PHYS101: Introduction to Mechanics

Example 4.3 Getting Up To Speed: Choosing the Correct System

A physics professor pushes a cart of demonstration equipment to a lecture hall, as seen in Figure 4.10. Her mass is 65.0 kg, the cart's is 12.0 kg, and the equipment's is 7.0 kg. Calculate the acceleration produced when the professor exerts a backward force of 150 N on the floor. All forces opposing the motion, such as friction on the cart's wheels and air resistance, total 24.0 N.

Figure 4.10 A professor pushes a cart of demonstration equipment. The lengths of the arrows are proportional to the magnitudes of the forces (except for $f$, since it is too small to draw to scale). Different questions are asked in each example; thus, the system of interest must be defined differently for each. System 1 is appropriate for this example, since it asks for the acceleration of the entire group of objects. Only $F_{floor}$ and $f$ are external forces acting on System 1 along the line of motion. All other forces either cancel or act on the outside world. System 2 is chosen for Example 4.4 so that $F_{prof}$ will be an external force and enter into Newton's second law. Note that the free-body diagrams, which allow us to apply Newton's second law, vary with the system chosen.

Strategy

Since they accelerate as a unit, we define the system to be the professor, cart, and equipment. This is System 1 in Figure 4.10. The professor pushes backward with a force $F_{foot}$ of 150 N. According to Newton's third law, the floor exerts a forward reaction force $F_{floor}$ of 150 N on System 1. Because all motion is horizontal, we can assume there is no net force in the vertical direction.

The problem is therefore one-dimensional along the horizontal direction. As noted, $f$ opposes the motion and is thus in the opposite direction of $F_{floor}$. Note that we do not include the forces $F_{prof}$ or $F_{cart}$ because these are internal forces, and we do not include $F_{foot}$ because it acts on the floor, not on the system. There are no other significant forces acting on System 1. If the net external force can be found from all this information, we can use Newton's second law to find the acceleration as requested. See the free-body diagram in the figure.

Solution

Newton's second law is given by

$a= \frac{F_{net}}{m}$.

The net external force on System 1 is deduced from Figure 4.10 and the discussion above to be

$F_{net}=F_{floor}−f = 150\ N − 24.0\ N = 126\ N$.

The mass of System 1 is

$m=(65.0 + 12.0 + 7.0)kg = 84\ kg$.

These values of $F_{net}$ and $m$ produce an acceleration of

$\begin{aligned}&a=\frac{F_{\text {net }}}{m} \\&a=\frac{126 \mathrm{~N}}{84 \mathrm{~kg}}=1.5 \mathrm{~m} / \mathrm{s}^{2}\end{aligned}$

Discussion

None of the forces between components of System 1, such as between the professor's hands and the cart, contribute to the net external force because they are internal to System 1. Another way to look at this is to note that forces between components of a system cancel because they are equal in magnitude and opposite in direction. For example, the force exerted by the professor on the cart results in an equal and opposite force back on her. In this case both forces act on the same system and, therefore, cancel. Thus internal forces (between components of a system) cancel. Choosing System 1 was crucial to solving this problem.

Example 4.4 Force on the Cart – Choosing a New System

Calculate the force the professor exerts on the cart in Figure 4.10 using data from the previous example if needed.

Strategy

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, $F_{prof}$, is an external force acting on System 2. $F_{prof}$ was internal to System 1, but it is external to System 2 and will enter Newton's second law for System 2.

Solution

Newton's second law can be used to find $F_{prof}$. Starting with

$a = \frac{F_{net}}{m}$

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

$F_{net}=F_{prof} − f$,

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

$F_{prof} = F_{net} + f$.

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

$F_{net}=ma$,

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

$F_{net}=ma$,

$F_{net}=(19.0\ kg)(1.5\ m/s^2)=29\ N$.

Now we can find the desired force:

$F_{prof}=F_{net}+f$,

$F_{prof}=29\ N + 24.0\ N = 53\ N$.

Discussion

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).