Chapter 9: Q78GP (page 230)

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.)

Short Answer

Expert verified

The mass of A is 0.249 kg, the mass of C is 0.194 kg, and the mass of D is \(5.54 \times {10^{ - 2}}\;{\rm{kg}}\).

Step by step solution

Identification of the given data

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\({x_{\rm{D}}} = 17.50\;{\rm{cm}}\).

Conditions for static equilibrium

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

Translational equilibrium: The net force acting on the system must be zero.

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

Rotational equilibrium: The net torque acting on the system about the pivot point must be zero.

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

Evaluation of the net torque on the bottom bar

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)

Evaluation of the net force on the bottom bar

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)

Evaluation of the net torque on the middle bar

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\(5.54 \times {10^{ - 2}}\;{\rm{kg}}\).

Evaluation of the mass C

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.

Evaluation of the net force on the middle bar

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)

Evaluation of the net torque on the top bar

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.