(I) Calculate the mass m needed in order to suspend the leg shown in Fig. 9–47. Assume the leg (with cast) has a mass of 15.0 kg, and its CG is 35.0 cm from the hip joint; the cord holding the sling is 78.0 cm from the hip joint.

The mass of the leg is\(M = 15\;{\rm{kg}}\).

The center of gravity is\({\rm{CG}} = 35\;{\rm{cm}}\).

The length of the sling is \({x_2} = 78\;{\rm{cm}}\).

Since the mass needed to suspend the leg is stationary, the tension in the sling would be mg. Also, consider the counterclockwise direction as positive.

The following is the free-body diagram of the tree.

The relation for the net torque is shown below.

\(\begin{array}{c}\sum \tau = 0\\\left( {mg \times {x_2}} \right) - \left( {Mg \times {x_1}} \right) = 0\end{array}\)

Here,\(g\)is the gravitational acceleration, and\({x_1}\)is the distance of the center of gravity with the value\({\rm{CG}} = 35\;{\rm{cm}}\).

Put the values in the above relation.

\(\begin{array}{c}\left( {m\left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {35\;{\rm{cm}}} \right)} \right) - \left( {\left( {15\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {78\;{\rm{cm}}} \right)} \right) = 0\\m = 6.7\;{\rm{kg}}\end{array}\)

Thus, \(m = 6.7\;{\rm{kg}}\) is the mass required to suspend the leg.