by the Earth. If we're dealing with objects at the Earth's surface, which we usually are, we call this acceleration g, which has a value of 9.8 m/s2. This value is determined by three things: the mass of the Earth, the radius of the Earth, and a number called the universal gravitational constant. We'll be dealing with all that later in the semester, though, so don't worry about it yet. For now, all you need to remember is that g is 9.8 m/s2 at the surface of the Earth, directed down.

A typical one-dimensional free fall question (free fall meaning that the only acceleration we have to worry about is g) might go like this.

You throw a ball straight up. It leaves your hand at 12.0 m/s.

(a) How high does it go?

(b) If, when the ball is on the way down, you catch it at the same height at which you let it go, how long was it in flight?

(c) How fast is it traveling when you catch it?

Origin = height at which it leaves your hand Positive direction = up

(a) At the very top of its flight, the ball has an instantaneous velocity of zero. We can plug v = 0 into the equation:

This gives:

0 = 144 + 2 (-9.8) x

Solving for x gives x = 7.35 m, so the ball goes 7.35 m high.

(b) To analyze the rest of the problem, it's helpful to remember that the down half of the trip is a mirror image of the up half. In other words, if, while going up, the ball passes through a particular height at a particular velocity (2 m/s up, for example), on its way down it will pass through that height at the same speed, with its velocity directed down rather than up. This means that the up half of the trip takes the same time as the down half of the trip, so we could just figure out how long it takes to reach its maximum height, and then double that to get the total time.

Another way to do it is simply to plug x = 0 into the equation:

This gives 0 = 0 + 12 t - 4.9 t2

A factor of t can be canceled out of both terms, leaving:

0 = 12 - 4.9 t, which gives a time of t = 12 / 4.9 = 2.45 s.

(c) The answer for part (c) has to be 12 m/s down, because of the mirror-image relationship between the up half of the flight and the down half. We could also figure it out using the equation:

v = vo + a t

which gives:

v = 12 - 9.8 (2.45) = -12 m/s.

Graphical analysis; and Vectors

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Sections 2.8 - 3.4

Graphs

Drawing good pictures can be the secret to solving physics problems. It's amazing how much information you can get from a diagram. We also usually need equations to find numerical solutions. Graphs are basically pictures of equations. Learning how to interpret these pictures can really help you understand physics.

Let's return to our last example, a ball thrown vertically upward with an initial speed of 12 m/s. The only acceleration we have to worry about is the acceleration due to gravity, 9.8 m/s2 down. This acceleration is constant, so it's easy to plot on a graph.

If the time T represents the time when the ball returns to your hand, the area under the curve must equal -24.0 m/s, because we know the velocity changes from 12.0 m/s to -12.0 m/s. This allows us to solve for T, using:

T = -24.0 / -9.8 = 2.45 s, agreeing with what we calculated previously.

What about the velocity graph? The equation for velocity is:

v = vo + at Plugging in the initial velocity and acceleration here gives v = 12.0 -9.8t

The velocity graph can give all sorts of information:

The slope of the velocity graph is the acceleration, while the area under the curve is the displacement.

For this example of a ball going up and then back down, the graph confirms that the time taken on the way up equals the time taken on the way down.

Calculating the area under the curve for the ball on the way up (the positive area on the graph) gives the maximum displacement. The area is just a triangle, with a a base T/2 = 1.2245 s and height of 12 m/s. This gives an area of 0.5(1.2245)(12) = 7.35 m. Again, this agrees with the maximum height calculated previously.

The positive and negative areas cancel each other out, meaning the net displacement is zero. This is correct, as the ball returns to its starting point.

The graph of position as a function of time is a plot of the equation:

x = xo + vot + 1/2 at2 In this case, that's x = 0 + 12t -4.9t2

The slope of the position graph gives the instantaneous velocity. This is positive but steadily decreasing on the way up, zero at the very top, and then becomes more and more negative on the way down.

Vectors and scalars

We'll move on from looking at motion in one dimension to motion in two or three dimensions. It's critical now to distinguish between two kinds of quantities, scalars and vectors. A scalar is something that's just a number with a unit, like mass ( 200 g ) or temperature ( -30°C). A vector has both a number and a direction, like velocity. If

you came to campus on the T today, at some point you may have been traveling 20 km/hr east. Velocity is a combination of a scalar (speed, 20 km/hr) and a direction (east).

Examples of scalars : mass, temperature, speed, distance

Examples of vectors : displacement, velocity, acceleration, force

One crucial difference between scalars and vectors involves the use of plus and minus signs. A scalar with a negative sign means something very different from a scalar with a plus sign; +30°C feels an awful lot different than -30°C, for example. With a vector the sign simply tells you the direction of the vector. If you're traveling with a velocity of 20 km/hr east, it means you're traveling east, and your speed is 20 km/hr. A velocity of -20 km/hr east also means that you're traveling at a speed of 20 km/hr, but in the direction opposite to east : 20 km/hr west, in other words. With a vector, the negative sign can always be incorporated into the direction.

Note that a vector will normally be written in bold, like this : A. A scalar, like the magnitude of the vector, will not be in bold face (e.g., A).

Components of a vector

A vector pointing in a random direction in the x-y plane has x and y components: it can be split into two vectors, one entirely in the x-direction (the x-component) and one entirely in the y-direction (the y-component). Added together, the two components give the original vector.

The easiest way to add or subtract vectors, which is often required in physics, is to add or subtract components. Splitting a vector into its components involves nothing more complicated than the trigonometry associated with a right-angled triangle.

Consider the following example. A vector, which we will call A, has a length of 5.00 cm and points at an angle of 25.0° above the negative x-axis, as shown in the diagram. The x and y components of A, Axand Ay are found by drawing right-angled triangles, as shown. Only one right-angled triangle is actually necessary; the two shown in the diagram are identical.

Knowing the length of A, and the angle of 25.0°, Ax and Ay can be found by re-arranging the expressions for sin and cos.

Note that this analysis, using trigonometry, produces just the magnitudes of the vectors Ax and Ay. The directions can be found by looking on the diagram. Usually, positive x is to the right of the origin; Axpoints left, so it is negative:

Ax = -4.53 cm in the x-direction (or, 4.53 cm in the negative x-direction)

Positive y is generally up; Ay is directed up, so it is positive:

Ay = 2.11 cm in the y-direction

Adding vectors

It's fairly easy to add vectors that are parallel to each other, or at right angles. In general, however, the angle between vectors being added (or subtracted) will be something other than 0, 90, or 180°. Consider the following example, where the vector C equals A + B.

A has a length of 5.00 cm and points at an angle of 25.0° above the negative x-axis. B has a length of 7.00 cm and points at an angle of 40.0° above the positive x-axis. If C = A + B, what is the magnitude and direction of C? There are basically two ways to answer this. One way is to draw a vector diagram, moving the tail of B to the head of A, or vice versa. The vector C will then extend from the origin to wherever the tip of the second vector is.

The second way to find the magnitude and direction of C we'll use a lot in this course, because we'll often have vector equations of the form C =A + B. The simplest way to solve any vector equation is to split it up into one-dimensional equations, one equation for each dimension of the vector. In this case we're working in two dimensions, so the one vector equation can be replaced by two equations:

In the x-direction : Cx = Ax + Bx

In the y-direction : Cy = Ay + By

In other words, to find the magnitude and direction of C, the vectors A and B are split into components. The components are:

Ax = -4.532 cm in the x-direction

Ay = 2.113 cm in the y-direction

Bx = 7.00 cos40 = 5.362 cm

By = 7.00 sin40 = 4.500 cm

Bx = 5.362 cm in the x-direction

By = 4.500 cm in the y-direction

The components of C are found by adding the components of A and B:

Cx = Ax + Bx = (-4.532 + 5.362) cm in the x-direction = 0.83 cm in the x-direction

Cy = Ay + By= (2.113 + 4.500) cm in the y-direction = 6.61 cm in the y-direction

The magnitude of C can be found from its components using the Pythagorean theorem:

The direction of C can be found by taking the inverse tangent of Cy/Cx:

inverse tan of 6.613 / 0.830 = 82.8°.

Combined, this gives C = 6.66 cm at an angle of 82.8° above the positive x-axis.

Note how the calculations and the diagrams go hand-in-hand. This will often be the case; it is always a good idea to draw diagrams as you go along.

Note also that we could have used the cosine law to get the length of C, and then applied the sine law, with a bit of geometry, to get the angle. It's worth trying that for yourself, just to convince yourself that the numbers come out the same.

Motion in two dimensions

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Sections 3.5 - 3.7

Extending things from 1 dimension

In 1 dimension, we wrote down some general equations relating velocity to displacement, and relating acceleration to the change in velocity. We also wrote down the four equations that apply in the special case where the acceleration is constant. We're going to do the same thing in 2 dimensions, and the equations will look similar; this shouldn't be surprising because, as we will see, a two (or three) dimensional problem can always be broken down into two (or three) 1-dimensional problems.

When we're dealing with more than 1 dimension (and we'll focus on 2D, but we could use these same equations for 3D), the position is represented by the vector r. The velocity will still be represented by vand the acceleration by a. In general, the average velocity will be given by:

The instantaneous velocity is given by a similar formula, with the condition that a very small time interval is used to measure the displacement.

A similar formula gives the average acceleration:

Again, the instantaneous acceleration is found by measuring the change in velocity over a small time interval.

The constant acceleration equations

When the acceleration is constant, we can write out four equations relating the displacement, initial velocity, velocity, acceleration, and time for each dimension. Like the 1D equations, these apply under the following conditions:

1. the acceleration is constant2. the motion is measured from t = 03. the equations are vector equations, but the variables are not normally written in

bold letters. The fact that they are vectors comes in, however, with positive and negative signs.

If we focus on two dimensions, we get four equations for the x direction and four more for the y direction. The four x equations involve only the x-components, while the four y equations involve only the y-components.

One thing to notice is that the time, t, is the only thing that doesn't involve an x or a y. This is because everything else is a vector (or a component of a vector, if you'd rather look at it that way), but time is a scalar. Time is the one thing that can be used directly

in both the x and y equations; everything else (displacement, velocity, and acceleration) has to be split into components.

This is important!

Something that probably can't be emphasized enough is that even though an object may travel in a two-dimensional path (often following a parabola, in the standard case of an object moving under the influence of gravity alone), the motion can always be reduced to two independent one-dimensional motions. The x motion takes place as if the y motion isn't happening, and the y motion takes place independent of whatever is happening in the x direction.

One good example of this is the case of two objects (e.g., baseballs) which are released at the same time. One is dropped so it falls straight down; the other is thrown horizontally. As long as they start at the same height, both objects will hit the ground at the same time, no matter how fast the second one is thrown.

What we're ignoring

We will generally neglect the effect of air resistance in most of the problems we do. In some cases that's just fine. In other cases it's not so fine. A feather and a brick dropped at the same time from the same height will not reach the ground at the same time, for example. This has nothing to do with the weight of the feather compared to the brick. It's simply air resistance; if we took away all the air and dropped the feather and brick, they would hit the ground at exactly the same time.

So, remember that we're often analyzing ideal cases, especially this early in the semester. In reality, things might be a little different because of factors we're neglecting at the moment.

An example

Probably the simplest way to see how to apply these constant acceleration equations is to work through a sample problem. Let's say you're on top of a cliff, which drops vertically 150 m to a flat valley below. You throw a ball off the cliff, launching it at 8.40 m/s at an angle of 20° above the horizontal. (a) How long does it take to reach the ground?(b) How far is it from the base of the cliff to the point of impact?

It's a good idea to be as systematic as possible when it comes to analyzing the situation. Here's an example of how to organize the information you're given. First, draw a diagram.

Then set up a table to keep track of everything you know. It's important to pick an origin, and to choose a coordinate system showing positive directions. In this case, the origin was chosen to be the base of the cliff, with +x being right and +y being up. You don't have to choose the origin or the positive directions this way. Pick others if you'd like, and stick with them (an origin at the top of the cliff, and/or positive y-direction down would be two possible changes).

Now that everything's neatly organized, think about what can be used to calculate what. You know plenty of y-information, so we can use that to find the time it takes to reach the ground. One way to do this (definitely not the only way) is to do it in two steps, first calculating the final velocity using the equation:

This gives vy2 = 2.8732 + 2 (-9.8) (-150) = 2948.3 m2 / s2 . Taking the square root

gives: vy = +/- 54.30 m/s.

Remember that the square root can be positive or negative. In this case it's negative, because the y-component of the velocity will be directed down when the ball hits the ground.

Now we can use another equation to solve for time:

So, -54.30 = 2.873 - 9.8 t, which gives t = 5.834 seconds. Rounding off, the ball was in the air for 5.83 s.

We can use this time for part (b), to get the distance traveled in the x-direction during the course of its flight. The best equation to use is:

 So, from the base of the cliff to the point of impact is 46.0 m.

A point about symmetry

At some point in its flight, the ball in the example above returned to the level of the top of the cliff (you threw it from the top of the cliff, it went up, and on its way down it passed through a point the same height off the ground as the top of the cliff). What was the ball's velocity when it passed this height? Its speed will be the same as the initial speed, 8.40 m/s, and its angle will be the same as the launch angle, only measured below the horizontal.

This is not just true of the initial height. At every height the ball passes through on the way up, there is a mirror-image point (at the same height, with the same speed, and the same angle, just down rather than up) on the downward part of the path.

Relative velocity

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Section 3.8

Relative velocity in 1 dimension

Most people find relative velocity to be a relatively difficult concept. In one dimension, however, it's reasonably straight-forward. Let's say you're walking along a road, heading west at 8 km/hr. A train track runs parallel to the road and a train is passing by, traveling at 40 km/hr west. There is also a car driving by on the road, going 30 km/hr east. How fast is the train traveling relative to you? How fast is the car traveling relative to you? And how fast is the train traveling relative to the car?

One way to look at it is this: in an hour, the train will be 40 km west of where you are now, but you will be 8 km west, so the train will be 32 km further west than you in an hour. Relative to you, then, the train has a velocity of 32 km/hr west. Similarly, relative to the train, you have a velocity of 32 km/hr east.

Using a subscript y for you, t for the train, and g for the ground, we can say this: the velocity of you relative to the ground =   = 8 km/hr west the velocity of the train relative to the ground =   = 40 km/hr west

Note that if you flip the order of the subscripts, to get the velocity of the ground relative to you, for example, you get an equal and opposite vector. You can write this equal and opposite vector by flipping the sign, or by reversing the direction, like this: the velocity of the ground relative to you =   = -8 km/hr west = 8 km/hr east

The velocity of the train relative to you,   , can be found by adding vectors appropriately. Note the order of the subscripts in this equation:

Rearranging this gives:

A similar argument can be used to show that the velocity of the car relative to you is 38 km/hr east. the velocity of you relative to the ground =   = 8 km/hr west = -8 km/hr east the velocity of the car relative to the ground =   = 30 km/hr east

The velocity of the train relative to the car is 70 km/hr west, and the velocity of the car relative to the train is 70 km/hr east.

Relative velocity in 2 dimensions

In two dimensions, the relative velocity equations look identical to the way they look in one dimension. The main difference is that it's harder to add and subtract the vectors, because you have to use components. Let's change the 1D example to 2D. The train still moves at 40 km/hr west, but the car turns on to a road going 40° south of east, and travels at 30 km/hr. What is the velocity of the car relative to the train now?

The relative velocity equation for this situation looks like this:

The corresponding vector diagram looks like this:

Because this is a 2-D situation, we have to write this as two separate equations, one for the x-components (east-west) and one for the y-components (north-south):

Now we have to figure out what the x and y components are for these vectors. The train doesn't have a y-component, because it is traveling west. So:

The car has both an x and y component:

Plugging these in to the x and y equations gives:

Combining these two components into the vector gives a magnitude of:

at an angle given by the inverse tangent of 19.3 / 63.0, which is 17 degrees. So, the velocity of the car relative to the train is 66 km/hr, 17 degrees south of east.

Newton's laws of motion

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Sections 4.1 - 4.5

Force

We've introduced the concept of projectile motion, and talked about throwing a ball off a cliff, analyzing the motion as it traveled through the air. But, how did the ball get its initial velocity in the first place? When it hit the ground, what made it eventually come to a stop? To give the ball the initial velocity, we threw it, so we applied a force to the ball. When it hit the ground, more forces came into play to bring the ball to a stop.

A force is an interaction between objects that tends to produce acceleration of the objects. Acceleration occurs when there is a net force on an object; no acceleration occurs when the net force (the sum of all the forces) is zero. In other words, acceleration occurs when there is a net force, but no acceleration occurs when the forces are balanced. Remember that an acceleration produces a change in velocity (magnitude and/or direction), so an unbalanced force will change the velocity of an object.

Isaac Newton (1642-1727) studied forces and noticed three things in particular about them. These are important enough that we call them Newton's laws of motion. We'll look at the three laws one at a time.

Newton's first law

The ancient Greeks, guided by Aristotle (384-322 BC) in particular, thought that the natural state of motion of an object is at rest, seeing as anything they set into motion eventually came to a stop. Galileo (1564-1642) had a better understanding of the situation, however, and realized that the Greeks weren't accounting for forces such as friction acting on the objects they observed. Newton summarized Galileo's thoughts about the state of motion of an object in a statement we call Newton's first law.

Newton's first law states that an object at rest tends to remain at rest, and an object in motion tends to remain in motion with a constant velocity (constant speed and direction of motion), unless it is acted on by a nonzero net force.

Note that the net force is the sum of all the forces acting on an object.

The tendency of an object to maintain its state of motion, to remain at rest or to keep moving at a constant velocity, is known as inertia. Mass is a good measure of inertia; light objects are easy to move, but heavy objects are much harder to move, and it is much harder to change their motion once they start moving.

A good question to ask is: do Newton's laws apply all the time? In most cases they do, but if we're trying to analyze motion in an accelerated reference frame (while we're spinning around would be a good example) then Newton's law are not valid. A

reference frame in which Newton's laws are valid is known as an inertial reference frame. Any measurements we take while we're not moving (while we're in a stationary reference frame, in other words) or while we're moving at constant velocity (on a train traveling at constant velocity, for example) will be consistent with Newton's laws.

Newton's second law

If there is a net force acting on an object, the object will have an acceleration and the object's velocity will change. How much acceleration will be produced by a given force? Newton's second law states that for a particular force, the acceleration of an object is proportional to the net force and inversely proportional to the mass of the object. This can be expressed in the form of an equation:

In the MKS system of units, the unit of force is the Newton (N). In terms of kilograms, meters, and seconds, 1 N = 1 kg m / s2.

Free-body diagrams

In applying Newton's second law to a problem, the net force, which is the sum of all the forces, often has to be determined so the acceleration can be found. A good way to work out the net force is to draw what's called a free-body diagram, in which all the forces acting on an object are shown. From this diagram, Newton's second law can be applied to arrive at an equation (or two, or three, depending on how many dimensions are involved) that will give the net force.

Let's take an example. This example gets us ahead of ourselves a little, by bringing in concepts we haven't talked about yet, but that's fine because (a) we'll be getting to them very shortly, and (b) there's a good chance you've seen them before anyway. Say you have a box, with a mass of 2.75 kg, sitting on a table. Neglect friction. There is a rope tied to the box and you pull on it, exerting a force of 20.0 N at an angle of 35.0¡ above the horizontal. A second rope is tied to the other side of the box, and your friend exerts a horizontal force of 12.0 N. What is the acceleration of the box?

The first step is to draw the free-body diagram, accounting for all the forces. The four forces we have to account for are the 20.0 N force you exert on it, the 12.0 N force your friend exerts, the force of gravity (the gravitational force exerted by the Earth on the box, in other words), and the support force provided by the table, which we'll call the normal force, because it is normal (perpendicular) to the surface the box sits on.

The free-body diagram looks like this:

We can apply Newton's second law twice, once for the horizontal direction, which we'll call the x-direction, and once for the vertical direction, which we'll call the y-direction. Let's take positive x to be right, and positive y to be up. The box accelerates across the table, so it has an acceleration in the x direction but not in the y direction (it doesn't accelerate vertically).

In the x direction, summing the forces gives:

The x-component of the force you exert is partly canceled by the force your friend exerts, but you win the tug-of-war and the box accelerates towards you. Solving for the horizontal acceleration gives: ax = 4.4 / 2.75 = 1.60 m/s2 to the right.

In the y direction, there is no acceleration, which means the forces have to balance. This allows us to solve for the normal force, because when we add up all the forces we get:

The gravitational force is often referred to as the weight. To remind you that this is actually a force, I'll generally refer to it as the force of gravity, or gravitational force, rather than the weight. The force of gravity is simply the mass times g, 2.75 x 9.8 = 26.95 N. Solving for the normal force gives:

FN = 26.95 - 20.0 sin35 = 26.95 - 11.47 = 15.5 N. In many problems the normal force will turn out to have the same magnitude as the force of gravity, but that is not always true, and it is not true in this case.

Newton's third law

A force is an interaction between objects, and forces exist in equal-and-opposite pairs. Newton's third law summarizes this as follows: when one object exerts a force on a second object, the second object exerts an equal-and-opposite force on the first object. Note that "equal-and-opposite" is the shortened form of "equal in magnitude but opposite in direction".

Consider the free-body diagram of the box in the example above. The box experiences 4 different forces, one from you, one from your friend, one from the Earth (the gravitational force) and one from the table. By Newton's law, the box also exerts 4 forces. If you exert a 20.0 N force on the box, the box exerts a 20.0 N force on you. Your friend exerts a 12.0 N force to the left, so the box exerts a 12.0 N force to the right on your friend. The table exerts an upward force on the box, the normal force, which is 15.5 N, so the box exerts a downward force of 15.5 N on the table. Finally, the Earth exerts a 26.95 N force down on the box, so the box exerts a 26.95 N force up on the Earth.

Although the forces between two objects are equal-and-opposite, the effect of the forces may or may not be similar on the two; it depends on their masses. Remember that the acceleration depends on both force and mass, and let's look at the force exerted by the Earth on a falling object. If we drop a 100 g (0.1 kg) ball, it experiences a downward acceleration of 9.8 m/s2, and a force of about 1 N, because it is attracted towards the Earth. The ball exerts an equal-and-opposite force on the Earth, so why doesn't the Earth accelerate upwards towards the ball? The answer is that it does, but because the mass of the Earth is so large (6.0 x 1024 kg) the acceleration of the Earth is much too small (about 1.67 x 10-25 m/s2) for us to notice.

In cases where objects of similar mass exert forces on each other, the fact that forces come in equal-and-opposite pairs is much easier to see.

Assorted forces, and applying Newton's laws

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Sections 4.6 - 4.7

Forces can come from various sources. Whenever two objects are touching, they usually exert forces on each other, Newton's third law reminding us that the forces are equal and opposite. Two objects do not have to be in contact to exert forces on each other, however. The force of gravity, for instance, is a good example of a force that can arise without two objects being in contact with each other.

Weight

Mass and weight are often used interchangeably, but they are quite different. The mass of an object is an intrinsic property of an object. If your mass is 50 kg, you have a mass of 50 kg no matter where you go: on Earth, on the Moon, in orbit, wherever. Your weight, on the other hand, will vary depending on where you are. Your weight is the magnitude of the gravitational force you experience, and it has units of force, Newtons. Because the gravitational force you experience on the Earth is different from that you'd experience on the Moon or in orbit, your weight would be different even though your mass remains constant.

We'll examine the force of gravity in more detail later in the course. For now, it's sufficient to remember that, at the surface of the Earth, the gravitational force on an object of mass m has a magnitude of mg. The direction is toward the center of the Earth.

The normal force

Many forces do come from objects being in contact with each other. A book rests on a table: the book exerts a downward force on the table, and the table exerts an equal-and-opposite force up on the book. We call this force the normal force, which doesn't mean that other forces are abnormal - "normal" is the technical physics word for perpendicular. We call it the normal force because the force is perpendicular to the interface where the book meets the table.

The normal force is just one component of the contact force between objects, the other component being the frictional force. The normal force is usually symbolized by   . In many cases the normal force is simply equal to the weight of an object, but that's because in many cases the normal force is the only thing counter-acting the weight...that is not always the case, however, and one should always be careful to calculate   by applying Newton's second law.

The tension force

Whenever we use a rope (or something equivalent, like a string) to exert a force on an object, we're creating tension in the rope that transmits the force we exert on the rope to the object at the other end of the rope. The tension force is usually labeled by T. To makes our lives simpler, we usually assume that the rope has no mass, and does not stretch. Using these assumptions, when we exert a certain T on our massless unstretchable rope, the rope exerts that same T on the object. The rope itself will feel like it's being pulled apart, because we'll be exerting a certain T in one direction and the object we're pulling on will be exerting an equal-and-opposite force at the other end.

One rule to remember - you can't push with a rope. The tension force always goes along the rope away from the object attached to it.

Applying Newton's laws

A good way to see how Newton's laws are applied is to do an example.

Example - A box of mass 3.60 kg rests on an inclined plane which is at an angle of 15.0° with respect to the horizontal. A string is tied to the box, and it exerts a force of 5.00 N up the ramp. What is the acceleration? Give the magnitude and direction.

The first step in solving this is to draw a picture, showing the box, inclined plane, and all the forces. Then choose a coordinate system, showing positive directions. One axis of your coordinate system should be aligned with the acceleration, which in this case is either up or down the ramp. Here, we've assumed that the box accelerates up the ramp, although we're not sure. This could be worked out by figuring out the component of the weight down the slope and comparing it to the tension force, but it's fine to guess. If we guess wrong we'll get a minus sign for the acceleration at the end which will tell us that the box accelerates down the slope...the magnitude will be correct, though.

The next step is to draw the free body diagram, which is similar to the first picture but without the inclined plane. Another difference is that the force of gravity has been split up into components parallel to the incline and perpendicular to the incline. In other words, its been split into its x and y components.

Applying Newton's second law in the y-direction gives:

The box is not accelerating in the y-direction, which is why the net force in that direction is zero. Solving for the normal force gives:

Applying Newton's second law in the x-direction gives:

Solving for the acceleration gives:

ax = T/m - g sin15° = 5.0/3.60 - 9.80 sin15° = -1.15 m/s2

So, the magnitude of the acceleration is 1.15 m/s2. The minus sign means that the box accelerates down the slope, rather than up as we'd assumed. This is because the component of the weight down the slope is larger than the tension.

The problem-solving process

If you apply a logical, step-by-step approach to these kinds of problems, you'll find them much easier to deal with. The example above should give you some idea of the steps to follow when trying to solve a typical problem. Summarizing them here:

1. Draw a diagram.2. Choose a coordinate system. Generally the problem will be easiest to solve if

you align one direction of your coordinate system with the acceleration of the object.

3. Draw a free-body diagram for each object, showing the forces applied to each object. Apply Newton's third law, which states that equal-and-opposite forces are exerted between interacting objects.

4. Apply Newton's second law, the sum of all the forces equals ma, for each direction for each free-body diagram. In other words, write out equations with the sum of all the forces in one direction on the left side, and the mass times the acceleration in that direction on the right side. Remember to sum the forces with the appropriate plus or minus signs, depending on their directions.

5. If you have the same number of equations as unknowns, solve. Otherwise, think about other principles of physics you can use to relate the unknown variables in the problem.

6. Always check to make sure your answers are reasonable. For example, are the numbers sensible? Do they have correct units?

7. Friction8. 9-24-99

9. Sections 4.8 - 4.910.The force of friction11.The normal force is one component of the contact force between two objects,

acting perpendicular to their interface. The frictional force is the other component; it is in a direction parallel to the plane of the interface between objects. Friction always acts to oppose any relative motion between surfaces.

12.For the simple example of a book resting on a flat table, the frictional force is zero. There is no force trying to move the book across the table, so there is no need for a frictional force because there is nothing for the frictional force to oppose. If we try to slide the book across the table, however, friction will come in to play.

13.Let's say it takes a force of 5 N to start the book moving. If we push on the book with a force of less than 5 N, the book won't move, because the frictional force will exactly balance the force we apply. If we push with a 1 N force, a 1 N frictional force opposes us. If we exert a 2 N force, the frictional force matches us at 2 N, and so on. When a frictional force exists but there is no relative motion of the surfaces in contact (e.g., the book isn't sliding across the table), we call it a static frictional force. The static frictional force is given by the equation:

14.15.The coefficient of friction (static or kinetic) is a measure of how difficult it is to

slide a material of one kind over another; the coefficient of friction applies to a pair of materials, and not simply to one object by itself.

16.Note that there is a less-than-or-equal-to sign in the equation for the static frictional force. The static force of friction has a maximum value, but when two surfaces are not moving relative to each other the static force of friction is always just enough to exactly balance any forces trying to produce relative motion.

17.What happens when one object is sliding over another, when there is relative motion between two surfaces? There will still be a frictional force, but because we're dealing with things in motion we call it the kinetic frictional force. There is a different coefficient of friction associated with kinetic friction, the kinetic coefficient of friction, which is always less than or equal to the static coefficient.

18.19.As with the static frictional force, the kinetic frictional force acts to oppose the

relative motion of the surfaces in contact. One important difference between the two is that the kinetic friction equation has an equals sign: the kinetic force of friction is always equal to the kinetic coefficient of friction times the normal force.

20.Applying Newton's laws

21.Let's try a couple of examples involving friction.22.Example 1 - A box of mass 3.60 kg travels at constant velocity down an

inclined plane which is at an angle of 42.0° with respect to the horizontal. A string tied to the box exerts a vertical force of 7.38 N. What is the kinetic coefficient of friction?

23.Again, start by drawing a picture. In this case, because the box is traveling down the ramp, we know the frictional force is kinetic. It opposes the motion, so the frictional force must be directed up the slope. One thing to keep in mind in this problem is that the velocity is constant - this means the acceleration is zero.

24.25.The free-body diagram is also shown, with the forces split into components

parallel and perpendicular to the inclined plane. Because there is no acceleration, any coordinate system is fine - a system parallel and perpendicular to the ramp is pretty convenient, though, because two of the forces are along those directions.

26.Applying Newton's second law to the forces in the y-direction :27.28.Again, we can use this equation to solve for the normal force:29.30.Applying Newton's second law in the x-direction gives:31.32.The box is moving at a constant velocity, so that means the acceleration is zero.

Solving for the kinetic force of friction gives:33.34.The coefficient of kinetic friction can be found from the normal force and the

frictional force:35.36.This is actually a relatively large value for the coefficient, so it's not easy to

move this box along this ramp.37.Example 238.Consider another example involving an inclined plane, only this time there will

be two boxes involved. This is quite a challenging example, so don't be too intimidated if it looks tricky. Start by seeing if you agree with the free-body

diagrams; if you understand those, you've made an important step in learning some physics.

39.Box 1, a wooden box, has a mass of 8.60 kg and a coefficient of kinetic friction with the inclined plane of 0.35. Box 2, a cardboard box, sits on top of box 1. It has a mass of 1.30 kg. The coefficient of kinetic friction between the two boxes is 0.45. The two boxes are linked by a rope which passes over a pulley at the top of the incline, as shown in the diagram. The inclined plane is at an angle of 38.0° with respect to the horizontal. What is the acceleration of each box?

40.The diagram for the situation looks like this:

41.42.The next step is to draw a free-body diagram of each box in turn. To draw

these, it helps to think about which way the boxes will accelerate. The two boxes are tied together, and the heavier box will win...in other words, it will accelerate down the slope and the lighter one, on top, will be accelerated up towards the pulley. The pulley, by the way, simply changes the direction of the tension force. We're assuming that the pulley is massless and frictionless, so both boxes feel the same tension force. It's important to know the direction of the acceleration (or, if you don't know, to guess) and apply what you figured out (or your guess) consistently to both boxes.

43.44.The free-body diagram of box 1 is relatively complicated, with a total of 6

forces appearing. The free-body diagram for box 2 is a little easier to deal with, having 4 forces, so that's a good place to start.

45.For box 2 - start by summing the forces in the y-direction, where there is no acceleration:

46.47.This can be solved to give the normal force:48.49.Now find the net force in the x direction, where there is an acceleration up the

slope:

50.51.There are two unknowns in this equation, the tension T and ax, the acceleration.

We can at least solve for T in terms of ax, like this:

52.53.Now move on to box 1. Going back to the free-body diagram and summing

forces in the y-direction gives:54.55.This equation can be solved to give the normal force associated with the

interaction between the inclined plane and box 1:56.57.Things are a little more complicated in the x-direction, but adding up the forces

gives:58.59.substituting in the expression we worked out for T, and what fB and fA are

gives:60.61.Moving the acceleration terms to the left side gives:62.63.Solving for the acceleration gives:

64.