11.1.4.1: Illustrations

Illustration 1: Newton's First Law and Reference Frames

On first glance it may seem like Newton's first law (an object at rest remains at rest and an object in motion remains in motion unless acted on by a net force) is contained within Newton's second law. This is actually not the case. The first law is also a statement regarding reference frames. This is the information NOT contained in the second law. Sometimes the first law is also called the law of inertia. It defines a certain set of reference frames in which the first law holds, and these reference frames are therefore called inertial frames of reference. Put another way, Newton's first law states that if the net force on an object is zero, it is possible to find at least one reference frame in which that object is stationary. There are many frames in which the object is moving with a constant velocity.

A ball popper on a cart (not shown to scale) is shown moving on a track in three different animations (position is given in meters and time is given in seconds). In each animation the ball is ejected straight up by the popper mechanism at \(t = 1\text{ s}\). Restart.

Let us first consider Animation 1. In this animation the cart is stationary. But is it really? We know that we cannot tell if we are stationary or moving at a constant velocity (in other words in an inertial reference frame). Recall that if we are moving relative to Earth at a constant velocity we are in an inertial reference frame. So how can we tell if we are moving? How about the cart? We cannot tell if there is motion as long as the relative motion with respect to Earth can be described by a constant velocity. In Animation 1 the cart could be stationary. In this case, we expect—and actually see—that the ball lands back in the popper. However, if the cart was moving relative to Earth and we were moving along with the cart, the motion of the ball and the cart would look exactly the same!

What would the motion of this ball and cart look like if the cart moved relative to our reference frame (or if we moved relative to its reference frame)? Animations 2 and 3 show the motion from different reference frames. What do these animations look like? Both animations resemble projectile motion. The motion of the ball looks like motion in a plane as opposed to motion on a line. Does the ball still land in the popper? Would you expect this? Sure. There is nothing out of the ordinary going on here. Since there are no forces in the x direction, the motion of the ball (and cart) should be described by constant velocity in that direction. Therefore the ball and the cart have the same constant horizontal velocity.

For more on reference frames and relative motion, see Chapter 1.9.

Illustration 2: Free-Body Diagrams

In Illustration 2 an \(8\text{-kg}\) block is pushed across the floor (position is given in centimeters and time is given in seconds). Along with the motion, several possible free-body diagrams are drawn for both the \(x\) and \(y\) components of the force. Note that only one of the possible free-body diagrams is correct for each component. Restart.

Look at the motion of the block by pressing "play." How do we analyze the motion of the block using forces? Well, the first thing we do is draw a picture that shows only the object and the direction of the forces. The picture we draw is called a free-body diagram. First we will analyze the forces in the \(x\) direction and then the forces in the \(y\) direction.

Consider the forces in the \(x\) direction (Free-Body \(x\)). What forces act? How big are they? How do we know? Click each of the four \(x\)-direction free-body diagrams. Which one do you think is correct? Usually we know all of the forces that act, but here we just know of the push that is shown in Free-Body 1x. Is that the only force acting in the \(x\) direction? Newton's second law says that a net force acting on an object means that the object must be accelerating (the object's velocity changes). Does the block's velocity change? No (You can tell either by looking at the block's motion or by calculating the velocity and showing that it does not change.); therefore there must be another force acting, that of friction that opposes the motion. This eliminates Free-Body 1x and Free-Body 3x because they show only one force. The second force not only opposes motion, but in this animation it is exactly the same size as the push. This means that Free-Body 2x is not correct either. Therefore, Free-Body 4x depicts the correct free-body diagram for the forces that act in the \(x\) direction. (The form of the frictional force will be considered in detail in Chapter 5.)

Now consider the forces in the \(y\) direction (Free-Body \(y\)). What forces act? How big are they? How do we know? Click each of the four \(y\)-direction free-body diagrams. Which one do you think is correct? Usually we know all of the forces that act, but here we just know of the force of gravity that is shown in Free-Body 1y. Is that the only force acting in the \(y\) direction? Since the block does not accelerate in the \(y\) direction, there must be another force acting. This eliminates Free-Body 1y and Free-Body 2y because they show only one force. The force that is missing is the so-called normal force (the force of the table acting on the block) that opposes gravity. The normal force not only opposes motion in the \(y\) direction but here is exactly the same size as the gravitational force, the object's weight. This means that Free-Body 3y cannot be correct either. Therefore, Free-Body 4y depicts the correct free-body diagram for the forces that act in the \(y\) direction. Note that the normal force is not always equal to the weight. If there is an acceleration in the \(y\) direction or if the block is on an incline, the normal is not equal to the weight.

Note that in order to solve the complete problem we would draw all of the forces on one free-body diagram. We have done the analysis here by breaking up the motion into components. Given what we have said above, what does the complete free-body diagram look like? The complete free-body diagram animation shows the combination of the forces in the \(x\) direction and the forces in the \(y\) direction.

Illustration 3: Newton's Second Law and Force

Although most physicists would agree that the concept of force is not as fundamental as the concept of a conservation law, it is still considered central to the study of physics. A force is a push, a pull, or any other interaction, exerted by one object on another object. We know from experience that a push or a pull often causes an object to move. This allows us to quantify the definition of force in terms of a quantity that was defined previously: acceleration. Restart.

Note

If an object's mass remains constant, the magnitude of a force exerted on an object is proportional to the time rate of change of the velocity (i.e., acceleration). Specifically, \(\sum\mathbf{F} = m\mathbf{a}\).

Use this definition as you consider the results of Illustration 3 (position is given in meters and time is given in seconds). Set the mass in the text box before you select the graph type, velocity or acceleration.

The two-handed image ("handy") interacts with the \(1.0\text{-kg}\) cart in the animation if the image is near the left-hand or right-hand end of the cart. The arrow below the cart shows the direction and strength of the force exerted on the cart. You will have to move the image to keep it behind the cart since the interaction changes direction if the image passes through the center of the cart. Start the animation and explore it for a few minutes. Reset the animation if the cart goes off the end of the track.

Now select velocity (and then acceleration). Drag the handy image to the left of the cart and try to apply the force for as brief a period of time as you can. This will result in a force applied to the cart only for a short period of time and then no force will act. What do the resulting velocity and acceleration graphs look like? The velocity graph should show an increasing velocity for the instant handy is acting on the cart; then it should have a slope of zero. The velocity only changes when the force is acting. The acceleration graph should give a spike during the application of the force and be zero otherwise. Repeat the same process when the image is to the right of the cart. What changes? Because force is a vector, the applied force is now in the negative \(x\) direction. Therefore, the velocity and the acceleration are now both negative as well.

Now select velocity (and then acceleration). Drag the handy image to the left of the cart and then keep dragging it to the right as the cart moves. This will result in a constant force applied to the cart. What do the resulting velocity and acceleration graphs look like? The velocity graph should have a constant slope upward while the acceleration graph should give a constant acceleration during the application of the force. Repeat the process when the image is to the right of the cart. What changes? Because force is a vector, the applied force is now in the negative \(x\) direction. Therefore, the velocity and the acceleration are now both negative as well.

What changes on the velocity and acceleration graphs will occur if the mass of the cart is doubled or decreased by a factor of two? Try it and find out. Since acceleration is equal to the force over the mass, an increase in mass means a smaller acceleration, and a decrease in mass means a larger acceleration.

Illustration 4: Mass on an Incline

A mass is on a frictionless incline as shown in the animation  (position is given in meters and time is given in seconds. You may adjust m, the mass of the block \(100\text{ grams }< m < 500\text{ grams})\), and \(\theta\), the angle of the incline \((10^{\circ} <\theta < 45^{\circ})\), and view how these changes affect the motion of the block. Restart.

One of the first things to stress about this type of problem is that, for a suitable set of coordinates, while it is a two-dimensional problem, the motion of the block is one dimensional. Since the motion of the block is down the incline, let's choose that direction for the \(x\) axis. Since coordinate axes are perpendicular, let's also call the direction normal to the incline, the \(y\) axis. This does two things for us: The net force (and therefore the acceleration) is now on axis (the \(x\) axis) and we do not need to decompose the normal force. Check the box and click the "register and play" button to see the free-body diagram for the block and the net force acting on the block.

What force determines the acceleration of the block? It is the part of the gravitational force that is down the incline (\(mg\sin\theta\)). Therefore, the other component of the gravitational force (\(mg\cos\theta\)) must be equal to the normal force since we do not see the block flying off of the incline. The acceleration of the block is \(g\sin\theta\) down the incline.

Now try changing the mass of the block. How do you think the block's acceleration will change as you change the mass?

Now change the angle of the incline. How do you think the angle of the incline affects the acceleration of the block. In the animation you are limited to \(10^{\circ}>\theta > 45^{\circ}\). Can you predict, from either the formula or the animation, what will happen to the normal force and the acceleration when \(\theta = 0^{\circ}\) and \(\theta = 90^{\circ}\)?

Illustration authored by Mario Belloni.
Script authored by Steve Mellema, Chuck Niederriter and modified by Mario Belloni.

Illustration 5: Pull Your Wagons

Two toy wagons, attached by a lightweight rope (of negligible mass), are pulled with a constant force using another lightweight rope (again of negligible mass) as shown in the animation (position is given in centimeters and time is given in seconds). Restart. The mass of the red wagon is \(2.0\text{ kg}\) and the mass of the blue wagon is \(1.2\text{ kg}\). What is the force of the hand on the rope, and what is the tension in the rope joining the two wagons? To answer these questions, you must apply Newton's second law. However, when applying Newton's second law, you must first define the system that you are considering. Let's answer each question separately.

What is the force of the hand on the rope? Begin by defining the system to which we will apply Newton's second law. Since we want to determine the force of the hand on the rope, we start by choosing the rope to be our system. What forces act on the rope? It may help to draw a free-body diagram. View a free-body diagram of the rope along with the animation.

Note that there are two forces on the rope, the force of the hand on the rope in the \(+x\) direction and the force of the red wagon on the rope in the \(-x\) direction. These two forces are equal in magnitude; therefore, the net force on the rope is zero. But how can it be zero if the rope's acceleration is NOT zero? Since the mass of the rope is negligible, we set the mass of the rope equal to zero, and because of Newton's second law, the net force on the rope is zero. Of course, the rope's mass, in reality, is not zero; however, it's close enough to zero that we can say that it's approximately zero. Ultimately, this means that the tension in the rope is constant.

What is the force of the red wagon on the rope? What system should we now consider? We have two choices: (1) Consider the red wagon as the system or (2) consider the blue wagon, the red wagon, and the rope between them as the system. Either choice can lead you to the answer, but choice (2) is the most direct and best choice in order to solve the problem most quickly.

Consider the two wagons and rope as the system as depicted in this animation. The gray box represents the system. Now, draw a free-body diagram for the system and then view the animation again in order to check your answer. Once you have drawn the free-body diagram and identified the forces, you can apply Newton's second law, determine the force of the red wagon on the rope, and solve for the force of the hand on the rope.

That answers the first question. Now for the second question: What is the tension in the rope joining the two wagons? You can answer this question by following a similar procedure. Identify your system, draw a free-body diagram, and apply Newton's second law.

Illustration authored by and placed into the public domain by Aaron Titus.

Illustration 6: Newton's Third Law, Contact Forces

Illustration 6 shows graphs of position, velocity, and acceleration vs. time for a \(2\text{-kg}\) red block (not shown to scale) pushed by a \(12\text{-N}\) force on a frictionless horizontal surface (position is given in meters and time is given in seconds). Restart. The red block is in contact with (and therefore pushes on) the green \(1\text{-kg}\) block (also not shown to scale). Click here to show and play the physical situation. Note that on the position vs. time graph each block's trajectory is shown in a color-coded \(x(t)\) function, while in the velocity and acceleration vs. time graphs, a single \(v(t)\) or \(a(t)\) is shown (the blocks move together and therefore must have the same velocity and acceleration). The blocks may not move together when you set the contact forces.

Now it is up to you to determine what contact forces are required to make the motion of the blocks physical. When you are ready, select the "set values and play" button with the default forces. What happens? The red block "moves through" the green one because the forces are not correct. The red block has the \(12\text{-N}\) force acting on it and the green block has no forces acting on it. Of course each object's weight and normal force act in the vertical direction, but they cancel for each object. Here we are just considering the horizontal forces that could give a net force.

Try some values for the forces and check to see if you can get the same motion of the blocks and the same graphs as the physical situation.

Were you able to get the motion correct?  Let us now go about it systematically instead of by exploring (or guessing). Show and play the physical situation with both masses as one system. If we look at things this way we have one object of mass \(3\text{ kg}\) and a net force of \(12\text{ N}\), which means an acceleration of \(4\text{ m/s}^{2}\) (this is borne out by the acceleration graph).

What next? We could analyze the forces acting on the first mass, but let's analyze the second mass since it has only the first mass pushing on it. Because it has an acceleration of \(4\text{ m/s}^{2}\) and a mass of \(1\text{ kg}\), it must experience a force of \(4\text{ N}\) from the push of the red mass. What about the red mass? Newton's third law says it must experience an equal and opposite force, here a force of \(-4\text{ N}\). Try these values out (\(-4\text{ N}\) for the force on the red block and \(4\text{ N}\) for the force on the green block) to see if you believe what Newton's third law says the forces should be.

Illustration authored by Anne J. Cox and Mario Belloni.
Script authored by Anne J. Cox.

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.