Question

(II) A 172-cm-tall person lies on a light (massless) board which is supported by two scales, one under the top of her head and one beneath the bottom of her feet (Fig. 9–64). The two scales read, respectively, 35.1 and 31.6 kg. What distance is the center of gravity of this person from the bottom of her feet?

Short answer

The center of gravity of this person from the bottom of her feet is \(90.5\;{\rm{cm}}\).

Step-by-step solution

Understating rotational equilibrium

For the system to be in rotational equilibrium, the addition of all torques about any axis acting on the system should be equal to zero.

Given information

Given data:

The height of the person is\(l = 172\;{\rm{cm}}\).

The reading of the first scale is\({m_1} = 35.1\;{\rm{kg}}\).

The reading of the second scale is \({m_2} = 31.6\;{\rm{kg}}\).

Calculate the center of gravity of this person from the bottom of her feet

The free body diagram of the person can be drawn as:

For the system to be in equilibrium, the net torque acting on the system should be equal to zero. Therefore, you can write:

\(\begin{array}{c}\sum {\tau _{{\rm{CG}}}} = 0\\{m_2}gx = {m_1}g\left( {l - x} \right)\\\left( {31.6\;{\rm{kg}}} \right)g\left( x \right) = \left( {35.1\;{\rm{kg}}} \right)g\left[ {\left( {172\;{\rm{cm}}} \right) - \left( x \right)} \right]\\\left( {31.6\;{\rm{kg}}} \right)\left( x \right) = \left( {6037.2\;{\rm{kg}} \cdot {\rm{cm}}} \right) - \left( {35.1\;{\rm{kg}}} \right)\left( x \right)\\x = 90.5\;{\rm{cm}}\end{array}\)

Thus, the center of gravity from the feet is \(90.5\;{\rm{cm}}\).