Figure 12-18 shows a mobile ofQ toy penguins hanging from a ceiling. Each crossbar is horizontal, has negligible mass, and extends three times as far to the right of the wire supporting it as to the left. Penguin 1 has mass ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{48}\mathbf{}\mathbf{kg}$. What are the masses of

(a) penguin 2,

(b) penguin 3, and

(c) penguin 4?

1. Mass of the first penguin is,$m_1=48\mathrm{kg}$
2. Each crossbar has negligible mass and extends three times as far to the right as to the left from the hanging point.

Using the conditions for equilibrium, we can write the force equations and torque equations at each point A, B, and C. Solving these equations, we will get the masses of the penguins.

Formulae:

The value of the net force at equilibrium, $F_{net}=0$ (i)

The value of the torque at equilibrium , ${\tau }_{net}=0$ (ii)

The force acting on a body due to the weight, F = mg (iii)

Free body diagram:

Consider the arrangement at point C. The hanging is in static equilibrium. Hence all the forces and torques about the point of rotation C are balanced. Here the forces are tension in the string $\left(T_c\right)$directed upwards and weights of the two penguins,$m_{3}g$and localid="1661148108879" $m_{4}g$. The angle between the moment arms r ( and 3r) and the weights $m_{3}g$(and $m_{4}g$) is $90°$.

Here, the force and torque equations using equations (i) and (ii) can be given at point C as:

$T_c-m_{3}g-m_{4}g=0$

And

$\left(r\right)m_3-\left(3r\right)m_4=0$

Solving these, we get,

$m_3=3m_4$

Now consider the arrangement at point B. The hanging is in static equilibrium. Hence all the forces and torques about point of rotation B are balanced. Here the forces are tension in the string $\left(T_B\right)$directing upwards and weight of the penguin 2, $m_{2}g$and$T_c$.

The angle between the moment arms r(and 3r) and the weights$m_{2}g$(and$T_c$) is$90°$.

Now, the force and torque equations using equations (i) and (ii) can be given at point B as:

$$\begin{gathered} T_B-m_{2}g-T_c=0 \\ {T}_B-m_{2}g-\left(m_3+m_4\right)g=0 \\ \left(r\right)m_2-\left(3r\right)T_c=0 \end{gathered}$$

Solving these, we get,

$$\begin{gathered} m_2=3T_c \\ {m}_2=3\left(m_3+m_4\right) \\ =3\left(4m_4\right) \\ =12m_4 \end{gathered}$$

Now consider the arrangement at point A. The hanging is in static equilibrium. Hence all the forces and torques about the point of rotation A are balanced. Here the forces are tension in the string $\left(T_A\right)$directed upwards and weight of the penguin 1 , $m_{1}g$and $T_B$. The angle between the moment arms (and) and the weights (and) is$90°$.

Hence the force and torque equations using equations (i) and (ii) can be given at point A as:

$$\begin{gathered} T_A-m_{1}g-T_B=0 \\ {T}_A-m_{1}g-\left(m_2+m_3+m_4\right)g=0 \\ \left(r\right)m_1-\left(3r\right)T_B=0 \end{gathered}$$

Solving these, we get,

$$\begin{gathered} 3T_B=m_1 \\ =48 \end{gathered}$$

(Since we are given that$m_1=48\mathrm{kg}$)

Thus, by solving the equations, we get,

$$\begin{gathered} 3\left(m_2+m_3+m_4\right)=m_1 \\ =48 \end{gathered}$$

$m_2=12m_4$

$m_3=3m_4$

We get the mass of all the penguins as:

$\begin{array}{l}{m}_2=12\mathrm{kg}\\ m_3=3\mathrm{kg}\\ m_4=1\mathrm{kg}\end{array}$

Thus, the value of the mass of the penguin 2 is 12 kg.

From part (a) calculations, we can see that the value of the penguin 3 is 3 kg.

From part (a) calculations, we can see that the value of the penguin 3 is 1 kg .