Force on the Cart—Choosing a New System

If we now define the system of interest to be the cart plus equipment (System 2 in Figure 4.10), then the net external force on System 2 is the force the professor exerts on the cart minus friction. The force she exerts on the cart, ${\mathbf{\text{F}}}_{\text{prof}}$, is an external force acting on System 2. ${\mathbf{\text{F}}}_{\text{prof}}$ was internal to System 1, but it is external to System 2 and will enter Newton’s second law for System 2.

Newton’s second law can be used to find ${\mathbf{\text{F}}}_{\text{prof}}$. Starting with

and noting that the magnitude of the net external force on System 2 is

we solve for $F_{\text{prof}}$, the desired quantity:

The value of $f$ is given, so we must calculate net $F_{\text{net}}$. That can be done since both the acceleration and mass of System 2 are known. Using Newton’s second law we see that

where the mass of System 2 is 19.0 kg ($m$= 12.0 kg + 7.0 kg) and its acceleration was found to be $a=1.5{\text{m/s}}^2$ in the previous example. Thus,

Now we can find the desired force:

It is interesting that this force is significantly less than the 150-N force the professor exerted backward on the floor. Not all of that 150-N force is transmitted to the cart; some of it accelerates the professor.

The choice of a system is an important analytical step both in solving problems and in thoroughly understanding the physics of the situation (which is not necessarily the same thing).