Newton's Laws in Three Dimensions
Article objectives
Forces have no perpendicular effects
Suppose you could shoot a rifle and arrange for a second bullet to be dropped from the same height at the exact moment when the first left the barrel. Which would hit the ground first? Nearly everyone expects that the dropped bullet will reach the dirt first, and Aristotle would have agreed. Aristotle would have described it like this. The shot bullet receives some forced motion from the gun. It travels forward for a split second, slowing down rapidly because there is no longer any force to make it continue in motion. Once it is done with its forced motion, it changes to natural motion, i.e. falling straight down. While the shot bullet is slowing down, the dropped bullet gets on with the business of falling, so according to Aristotle it will hit the ground first.
Figure a: A bullet is shot from a gun, and another bullet is simultaneously dropped from the same height. 1. Aristotelian physics says that the horizontal motion of the shot bullet delays the onset of falling, so the dropped bullet hits the ground first. 2. Newtonian physics says the two bullets have the same vertical motion, regardless of their different horizontal motions.
Luckily, nature isn't as complicated as Aristotle thought! To convince yourself that Aristotle's ideas were wrong and needlessly complex, stand up now and try this experiment. Take your keys out of your pocket, and begin walking briskly forward. Without speeding up or slowing down, release your keys and let them fall while you continue walking at the same pace.
You have found that your keys hit the ground right next to your feet. Their horizontal motion never slowed down at all, and the whole time they were dropping, they were right next to you. The horizontal motion and the vertical motion happen at the same time, and they are independent of each other. Your experiment proves that the horizontal motion is unaffected by the vertical motion, but it's also true that the vertical motion is not changed in any way by the horizontal motion. The keys take exactly the same amount of time to get to the ground as they would have if you simply dropped them, and the same is true of the bullets: both bullets hit the ground simultaneously.
These have been our first examples of motion in more than one dimension, and they illustrate the most important new idea that is required to understand the three-dimensional generalization of Newtonian physics:
Forces have no perpendicular effects.
When a force acts on an object, it has no effect on the part of the object's motion that is perpendicular to the force.
In the examples above, the vertical force of gravity had no effect on the horizontal motions of the objects. These were examples of projectile motion, which interested people like Galileo because of its military applications. The principle is more general than that, however. For instance, if a rolling ball is initially heading straight for a wall, but a steady wind begins blowing from the side, the ball does not take any longer to get to the wall. In the case of projectile motion, the force involved is gravity, so we can say more specifically that the vertical acceleration is 9.8 m/s\(^{2}\) , regardless of the horizontal motion.
Relationship to relative motion
These concepts are directly related to the idea that motion is relative. Galileo's opponents argued that the earth could not possibly be rotating as he claimed, because then if you jumped straight up in the air you wouldn't be able to come down in the same place. Their argument was based on their incorrect Aristotelian assumption that once the force of gravity began to act on you and bring you back down, your horizontal motion would stop. In the correct Newtonian theory, the earth's downward gravitational force is acting before, during, and after your jump, but has no effect on your motion in the perpendicular (horizontal) direction.
If Aristotle had been correct, then we would have a handy way to determine absolute motion and absolute rest: jump straight up in the air, and if you land back where you started, the surface from which you jumped must have been in a state of rest. In reality, this test gives the same result as long as the surface under you is an inertial frame. If you try this in a jet plane, you land back on the same spot on the deck from which you started, regardless of whether the plane is flying at 500 miles per hour or parked on the runway. The method would in fact only be good for detecting whether the plane was accelerating.
Figure b: This object experiences a force that pulls it down toward the bottom of the page. In each equal time interval, it moves three units to the right. At the same time, its vertical motion is making a simple pattern of +1, 0, -1, -2, -3, -4, ... units. Its motion can be described by an x coordinate that has zero acceleration and a y coordinate with constant acceleration. The arrows labeled x and y serve to explain that we are defining increasing x to the right and increasing y as upward.
Coordinates and components
How do we convert these ideas into mathematics? Figure b shows a good way of connecting the intuitive ideas to the numbers. In one dimension, we impose a number line with an x coordinate on a certain stretch of space. In two dimensions, we imagine a grid of squares which we label with x and y values, as shown in figure b.
But of course motion doesn't really occur in a series of discrete hops like in chess or checkers. Figure c shows a way of conceptualizing the smooth variation of the x and y coordinates. The ball's shadow on the wall moves along a line, and we describe its position with a single coordinate, y, its height above the floor. The wall shadow has a constant acceleration of -9.8 m/s\(^{2}\) . A shadow on the floor, made by a second light source, also moves along a line, and we describe its motion with an x coordinate, measured from the wall.
The velocity of the floor shadow is referred to as the x component of the velocity, written \(v_{x}\). Similarly we can notate the acceleration of the floor shadow as \(a_{x}\). Since \(v_{x}\) is constant, \(a_{x}\) is zero.
Similarly, the velocity of the wall shadow is called \(v_{y}\), its acceleration \(a_{y}\). This example has \(a_{y}\)=-9.8 m/s\(^{2}\) .
Because the earth's gravitational force on the ball is acting along the y axis, we say that the force has a negative y component, \(F_{y}\), but \(F_{x}\)=\(F_{z}\)=0.
The general idea is that we imagine two observers, each of whom perceives the entire universe as if it was flattened down to a single line. The y-observer, for instance, perceives y, \(v_{y}\), and \(a_{y}\), and will infer that there is a force, \(F_{y}\), acting downward on the ball. That is, y component means the aspect of a physical phenomenon, such as velocity, acceleration, or force, that is observable to someone who can only see motion along the y axis.
All of this can easily be generalized to three dimensions. In the example above, there could be a z-observer who only sees motion toward or away from the back wall of the room.
Figure c: The shadow on the wall shows the ball's y motion, the shadow on the floor its x motion.
Example 1: A car going over a cliff
◊ The police find a car at a distance w=20 m from the base of a cliff of height h=100 m . How fast was the car going when it went over the edge? Solve the problem symbolically first, then plug in the numbers.
◊ Let's choose y pointing up and x pointing away from the cliff. The car's vertical motion was independent of its horizontal motion, so we know it had a constant vertical acceleration of a=-g=-9.8 m/s\(^{2}\) . The time it spent in the air is therefore related to the vertical distance it fell by the constant-acceleration equation
$$\Delta{y} = 1/2 a_y \Delta{} t^2$$ $$\text{or} \; -h = 1/2 (-g) \Delta{} t^2$$
Solving for Δ t gives
$$Δt=2hg.$$
Since the vertical force had no effect on the car's horizontal motion, it had a\(_{x}\)=0, i.e., constant horizontal velocity. We can apply the constant-velocity equation
$$v_x = \Delta{x} / \Delta{t}$$ $$\text{i.e.,} \; v_x = w / \Delta{t}$$
We now substitute for Δ t to find
$$v_x=w / 2hg,$$
Plugging in numbers, we find that the car's speed when it went over the edge was 4 m/s, or about 10 mi/hr.
Projectiles move along parabolas.
What type of mathematical curve does a projectile follow through space? To find out, we must relate x to y, eliminating t. The reasoning is very similar to that used in the example above. Arbitrarily choosing x=y=t=0 to be at the top of the arc, we conveniently have x=Δ x, y=Δ y, and t=Δ t, so
$$y =1/2 a_y t^2 (a_y<0) x =vxt$$
We solve the second equation for t=x/v\(_{x}\) and eliminate t* in the first equation:
$$y=1/2 a_y (x/v_x)^2.$$
Since everything in this equation is a constant except for x and y, we conclude that y is proportional to the square of x. As you may or may not recall from a math class, y∝ x\(^{2}\) describes a parabola.
Figure e: A parabola can be defined as the shape made by cutting a cone parallel to its side. A parabola is also the graph of an equation of the form y∝ x\(^{2}\).
Figure f: Each water droplet follows a parabola. The faster drops' parabolas are bigger.
Newton's laws in three dimensions
It is now fairly straightforward to extend Newton's laws to three dimensions:
Newton's first law
If all three components of the total force on an object are zero, then it will continue in the same state of motion.
Newton's second law
The components of an object's acceleration are predicted by the equations
$$a_x =F_{x,total}/m, \; \; a_y =F_{y,total}/m, \; \; and \; a_z =F_{z,total}/m.$$
Newton's third law
If two objects A and B interact via forces, then the components of their forces on each other are equal and opposite:
$$FA \; \text{on}\; B,x =-FB \; \text{on}\; A,x$$ $$FA \; \text{on}\; B,y =-FB \; \text{on}\; A,y$$ $$\text{and} \; FA \; \text{on}\; B,z =-FB \; \text{on}\; A,z.$$
Example 2: Forces in perpendicular directions on the same object
◊ An object is initially at rest. Two constant forces begin acting on it, and continue acting on it for a while. As suggested by the two arrows, the forces are perpendicular, and the rightward force is stronger. What happens?
◊ Aristotle believed, and many students still do, that only one force can “give orders” to an object at one time. They therefore think that the object will begin speeding up and moving in the direction of the stronger force. In fact the object will move along a diagonal. In the example shown in the figure, the object will respond to the large rightward force with a large acceleration component to the right, and the small upward force will give it a small acceleration component upward. The stronger force does not overwhelm the weaker force, or have any effect on the upward motion at all. The force components simply add together:
$$F_{x,total} =F_{1,x} + t_o F_{2,x} \; \; F_{y,total} =t_o F_{1,y}+F_{2,y}$$