New Way to Approach Newton's Second Law (Physics Education)

by John R. Walkup, Ph.D.

Most students solve physics problems involving forces by initially drawing a free-body diagram for the body in question, and then vector summing the forces acting on the body.

Based on my own anecdotal experience teaching introductory physics, this traditional approach suffers two significant disadvantages:

1. Once students have summed the forces, they are often at a loss as to how to finish the problem.
2. Little connects the solutions to problems involving linear, radial, and angular acceleration.

These disadvantages appear because the traditional approach establishes no explicit connection between the motion of a body and the forces that appear in the free-body diagram. Even in situations where sufficient information exists for students to solve the problem through kinematic variables alone, students will still draw force vectors and sum them.

Rather, I suggest centering the solution process on the acceleration of the body because (a) acceleration appears in both Newton’s second law and all fundamental equations of motion and (b) embodied within the acceleration is all the information that is knowable about the subsequent motion of the body.

As a result, the acceleration is loaded with physical insight; knowledge of the acceleration is paramount to seeking a solution to a physical problem.  In this sense, it serves as the state vector for classical physics problems involving forces.

This approach naturally ties into the archetypal problem in introductory physics:

• “Given the forces acting on a body, find its subsequent motion.”

This statement is simply a recasting of the relationship between cause and effect in terms of the language of physics. Its corollary,

• “Given the motion of a body, find all that is knowable about the forces acting on the body,”

is not as strong a statement, since the relationship between an effect and its causes is not nearly as easy to establish.

Since acceleration links the forces acting on a body to its motion, then the key to solving the archetypal problem in introductory physics is to find the acceleration. Fortunately, there are only two ways of doing so — either use the equations of motion or apply Newton’s second law.

Finding acceleration

Let’s define NSL for Newton’s second law and EOM for equations of motion. We can then summarize the archetypal problem and its corollary as the following:

• “Given the forces acting on a body, find its motion,” that is, NSL → a → EOM.
• “Given the motion of a body, analyze the forces acting on the body,” that is, NSL ← a ← EOM.

Therefore, nearly the entirety of Newton’s second law problems involves either driving the process to the right (that is, NSL → a → EOM) or to the left (NSL ← a ← EOM).

Examples

Naturally, we need examples illustrating this method for all three distinct types of acceleration, that is, linear, radial, and angular acceleration.  In this blog post, I will concentrate on linear acceleration, with subsequent blogs devoted to the other two types.

Example 1:  NSL  → a → EOM

For the first example, we choose a problem that often gives introductory physics students fits — the accelerometer.

Picture a rock hanging by a string from the rear view mirror of an accelerating car. If the string forms an angle θ = 30^o with respect to the vertical, how long would it take for the car to accelerate from rest a distance of 400 meters?

As always, the key is to find the acceleration. We do not know enough about the motion of the object (the rock) to find its acceleration using the equations of motion, therefore we instead apply Newton’s second law. In other words, we are driving the problem-solving process to the right, NSL → a.

Summing the forces acting along the horizontal direction, we find N sin(30^o) = ma_x. Therefore, a_x = N sin(30^o)/m. Clearly we need an expression for the normal force. Summing the forces along the vertical, we find Ncos(30^o) − mg = ma_y = 0. This means that N = 11.32 m and, therefore, a_x = 5.7 m/s².

Knowing the acceleration, we finish the problem by performing the last step in the method – applying the equations of motion, that is a → EOM. The equation of motion for displacement along the horizontal direction, x = v_0xt + (1/2)a_xt², becomes 400 = (1/2)(5.7)t². Therefore, the time it takes for the car to travel 400 meters is t = 11.8 seconds.

Example 2:  NSL ← a ← EOM

Now consider the same vehicle, but with the problem statement given as:

A vehicle accelerates from rest over a distance of 400 meters in 11.8 seconds. Hanging from the rear-view mirror of the vehicle is a rock. Calculate the angle the string makes with the vertical.

In this case, we have sufficient information to find the acceleration through the equation of motion, so we apply a ← EOM.

The equation of motion for displacement, x = v_0xt + (1/2)a_xt², becomes 400 = (1/2)a_x(11.8)², which produces an acceleration a_x = 5.7 m/s².

Now that we have the acceleration, we finish the problem through the step NSL ← a.

Summing the forces acting along the horizontal direction, we find N sin(θ) = m (5.7).  Summing the forces along the vertical, we find N cos(θ) − mg = ma_y = 0. This means that tan(θ) = 5.7/9.8 and, therefore, θ = 30^o.

Summary

The vast majority of physics problems involving forces and motion correspond to one of two scenarios:

• NSL → a → EOM 

or

• NSL ← a ← EOM

Solving introductory physics problems therefore becomes a matter of examining the information given by the problem statement and driving the solution process to the right or to the left.

In future posts, I will show how the same method applies to rotational and radial acceleration. We will see that rotational acceleration is no harder than linear acceleration and that radial acceleration is actually easier.

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