91Ó°ÊÓ

$$\begin{gathered} m_1=3.5\mathrm{kg} \\ {m}_2=4.5\mathrm{kg} \\ {m}_3=4.8\mathrm{kg} \\ {m}_5=5.5\mathrm{kg} \end{gathered}$$

The tension in the top cord is 199 N.

In this problem we have mobiles consisting of masses connected by cords. We apply Newton’s second law to calculate the tensions in the cords.

The free-body diagrams for$m_1$and$m_2$for part (a) are shown to the right.

The bottom cord is only supporting $m_2=4.5\mathrm{kg}$against gravity, so its tension is $T_2=m_{2}g$.

On the other hand, the top cord is supporting a total mass of

$$\begin{gathered} m_1+m_2=\left(3.5\mathrm{kg}+4.5\mathrm{kg}\right) \\ =8.0\mathrm{kg} \end{gathered}$$

against gravity. Applying Newton’s second law gives

$T_1-T_2-m_{1}g=0$

So the tension is

$T_1=m_{1}g+T_2=(m_1+m_2)g$

From the equations above, we find the tension in the bottom cord to be

$\begin{array}{l}{T}_2=m_{2}g\\ =\left(4.5\mathrm{kg}\right)(9.8\mathrm{m}/{\mathrm{s}}^2)\\ =44\mathrm{N}\end{array}$

Hence the solution is$T_2=44\mathrm{N}$

Similarly, the tension in the top cord is

$$\begin{gathered} T_1=\left(m_1+m_2\right)g \\ =\left(8.0\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =78\mathrm{N} \end{gathered}$$

Hence the solution is$T_1=78\mathrm{N}$

The free-body diagrams for$m_3,m_4\mathrm{and}{m}_5$for part (b) are shown below

From the diagram, we see that the lowest cord supports a mass of$m_5=5.5\mathrm{kg}$against gravity and consequently has a tension of

$$\begin{gathered} T_5=m_{5}g \\ =\left(5.5\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =54\mathrm{N} \end{gathered}$$

Hence the solution is $T_5=54\mathrm{N}$

The top cord, as we are told, has a tension$T_3=199\mathrm{N}$which supports a total of $\left(199\mathrm{N}\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)=20.3\mathrm{kg}$, 10.3 kg of which is already accounted for in the figure. Thus, the unknown mass in the middle must be

$$\begin{gathered} {\mathrm{m}}_4=20.3\mathrm{kg}-10.3\mathrm{kg} \\ =10.0\mathrm{kg} \end{gathered}$$,

and the tension in the cord above it must be enough to support

$$\begin{gathered} m_4+m_5=(10.0\mathrm{kg}+5.50\mathrm{kg}) \\ =15.5\mathrm{kg} \end{gathered}$$

So

$T_4=(15.5\mathrm{kg})\left(9.8{\mathrm{mm}}^2\right)$

Hence the solution is$T_4=152\mathrm{N}$