(II) A man doing push-ups pauses in the position shown in Fig. 9–65. His mass \({\bf{m = 68}}\;{\bf{kg}}\). Determine the normal force exerted by the floor (a) on each hand; (b) on each foot.

When the value of the resultant force working on an object equals zero, the object is supposed to be in mechanical equilibrium. In mathematical form, it can be expressed as:

\(\sum {F_{{\rm{net}}}} = 0\)

Given data:

The mass of the person is\(m = 68\;{\rm{kg}}\).

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

Here, \({F_{\rm{h}}}\) is the force on each hand and \({F_{\rm{f}}}\) is the force on each foot, \({d_1} = 42\;{\rm{cm}}\) is the distance between the force \({F_{\rm{h}}}\) and force \(mg\), and \({d_2} = 95\;{\rm{cm}}\) is the distance between the force \(mg\) and force \({F_{\rm{f}}}\).

Now, take the torques about the point where the foot touches the ground, with the counterclockwise, as positive.

\(\begin{array}{c}\sum \tau = 0\\\frac{1}{2}mg{d_2} - {F_{\rm{h}}}\left( {{d_1} + {d_2}} \right) = 0\\\frac{1}{2}\left( {68\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {95\;{\rm{cm}}} \right) - {F_{\rm{h}}}\left[ {\left( {42\;{\rm{cm}}} \right) + \left( {95\;{\rm{cm}}} \right)} \right] = 0\\{F_{\rm{h}}} = 231\;{\rm{N}}\end{array}\)

Thus, the normal force exerted by the floor on each hand is \(231\;{\rm{N}}\).

Apply the force equilibrium condition along the vertical direction.

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\2{F_{\rm{h}}} + 2{F_{\rm{f}}} - mg = 0\\2\left( {231\;{\rm{N}}} \right) + 2{F_{\rm{f}}} - \left( {68\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right) = 0\\{F_{\rm{f}}} = 102.2\;{\rm{N}}\end{array}\)

Thus, the normal force exerted by the floor on each foot is \(102.2\;{\rm{N}}\).