In Fig. 5-43, a chain consisting of five links, each of mass0.100 kg, is lifted vertically with constant acceleration of magnitude$\mathbf{a}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{2}$. Find the magnitudes of (a) the force on link 1 from link 2, (b) the force on link 2 from link 3, (c) the force on link 3 from link 4, and (d) the force on link 4 from link 5. Then find the magnitudes of (e) the force$\overrightarrow{\mathbf{F}}$on the top link from the person lifting the chain and (f) the net force accelerating each link

$$\begin{gathered} \mathrm{m}=0.100\mathrm{kg} \\ \mathrm{a}=2.50\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

In solving this problem, we can use both Newton’s second and third laws. Each pair of links constitutes a third-law force pair, with $\overrightarrow{{\mathbf{F}}_{\mathbf{1}\mathbf{}\mathbf{o}\mathbf{n}\mathbf{}\mathbf{j}}}\mathbf{=}\mathbf{-}\overrightarrow{{\mathbf{F}}_{\mathbf{j}\mathbf{}\mathbf{o}\mathbf{n}\mathbf{}\mathbf{j}}}$.

The links are numbered from bottom to top. The forces on the first link are the force of gravity $\overrightarrow{mg}$, downward, and the force $\overrightarrow{F_{2on1}}$of link 2, upward, as shown in

the free-body diagram below (not drawn to scale). Take the positive direction to be

upward. Then Newton’s second law for the first link is $F_{2on1}-m_{1}g=m_{1}a$.

The equations for the other links can be written in a similar manner (see below).

Given that $a=2.50m/s$, from $F_{2on1}-m_{1}g=m_{1}a$, the force exerted by link 2 on link 1 is

$$\begin{gathered} F_{2on1}=m_1\left(a+g\right) \\ =\left(0.100kg\right)\left(2.5m/s^2+9.80m/s^2\right) \\ =1.23N \end{gathered}$$

From the free-body diagram above, we see that the forces on the second link are the force of gravity $\overrightarrow{m_{2}g}$, downward, the force$\overrightarrow{F_{2on1}}$ on of link 1, downward, and the force $\overrightarrow{F_{3on2}}$of link 3, upward. According to Newton’s third law$\overrightarrow{F_{1on2}}$ has the same magnitude as $\overrightarrow{F_{2on1}}$. Newton’s second law for the second link is

$F_{3on2}-F_{1on2}-m_{2}g=m_{2}a$

So

$$\begin{gathered} F_{3on2}=m_2\left(a+g\right)+F_{1on2} \\ =\left(0.100kg\right)\left(2.50m/s^2+9.80m/s^2\right)+1.23N \\ =2.46N \end{gathered}$$

Newton’s second law equation for link 3 is $F_{4on3}-F_{2on3}-m_{3}g=m_{3}a$, so

$$\begin{gathered} F_{4on3}=m_3\left(a+g\right)+F_{2on3} \\ =\left(0.100N\right)\left(2.50m/s^2+9.80m/s^2\right)+2.46N \\ =3.69N, \end{gathered}$$

Where Newton’s third law implies$F_{2on3}=F_{3on2}$ (since these are magnitudes of the force vectors).

Newton’s second law for link 4 is

$F_{5on4}-F_{3on4}-m_{4}g=m_{4}a$

So

$$\begin{gathered} F_{5on4}=m_4\left(a+g\right)+F_{3on4} \\ =\left(0.100kg\right)\left(2.50m/s^2+9.8m/s^2\right)+3.69N \\ =4.92N \end{gathered}$$

Where Newton’s third law implies$F_{3on4}=F_{4on3}$

Newton’s second law for the top link is $F-F_{4on5}-m_{5}g=m_{5}a$so

$$\begin{gathered} F=m_5\left(a+g\right)+f_{4on5} \\ =\left(0.100kg\right)\left(2.50m/s^2+9.80m/s^2\right)+4.92N \\ =6.15N \end{gathered}$$

Where$F_{4on5}=F_{5on4}$ by Newton’s third law.

Each link has the same mass $\left(m_1=m_2=m_3=m_4=m_5=m\right)$and the same acceleration, so the same net force acts on each of them:

$$\begin{gathered} F_{net}=ma \\ =\left(0.100kg\right)\left(2.50m/s^2\right) \\ =0.250N \end{gathered}$$