You might have already learned about Newton's second law of motion. But what good is it anyway? In this article, we'll convince you that this law of motion is not an abstract mathematical statement detached from reality. Instead, Newton's second law correctly describes many phenomena, many of which you encounter daily.
We'll begin with an illustration of Newton's second law of motion, in which we'll review the meaning of the symbols in the force law equation and discuss how these quantities are related. Next, we'll list 10 applications of Newton's second law of motion to everyday phenomena before selecting two examples for a more in-depth semi-quantitative analysis.
Illustration of Newton's Second Law of Motion
A dynamic quantity is a quantity whose value can change with time
If we want to describe the motion of a physical system, we must know its dynamic quantities first: displacement, velocity, and acceleration. For the case of linear motion, which describes the motion of an object in one spatial dimension only, along a straight line, applying Newton's second law gives us all the information we need about the change of the object's motion with time. In this case, we'd write:
\[\vec{a} = \frac{\sum \vec{F}}{m}.\]
Recall that the little arrows above the acceleration \(\vec{a}\) and the force \(\vec{F}\) denote vector quantities. It's essential to stick to vector notation because a system may accelerate along one spatial dimension but not along another dimension. For example, an apple falling from a tree only accelerates along the vertical dimension.
Moreover, since acceleration and force are the only vector quantities in Newton's second law of motion (mass doesn't have direction since it's a scalar), the acceleration of the center of mass of a system will always be in the same direction as the net force acting on it.
The symbol before \(\vec{F}\) is the Greek letter Sigma, which we use to denote a sum in mathematics. That is,
\[\vec{F}_{\text{net}} = \sum \vec{F} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \dots\]
Each of the terms on the right-hand side is a force acting on the system's center of mass. Since all the external forces acting on a system may come from different directions, it's vital not to drop vector notation when we're writing the net force! If the forces are unbalanced (meaning the net forces on the object is not zero) then the object accelerates.
Contrast this with the mass, which sits below the division line on the right-hand side. If the center of mass of the system increased by a factor of ten, we'd be dividing the net force by a value ten times larger than the original value. Consequently, the acceleration would decrease by a factor of ten. Therefore, the acceleration of the center of mass of a system is inversely proportional to the system's mass.
The figure below illustrates what we discussed above. Make sure you understand it before reading ahead.
Suppose we know the mass of a system and have computed its acceleration from the net force. How does this help us find the other dynamic quantities: velocity and displacement? We arrive at the answer from the definition of acceleration.
Acceleration is the rate of change of velocity with respect to time.
Written as an equation we have
\[\vec{a}_{\text{avg}}=\frac{\Delta \vec{v}}{\Delta t},\]
where \(\Delta \vec{v} = \vec{v} - \vec{v_0}\) denotes the difference between the final and initial velocity. You may recognize from previous math studies that the right-hand side looks like the slope of a curve. Indeed, this is what we mean by the phrase rate of change in the above definition. But why did we write \(\vec{a}_{\text{avg}}\) just now instead of just \(\vec{a}\)?
We do this to specify that the above formula is used to calculate the average acceleration of an object. The meaning of average acceleration, as opposed to instantaneous acceleration, is that we compute \(\Delta \vec{v}\) as the difference between the final and initial velocity over a fixed time interval \(\Delta t\). In doing so, however, we neglect any intermediate values the velocity may assume. Hence, the value is only an average.
Velocity is the rate of change of displacement with respect to time.
Similarly, the average velocity on a time interval satisfies the equation
\[\vec{v}_{\text{avg}}=\frac{\Delta \vec{x}}{\Delta t}.\]
where \(\Delta \vec{x} = \vec{x} - \vec{x_0}\) denotes the difference between the final and initial displacement. Working with average values for acceleration and velocity is enough to solve most problems you'll encounter in algebra-based physics courses. But you may be curious about how to make an average value, which is only an approximation, more exact. Doing so involves the concept of instantaneous acceleration we mentioned above.
The smaller the interval of time over which the velocity varies, the more accurate the value of the system's center of mass acceleration will be. To make the denominator arbitrarily small without the expression for acceleration becoming undefined, we have to borrow an idea from calculus, which we call a limit. Specifically, we say that in the limit that the time interval goes to zero, the resulting expression that we get is that of the derivative of velocity with respect to time.
Just for fun, this is how acceleration as the instantaneous rate of change of velocity with respect to time looks like in a calculus-based course like AP Physics C:
\[\vec{a} = \frac{\mathrm{d}\vec{v}}{\mathrm{d}t}\]
You don't have to worry about learning this equation. When you study calculus, you'll understand the meaning of replacing \(\Delta \vec{v}\) and \(\Delta t\) with \(\mathrm{d}\vec{v}\) and \(\mathrm{d}t\).
10 Applications of Newton's Second Law of Motion
Below is a list of 10 applications of Newton's second law of motion.
- A golfer strikes a golf ball with a club.
- A boy lets go of a rope swing to dive into a lake.
- A construction crane lifts a steel beam from the ground.
- A space shuttle uses its throttles to adjust its course in space.
- A sliding hockey puck slows down to a halt.
- A girl tosses a ball to her mother.
- Two dogs compete at tug-of-war over their favorite chew toy.
- A car crashes into a parked car and drags it along.
- The Earth-Moon system moves in a fixed orbit around the Sun.
- Two workers push a crate up a ramp.
Here's a question for you. Did you passively note what each situation was describing or did you try to think about the dynamic quantities involved in each scenario? Physics, just like mathematics, is not a spectator sport. You have to actively think about what you're reading to solidify your understanding. Asking yourself questions such as the ones below is a great way to accomplish this.
- What are the interacting systems in this scenario?
- What are the component parts of each system, if any?
- Which system's velocity is changing?
- What are the external forces acting on the moving system?
- What would a free-body diagram of this scenario look like?
The example below is meant to guide you on how to answer these questions effectively.
Answer the questions for the case of a construction crane lifting a steel beam from the ground.
- What are the interacting systems in this scenario?
- The interacting systems are the construction crane and the steel beam.
- What are the component parts of each system, if any?
- The steel beam is a rigid solid object without component parts. On the other hand, the crane has many component parts: a motor with counterweights, the operator's cabin, the working arm, a trolley with a hook block, and a system of wires connecting everything. However, the only component of the crane we need to consider to analyze the motion of the steel beam is the wire from the working arm to the hook block.
- Which system's velocity is changing?
- The steel beam's velocity is changing from rest to a non-zero value pointing upwards as it is pulled by the crane.
- What are the external forces acting on the moving system?
- The moving system is the steel beam and the external forces acting on it are gravity and the tension from the hook block.
- What would a free-body diagram of this scenario look like?
- A free-body diagram of this scenario appears below. Note that we've omitted all details except the two forces acting on the steel beam: the applied tension force from the crane's chain pulling upward and the force of gravity pulling the beam downward. Since the beam had a positive velocity pointing upwards, the applied force is bigger in magnitude than the force of gravity opposing, as depicted by the arrows.
Before moving on to the next section, try repeating the above exercise with another of the 10 applications of Newton's second law we listed above.
Center of mass
With these applications of Newton's second law of motion fresh in your mind, now is an excellent opportunity to clarify why we kept using the phrase center of mass in the previous section. A system can consist of many moving parts or have objects inside it interacting with each other. However, if the interactions among the moving parts do not affect the outward motion of the system as a whole, we may neglect them.
For example, astronauts may move about inside the shuttle but, viewed from the outside, whatever they are doing has no effect on the overall motion of the ship. Thus, we say that the variables \(\vec{x}\), \(\vec{v}\), and \(\vec{a}\) always refer to the center of mass quantities to highlight that we only care about the overall motion of the system.
2 examples of Newton's Second Law of Motion
Now that we've seen some applications of Newton's second law of motion, it's time to analyze two examples in greater detail. Let's pick two from the list above: the girl tossing a ball to her mother and the Earth-Moon system. We've chosen these two because the first is an example of projectile motion, while the second is an example of circular motion. Both topics appear in the AP Physics 1 exam, so you need to familiarize yourself with these ideas!
Example of Projectile Motion: Tossing a Ball
The characteristic feature of projectile motion is that it traces out a parabola. In fact, launching a projectile in any direction other than directly vertically upward will result in parabolic motion; changing the launch angle will only affect the range of the projectile. For the sake of simplicity, we will neglect air resistance. Hence, the net force acting on the ball at every point along its flight will be the force of gravity. The figure below depicts the ball at the highest point in its trajectory. At this exact point in this path, the ball's vertical velocity is zero, which is why we've omitted the vector along that direction.
A ball thrown at a 45 °angle will trace out a parabola, Vaia Originals.
Knowing the angle and initial velocity at which the ball was tossed, it might be interesting to find out the maximum vertical displacement it reaches. The following example will help us quantitatively determine the answer to this question.
Question
A girl tosses a ball at a velocity of \(5\,\mathrm{\frac{m}{s}}\) towards her mother. If the angle at which she tosses the ball is \(45^\circ\), what is the maximum height the ball reaches along its path? Assume the girl is \(1.40 \, \mathrm{m}\) tall. Ignore the effects of air resistance.
Solution
First of all, note that the only force acting on the ball during its flight is the force of gravity. Because this force accelerates all falling objects uniformly, we are in a situation of constant acceleration. This means we can use the kinematic equations. Moreover, we are not interested in the ball's horizontal displacement, so we need only consider its motion along the vertical direction.
We proceed by invoking the time-independent kinematic equation:
\[v_y^2 = v_{0y}^2 + 2a_y(y - y_0).\]
The ball's vertical velocity will be zero at the highest point in its flight, so we can set \(v_y = 0\,\mathrm{\frac{m}{s}}\). With this information, we can rearrange the equation to solve for the quantity we're interested in:
\begin{align} 0 &= v_{0y}^2 + 2a_yy - 2a_yy_0,\\ -2a_yy &= v_{0y}^2 - 2a_yy_0,\\ y &= -\frac{v_{0y}^2}{2a_y} + y_0.\end{align}
where \(y\) denotes the ball's maximum height. Now, we know that the vertical acceleration is the acceleration due to gravity, so \(a_y = -9.8\,\mathrm{\frac{m}{s^2}}\). Moreover, \(y_0 = 1.40 \, \mathrm{m}\) corresponds to the girl's height. What about the ball's initial velocity along the vertical direction?
Here we need to use some basic trigonometry:
\begin{align} v_{y0} &= \vec{v}_0\sin(\theta)\\ &= 5\,\mathrm{\frac{m}{s}}\sin(45^\circ)\\ &= 5\,\mathrm{\frac{m}{s}} \times \frac{\sqrt{2}}{2} \\ &= 3.54 \,\mathrm{\frac{m}{s}}.\end{align}
We are now in a position to substitute in all known quantities:
\begin{align} y &= -\frac{v_{0y}^2}{2a_y} + y_0,\\ &= \frac{-\left(3.54 \,\mathrm{\frac{m}{s}}\right)^2}{2\left(-9.8\,\mathrm{\frac{m}{s^2}}\right)} + 1.40 \, \mathrm{m},\\ &= 2.04 \, \mathrm{m}.\end{align}
Thus, the ball reaches a maximum height of \(2.04 \, \mathrm{m}\).
We solved the problem above with the help of the time-independent kinematic equation. However, this is not the only way of arriving at the answer. Can you think of another method?
Hint: Use \(v_y = v_{0y} + a_yt\) to solve for the time required by the ball to reach its maximum height and then plug this result into the quadratic kinematic equation.
Examples of Circular Motion: The Earth-Moon System
The figure below is a free-body diagram of an object orbiting a central-force field. At the center of the diagram, we have the gravitational source, while on the right side, we have the object in orbit. Since the Sun keeps the Earth-Moon system in orbit around it due to gravity, and gravity is a force, this is an example of Newton's second law. The yellow dot is the Sun, while the blue dot with the grey dot around it is the Earth-Moon system.
Why are we ignoring the Moon's motion in this example? While it does feel a gravitational attraction to the Sun, its gravitational interaction with the Earth is much stronger. Indeed, it is incorrect to say that the Earth orbits the Sun. Rather, it's the center of mass of the Earth-Moon system which orbits the Sun. Put another way, the interaction between the Earth and the Moon does not contribute to their center of mass's acceleration due to the external force exerted by the Sun.
According to Kepler's Laws, planetary orbits are, in fact, ellipses, meaning that they are not perfectly circular. However, in many cases such as this one, it's okay to approximate the motion of the Earth-Moon system around the Sun as circular.
Applications of Second Law - Key takeaways
- The displacement, velocity, and acceleration of the center of mass of a system describe its linear motion.
- The acceleration of the center of mass of a system will always be in the same direction as the net force acting on it.
- The acceleration of the center of mass of a system is directly proportional to the net force acting on it and inversely proportional to its mass.
- Acceleration \(\vec{a}\) is equal to the rate of change of velocity with respect to time.
- Velocity \(\vec{v}\) is equal to the rate of change of position with respect to time.
- The variables \(\vec{x}\), \(\vec{v}\), and \(\vec{a}\) all refer to the center of mass quantities.
- If the interacting objects are parts of the same system, there will be no change in that system's center of mass velocity.
References
- Fig.1 - Acceleration of a falling apple, Vaia Originals
- Fig.2 - Free-body diagram of a steel beam, Vaia Originals
- Fig.3 - A thrown ball tracing a parabola, Vaia Originals
- Fig.4 - The Earth-Moon system orbiting the Sun, Vaia Originals
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