The mobile in Fig. 9–90 is in equilibrium. Object B has mass of 0.748 kg. Determine the masses of objects A, C, and D. (Neglect the weights of the crossbars.)

The mass of object B is\({m_{\rm{B}}} = 0.748\;{\rm{kg}}\).

The distance of object A from the suspension point is\({x_{\rm{A}}} = 30\;{\rm{cm}}\).

The distance of object B from the suspension point is\({x_{\rm{B}}} = 5\;{\rm{cm}}\).

The distance of object C from the suspension point is\({x_{\rm{C}}} = 5\;{\rm{cm}}\).

The distance of object D from the suspension point is\({x_{\rm{D}}} = 17.50\;{\rm{cm}}\).

The equilibrium is achieved when the following two conditions are satisfied:

\(\sum {\vec F} = 0\)

\(\sum {\vec \tau } = 0\)

The free body diagram of the bottom crossbar CD is drawn below.

The net torque on the bottom bar about the point of suspension is zero because the crossbar CD is in equilibrium. Consider the anti-clockwise torques to be positive.

The net torque acting on the bottom crossbar is as follows:

\(\begin{array}{c}{\sum \tau _{{\rm{bottom}}}} = \left( {{m_{\rm{D}}}g} \right){x_{\rm{D}}} - \left( {{m_{\rm{C}}}g} \right){x_{\rm{C}}}\\0 = \left( {{m_{\rm{D}}}g} \right){x_{\rm{D}}} - \left( {{m_{\rm{C}}}g} \right){x_{\rm{C}}}\\\left( {{m_{\rm{D}}}g} \right){x_{\rm{D}}} = \left( {{m_{\rm{C}}}g} \right){x_{\rm{C}}}\\{m_{\rm{C}}} = \left( {\frac{{{x_{\rm{D}}}}}{{{x_{\rm{C}}}}}} \right){m_{\rm{D}}}\end{array}\)

Substitute the known numerical values in the above equation.

\(\begin{array}{c}{m_{\rm{C}}} = \left( {\frac{{17.50\;{\rm{cm}}}}{{5\;{\rm{cm}}}}} \right){m_{\rm{D}}}\\ = 3.50{m_{\rm{D}}}\end{array}\) … (i)

The net vertical force on the bottom bar must be zero because the crossbar is in static equilibrium. Therefore, the net force on CD crossbar will be,

\(\begin{array}{c}\sum {{F_{\rm{y}}}} = {F_{{\rm{CD}}}} - {m_{\rm{C}}}g - {m_{\rm{D}}}g\\0 = {F_{{\rm{CD}}}} - {m_{\rm{C}}}g - {m_{\rm{D}}}g\\{F_{{\rm{CD}}}} = \left( {{m_{\rm{C}}} + {m_{\rm{D}}}} \right)g\end{array}\)

Substitute equation (i) in the above expression.

\(\begin{array}{c}{F_{{\rm{CD}}}} = \left( {\left( {3.50{m_{\rm{D}}}} \right) + {m_{\rm{D}}}} \right)g\\ = \left( {4.50{m_{\rm{D}}}} \right)g\end{array}\) … (ii)

The free body diagram of the middle crossbar is drawn below.

The net torque on the middle bar about the point of suspension is zero because the crossbar is in static equilibrium. Therefore, the net torque on the middle crossbar is as follows:

\(\begin{array}{c}{\sum \tau _{{\rm{middle}}}} = {F_{{\rm{CD}}}}{x_{{\rm{CD}}}} - \left( {{m_{\rm{B}}}g} \right){x_{\rm{B}}}\\0 = {F_{{\rm{CD}}}}{x_{{\rm{CD}}}} - \left( {{m_{\rm{B}}}g} \right){x_{\rm{B}}}\\{F_{{\rm{CD}}}}{x_{{\rm{CD}}}} = \left( {{m_{\rm{B}}}g} \right){x_{\rm{B}}}\\{F_{{\rm{CD}}}} = \left( {\frac{{{x_{\rm{B}}}}}{{{x_{{\rm{CD}}}}}}} \right){m_{\rm{B}}}g\end{array}\)

Substitute equation (ii) in the above equation.

\(\begin{array}{c}\left( {4.50{m_{\rm{D}}}} \right)g = \left( {\frac{{{x_{\rm{B}}}}}{{{x_{{\rm{CD}}}}}}} \right){m_{\rm{B}}}g\\{m_{\rm{D}}} = \left( {\frac{{{x_{\rm{B}}}}}{{{x_{{\rm{CD}}}}}}} \right)\frac{{{m_{\rm{B}}}}}{{4.50}}\end{array}\)

Substitute the know numerical values in the above expression for\({m_{\rm{D}}}\).

\(\begin{array}{c}{m_{\rm{D}}} = \left( {\frac{{5\;{\rm{cm}}}}{{15\;{\rm{cm}}}}} \right)\frac{{0.748\;{\rm{kg}}}}{{4.50}}\\ = 0.05541\;{\rm{kg}}\\ = 5.54 \times {10^{ - 2}}\;{\rm{kg}}\end{array}\)

Thus, the mass of object D is\(5.54 \times {10^{ - 2}}\;{\rm{kg}}\).

Substitute the mass of D in equation (i).

\(\begin{array}{c}{m_{\rm{C}}} = 3.50{m_{\rm{D}}}\\ = 3.50\left( {5.54 \times {{10}^{ - 2}}\;{\rm{kg}}} \right)\\ = 0.194\;{\rm{kg}}\end{array}\)

Thus, the mass of object C is 0.194 kg.

The net vertical force on the middle bar must be zero because the crossbar is in static equilibrium. Therefore, the net force on the middle crossbar is as follows:

\(\begin{array}{c}\sum {{F_{\rm{y}}}} = {F_{{\rm{BCD}}}} - {F_{{\rm{CD}}}} - {m_{\rm{B}}}g\\0 = {F_{{\rm{BCD}}}} - {F_{{\rm{CD}}}} - {m_{\rm{B}}}g\\{F_{{\rm{BCD}}}} = {F_{{\rm{CD}}}} + {m_{\rm{B}}}g\\ = \left( {4.50{m_{\rm{D}}} + {m_{\rm{B}}}} \right)g\end{array}\) … (iii)

The free body diagram of the top crossbar is drawn below.

The net torque on the top bar about the point of suspension is zero because the crossbar is in static equilibrium. Therefore, the net torque on the middle crossbar is as follows:

\(\begin{array}{c}{\sum \tau _{{\rm{top}}}} = \left( {{m_{\rm{A}}}g} \right){x_{\rm{A}}} - {F_{{\rm{BCD}}}}{x_{{\rm{BCD}}}}\\0 = \left( {{m_{\rm{A}}}g} \right){x_{\rm{A}}} - {F_{{\rm{BCD}}}}{x_{{\rm{BCD}}}}\\\left( {{m_{\rm{A}}}g} \right){x_{\rm{A}}} = {F_{{\rm{BCD}}}}{x_{{\rm{BCD}}}}\\{m_{\rm{A}}} = \frac{{{F_{{\rm{BCD}}}}{x_{{\rm{BCD}}}}}}{{g{x_{\rm{A}}}}}\end{array}\)

Substitute equation (3) in the above expression.

\(\begin{array}{c}{m_{\rm{A}}} = \frac{{\left( {\left( {4.50{m_{\rm{D}}} + {m_{\rm{B}}}} \right)g} \right){x_{{\rm{BCD}}}}}}{{g{x_{\rm{A}}}}}\\ = \frac{{\left( {4.50{m_{\rm{D}}} + {m_{\rm{B}}}} \right){x_{{\rm{BCD}}}}}}{{{x_{\rm{A}}}}}\end{array}\)

Substitute the known numerical values in the above equation for the mass of object A.

\(\begin{array}{c}{m_{\rm{A}}} = \frac{{\left( {4.50\left( {5.54 \times {{10}^{ - 2}}\;{\rm{kg}}} \right) + 0.748\;{\rm{kg}}} \right)\left( {7.50\;{\rm{cm}}} \right)}}{{30\;{\rm{cm}}}}\\ = 0.249\;{\rm{kg}}\end{array}\)

Thus, the mass of object A is 0.249 kg.