# Video: Determining the Action and Reaction Forces Associated with an Accelerating Object

A 1100.0-kg artillery shell is fired from a battleship, and the shell is accelerated at 2.40 × 10⁴ m/s²? What is the magnitude of the net external force exerted on the shell? What is the magnitude of the force exerted on the ship by the shell’s acceleration?

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### Video Transcript

A 1100.0-kilogram artillery shell is fired from a battleship, and the shell is accelerated at 2.40 times 10 to the fourth meters per second squared. What is the magnitude of the net external force exerted on the shell? What is the magnitude of the force exerted on the ship by the shell’s acceleration?

In this problem statement, we see some given information: the mass and acceleration of the shell. And we also see a force pair: there’s the force on the shell and there’s the force on the ship from the shell.

Let’s use this information to solve the problem. Now if we call 𝑚 the mass of the shell that’s given as 1100.0 kilograms and we call 𝑎 the acceleration the shell experiences that’s 2.40 times 10 to the fourth meters a second squared.

And the first thing we want to figure out here is the magnitude of the net external force exerted on the shell. And we can write that force as 𝐹 sub shell. And then the vertical lines on either side of that represent the magnitude of this term, so we’re not solving for a vector but for a scalar quantity.

Now to figure that out, let’s recall Newton’s second law. Newton’s second law tells us that the net force acting on an object is equal to that object’s mass multiplied by its acceleration. Now that’s just the information we have about the shell, so we can move forward solving for the force on the shell.

When we plug in these given values, we see that the magnitude of the force exerted on the shell is equal to 1100.0 kilograms multiplied by the acceleration of 2.40 times 10 to the fourth meters a second squared. Multiplying those together, we find a total force magnitude exerted on the shell of 2.64 times 10 to the seventh newtons, which is the same as 26400 kilonewtons.

So that number is equal to the total external force exerted on the shell. Now we move on to the second part of our problem, which asks what is the magnitude of the force exerted on the ship because of the shell’s acceleration? We can call that force magnitude 𝐹 sub ship.

And again we have absolute value bars around that to represent that we’re solving for magnitude. Now if we dropped the absolute value bars on both 𝐹 sub ship and 𝐹 sub shell, then we know something about how they relate.

These two forces are an action–reaction pair. This means the force of one on the other is equal and opposite to the force of the other on the one.

Now because in our case we’re working with force magnitudes rather than force vectors, we can write an action–reaction pair this way. The magnitude of 𝐹 sub ship, the force exerted on the ship due to the shell, is equal to the magnitude of the force exerted on the shell.

In other words, the magnitude of 𝐹 sub ship is equal also to 26400 kilonewtons. This is because these forces are in equal and opposite directions, but our signs are the same in this case because we’re solving for force magnitude.