11.1.5.1: Illustrations

Illustration 1: Static and Kinetic Friction 

How does friction change our analysis? Well, friction is due to the fact that objects, even smooth ones, are actually very bumpy on a microscopic level. Because of this, atoms from each surface are in such close contact that they form chemical bonds with each other. These bonds need to be broken in order for motion to occur. Restart.

The direction of the frictional force is opposite to the direction of motion, and the magnitude of the frictional force is proportional to the magnitude of the normal force, \(\text{N}\). There are two kinds of frictional forces: static and kinetic friction.

Static friction is the force when there is no relative motion between surfaces in contact. Conversely, kinetic friction is the force when there is relative motion between surfaces in contact. We represent the coefficient of friction by the Greek letter \(\mu\), and use subscripts corresponding to static and kinetic friction, respectively: \(\mu_{s}\) and \(\mu_{k}\). It is always the case that \(\mu_{s} >\mu_{k}\).

Consider what happens when you pull on a stationary block up to the point where there is still no motion. Set the mass to \(100\text{ kg}\) and vary \(F_{\text{applied}}\). The frictional force \(f_{s}\) matches \(F_{\text{applied}}\) up to the frictional force's maximum value, \(f_{\text{s; max}} = \mu_{s}\) \(N = 392\text{ N}\). After that the frictional force dramatically decreases, the block accelerates, and the frictional force becomes the kinetic friction of motion, \(f_{k} = \mu_{k}\text{ N}\).

So what are \(\mu_{s}\) and \(\mu_{k}\)? Well, given that there is no motion until when \(F_{\text{applied}} = 392\text{ N}\), \(f_{\text{s; max}}\) is approximately \(392\text{ N}\). Therefore, given that the normal force is \(980\text{ N}\), \(\mu_{s} = 0.4\). Now there is an acceleration when \(F_{\text{applied}} = 392\text{ N}\). Since the change in \(v = 9.8\text{ m/s}\) in \(5\) seconds, \(a = 1.96\text{ m/s}^{2}\). Given this, \(ma = F_{\text{applied}} - f_{k} = F_{\text{applied}} - \mu_{k}\text{ N}\). Therefore, after some algebra, \(\mu_{k} = 0.2\).

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

Illustration 2: Uniform Circular Motion: \(F_{c}\) and \(a_{c}\)

Uniform circular motion is an interesting mixture of one- and two-dimensional concepts. During uniform circular motion, the speed of the object must be constant. This is the uniform in uniform circular motion. So is an object moving in a circle with a constant speed accelerating?  Yes!  Why? The velocity is changing with time. Watch the animation (position is shown in meters and time is shown in seconds). The animation depicts an object moving in a circle at a constant speed. To determine the acceleration we need to consider the change in velocity for a change in time. Restart.

Since the speed does not change in time, what does change in time? It is the direction that changes with time. Draw two velocity vectors to convince yourself that the direction of the velocity changes with time. Recall that velocity has a direction (which always points tangent to the path, the so-called tangential direction) and a magnitude, and either or both can change with time. In what direction does the change in velocity point? Calculate the acceleration. It points toward the center of the circle. Since the object is accelerating, this motion must be due to a force (or a set of forces, a net force) that points solely toward the center of the circle. (Note: if the motion is nonuniform circular motion, the net force can point in another direction.) This direction—toward the center of the circle—is called the centripetal or center-seeking direction. It is often also called the radial direction, since the radius points from the center of the circle out to the object (the net force points in the opposite direction).

Therefore, for uniform circular motion, the acceleration always points toward the center of the circle. This is despite the fact that the velocity and the acceleration point in changing directions as time goes on. However, we get around this apparent difficulty in describing direction by defining the centripetal or radial direction and the tangential direction (the direction tangent to the circle). These directions change, but the velocity is always tangent to the circle, and the net force is always pointing toward the center of the circle. The following animation shows velocity and acceleration as the object undergoes uniform circular motion.

Illustration 3: The Ferris Wheel

So how does our analysis of Newton's laws change when an object is moving in uniform circular motion? There are really only two things to remember.

First, the net force is always toward the center of the circle for uniform circular motion. This net force is responsible for the acceleration toward the center of the circle, the centripetal acceleration we saw in Illustration 5.2.

Second, because the centripetal acceleration is a positive number, \(v^{2}/r\), it can never be negative. So unlike linear motion in which you have a choice of where to place the coordinate axes (to make life easier or more difficult), the choice here is critical. Your choice of coordinates must have one axis with its positive direction pointing toward the center of the circle.

In the animation, a Ferris wheel rotates at constant speed as shown (position is shown in meters and time is shown in minutes). Each square represents a chair on the Ferris wheel. Restart.

Consider a rider at point (a). What does the free-body diagram for a chair on the Ferris wheel look like at this point? To answer this question we must determine the applied forces that act on a rider when the rider is at point (a). At point (a) there are the normal force and the weight acting in opposite directions. Are the forces the same size or different? They must be different and the normal force must be bigger. Why? We know that the net force must point toward the center of the circle and that the net force is \(ma = m v^{2}/r\) for uniform circular motion.

What is the acceleration of the rider when the rider is at point (a)? As stated above we know the acceleration must be \(v^{2}/r\), where \(v = 2\pi r/T\), where \(T\) is the period of one revolution.

What about the answers to these questions when the rider is at points (b), (c), and (d)? Well, the forces may be different or point in different directions, but the results are the same. The net force must be toward the center of the circle and be \(m v^{2}/r\).

Illustration authored by Mario Belloni.
Script authored by Aaron Titus.

Illustration 4: Springs and Hooke's Law

Springs are interesting objects that for a range of stretching and compression follow Hooke's law. Hooke's law states that the force that the spring exerts is \(F = -k \Delta x\), where \(k\) is the spring constant and \(\Delta x\) is the displacement of the spring from its equilibrium position. In this Illustration the spring can be stretched by click-dragging the blue ball (position is given in centimeters and force is given in newtons). Slowly drag the spring back and forth and watch the graph. Restart.

Where is the equilibrium position of the spring? Given that \(\Delta x\) is measured from the equilibrium of the spring, look for the position where \(F = 0\text{ N}\). This occurs at \(x = 30\text{ cm}\). This is the equilibrium position.

What is the spring constant of the spring? It is not the force shown in the table divided by the position shown in the table. Why not? Recall that the "\(\Delta x\)" in the spring force equation is measured from equilibrium. Therefore, at maximum extension, \(x = 20\text{ cm}\) and the force is \(-160\text{ N}\). Therefore, \(k = 800\text{ N/m}\). Given that the negative of the slope of the line on the graph should also be \(k\), we can measure the spring constant by finding the slope and we get the same result.

The fact that spring forces are variable with position means that while we can determine the force, we cannot (given what we currently know) determine the velocity and position vs. time for an object attached to a stretched spring. Why? The force is not constant (it varies with position), and therefore the acceleration is not constant. This means we cannot use the kinematic equations for constant acceleration. What can we do? We can use concepts that you will learn about in Chapters 6 and 7.

Illustration 5: Air Friction

In this Illustration we compare the motion of a red projectile launched upward to that of an identical green projectile launched upward but subjected to the force of air friction. To make the motion easier to see, we have given both projectiles a slight horizontal velocity and do not consider frictional effects in this direction either. In addition, we show the free-body diagrams for each projectile (the force of gravity is drawn with a fatter vector so it is easier to see). Restart.

Watch the Position Graph animation and look at the free-body diagram. First, what is the direction of the force of air friction? It opposes motion, just as static and kinetic friction do. Consider the Velocity Graph animation. If we look at the motion on the way up, the velocity is positive, and therefore the force of friction opposes the motion and is downward, hence \(|a_{y}| > g\) on the way up. At the top of the arc, the velocity is zero, and hence \(|a_{y}| = g\). On the descent, the velocity is downward, and the force of air friction is therefore upward and hence \(|a_{y}| < g\). Therefore, |a_y| is greater on the way up! This is borne out by the Acceleration Graph. At some point, the frictional force has exactly the same size as the force of gravity. When this occurs there is no longer a net force, and the acceleration of the projectile is zero. The velocity corresponding to this situation is called the terminal velocity.

These animations are valid at low speeds. We can experimentally determine that the force of air friction is proportional to the velocity at low speeds, with \(\mathbf{R} = -b\:\mathbf{v}\), where \(\mathbf{R}\) is the resistive or drag force and b is a constant that depends on the properties of the air and the size and shape of the object. One benefit of this model is that the mathematics is a little easier to handle than for the high-speed case.

For massive small objects at high speeds (not depicted, but you can look at Exploration 5.6 to view this model) we can experimentally determine that the force of air friction is proportional to the velocity squared. The magnitude of the drag force can be represented as \(R = 1/2 D\rho Av^{2}\), where \(\rho\) is the density of air (mass/volume), \(A\) is the cross sectional area of the object, \(v\) is the magnitude of the velocity, and \(D\) is the drag coefficient \((0.2–2.0)\). Sometimes the drag force is written as \(bv^{2}\) with the assignment that \(b = 1/2D\rho A\). We can solve for the velocity as a function of time, but it is harder. We must be careful in this model if we have two-dimensional motion, since the \(x\) and \(y\) motions are no longer independent.

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