Figure 5-27 shows three blocks being pushed across a frictionless floor by horizontal force$\overrightarrow{\mathbf{F}}$. What total mass is accelerated to the right by (a) force$\overrightarrow{\mathbf{F}}$, (b) force${\overrightarrow{\mathbf{F}}}_{\mathbf{21}}$on block 2 from block 1, and (c) force${\overrightarrow{\mathbf{F}}}_{\mathbf{32}}$on block 3 from block 2? (d) Rank the blocks according to their acceleration magnitudes, greatest first. (e) Rank forces$\overrightarrow{\mathbf{F}}\mathbf{,}{\overrightarrow{\mathbf{F}}}_{\mathbf{21}}\mathbf{}\mathbf{and}\mathbf{}{\overrightarrow{\mathbf{F}}}_{\mathbf{32}}$, and according to magnitude, greatest first.

$$\begin{gathered} {\mathrm{m}}_1=5\mathrm{kg} \\ {\mathrm{m}}_2=2\mathrm{kg} \\ {\mathrm{m}}_3=10\mathrm{kg} \end{gathered}$$

The problem is based on Newton’s second law of motion which states that the acceleration of an object is dependent on the net force acting upon the object and the mass of the object. Using a free body diagram, how much mass the respective force is accelerating can be calculated. Using Newton's 2^nd law of motion, the block’s acceleration and the forces can be ranked.

Formula:

${\mathbf{F}}_{\mathbf{net}}\mathbf{=}\mathbf{ma}$

$\overrightarrow{\mathrm{F}}$is pushing all the three blocks, so we conclude that,

role="math" localid="1657009110359" $$\begin{gathered} \mathrm{Total}\mathrm{mass}\mathrm{accelerated}\mathrm{by}\overrightarrow{F}is={\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3 \\ =5\mathrm{kg}+2\mathrm{kg}+10\mathrm{kg} \\ =17\mathrm{kg} \end{gathered}$$

${\overrightarrow{F}}_{21}$is pushing all the three blocks, so we conclude that,

role="math" localid="1657009265260" $$\begin{gathered} \mathrm{Total}\mathrm{mass}\mathrm{accelerated}\mathrm{by}\overrightarrow{{\mathrm{F}}_{21}}={\mathrm{m}}_2+{\mathrm{m}}_3 \\ =2\mathrm{kg}+10\mathrm{kg} \\ =12\mathrm{kg} \end{gathered}$$

As we can see in the diagram,

${\overrightarrow{\mathrm{F}}}_{32}$is pushing all the three blocks, so we conclude that,

$$\begin{gathered} \mathrm{Total}\mathrm{mass}\mathrm{accelerated}\mathrm{by}\overrightarrow{{\mathrm{F}}_{32}}={\mathrm{m}}_3 \\ =10\mathrm{kg} \end{gathered}$$

The value for the acceleration for all three situations will be the same.

As the acceleration is the same for all three situations, the magnitude of the force will be based on the mass it is accelerating.

So with this newton’s 2^nd law for each force will be,

$$\begin{gathered} \overrightarrow{F}=\mathrm{ma} \\ \overrightarrow{\mathrm{F}}=17\mathrm{a} \\ \overrightarrow{{\mathrm{F}}_{21}}=\mathrm{ma} \\ \overrightarrow{{\mathrm{F}}_{21}}=12\mathrm{a} \\ \overrightarrow{{\mathrm{F}}_{32}}=\mathrm{ma} \\ \overrightarrow{{\mathrm{F}}_{32}}=10\mathrm{a} \end{gathered}$$

This gives us the relation,

$\overrightarrow{\mathrm{F}}>{\overrightarrow{\mathrm{F}}}_{21}>{\mathrm{F}}_{32}$