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Magazine Chapter 9: Q2P (page 230)
(I) Calculate the mass m needed in order to suspend the leg shown in Fig. 9鈥�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.
Short Answer
The mass m needed to suspend the leg is \(m = 6.7\;{\rm{kg}}\).
Step by step solution
Given Data
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 \({x_2} = 78\;{\rm{cm}}\).
Evaluate the tension on the sling
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.
Calculate the mass needed to overhang the leg
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.
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