Redo Example 9–9, assuming now that the person is less bent over so that the 30° in Fig. 9–14b is instead 45°. What will be the magnitude of\({F_v}\) on the vertebra?

FIGURE 9–14 (b) Forces on the back exerted by the back muscles and by the vertebrae when a person bends over. (c) Finding the lever arm for FB

The given data can be listed below as:

• The force due to the bicep muscle is\({F_{\rm{M}}}\).
• The force acting on the fifth lumbar vertebra is\({F_{\rm{V}}}\).
• The weight of the head for the lever arm is\({w_{\rm{H}}} = 0.07w\).
• The weight of two arms for the lever arm is\({w_{\rm{A}}} = 0.12w\).
• The weight of the trunk for the lever arm is\({w_{\rm{T}}} = 0.46w\).
• The total weight of the person is w.
• The acceleration due to gravity is\(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

The weights of the lever arm are acting in the downward direction.At equilibrium, the net torque acting about point S (at the spinal base) becomes zero. The resolve force is acting on the back due to the back muscles.

This back muscle force will rotate the trunk of the person in the anticlockwise direction. In contrast, the weights of the lever arm rotate it in the clockwise direction. Resolve the vertebra forces in the x and y components. Then, evaluate the resultant force of the vertebra.

Forces on the back exerted by the back muscles and the vertebrae when a person bends over

Finding the lever arm for FB

Here, the inclination of the lever arm with the horizontal axis is\(\left( {45^\circ - 12^\circ } \right)\), which is equal to \(33^\circ \).

At equilibrium, the net torque acting on the back about point S becomes zero.

From the above figure (a), the torques equation can be expressed as:

\(\begin{array}{c}\sum {{T_{net}}} = 0\\\left\{ \begin{array}{l}48{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)\sin 12^\circ \times {F_{\rm{M}}} - 72{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)\sin 45^\circ \times {w_{\rm{H}}}\\ - 48{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)\sin 45^\circ \times {w_{\rm{A}}} - 36\,{\rm{cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)\sin 45^\circ \times {w_{\rm{T}}}\end{array} \right\} = 0\\0.48{\rm{ m}} \times \sin 12^\circ \times {F_{\rm{M}}} = \left\{ \begin{array}{l}0.72{\rm{ m}} \times \sin 45^\circ \times {w_{\rm{H}}}\\ + 0.48{\rm{ m}} \times \sin 45^\circ \times {w_{\rm{A}}}\\ + 0.36\,{\rm{m}} \times \sin 45^\circ \times {w_{\rm{T}}}\end{array} \right\}\\{F_{\rm{M}}} = \frac{{\left\{ \begin{array}{l}0.72{\rm{ m}} \times \sin 45^\circ \times {w_{\rm{H}}}\\ + 0.48{\rm{ m}} \times \sin 45^\circ \times {w_{\rm{A}}}\\ + 0.36\,{\rm{m}} \times \sin 45^\circ \times {w_{\rm{T}}}\end{array} \right\}}}{{\left( {0.48{\rm{ m}} \times \sin 12^\circ } \right)}}\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}{F_{\rm{M}}} = \frac{{\left\{ \begin{array}{l}0.72{\rm{ m}} \times \sin 45^\circ \times 0.07w\\ + 0.48{\rm{ m}} \times \sin 45^\circ \times 0.12w\\ + 0.36\,{\rm{m}} \times \sin 45^\circ \times 0.46w\end{array} \right\}}}{{\left( {0.48{\rm{ m}} \times \sin 12^\circ } \right)}}\\ = \frac{{0.193w{\rm{ m}}}}{{0.099{\rm{ m}}}}\\ = 1.95w\end{array}\)

At equilibrium, the forces along the vertical direction can be equated to zero. From figure (b), the y-component of force on the vertebra can be expressed as:

\(\begin{array}{c}\sum {{F_{\rm{y}}}} = 0\\{F_{{\rm{vy}}}} - {F_{\rm{M}}}\sin 33^\circ - {w_{\rm{H}}} - {w_{\rm{A}}} - {w_{\rm{T}}} = 0\\{F_{{\rm{vy}}}} = {F_{\rm{M}}}\sin 33^\circ + {w_{\rm{H}}} + {w_{\rm{A}}} + {w_{\rm{T}}}\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}{F_{{\rm{vy}}}} = 1.95w \times \sin 33^\circ + 0.07w + 0.12w + 0.46w\\ = 1.95w \times \sin 33^\circ + 0.65w\\ = 1.71w\end{array}\)

At equilibrium, the forces along the horizontal direction can be equated to zero.

From figure (b), the x-component of force on the vertebra can be expressed as:

\(\begin{array}{c}\sum {{F_x}} = 0\\{F_{{\rm{vx}}}} - {F_{\rm{M}}}\cos 33^\circ = 0\\{F_{{\rm{vx}}}} = {F_{\rm{M}}}\cos 33^\circ \end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}{F_{{\rm{vx}}}} = 1.95w \times \cos 33^\circ \\ = 1.95w \times \cos 33^\circ \\ = 1.63w\end{array}\)

The resultant force on vertebra can be expressed as:

\({F_{\rm{v}}} = \sqrt {F_{{\rm{vx}}}^{\rm{2}} + F_{{\rm{vy}}}^{\rm{2}}} \)

Substitute the values in the above equation.

\(\begin{array}{c}{F_{\rm{v}}} = \sqrt {{{\left( {1.63w} \right)}^2} + {{\left( {1.71w} \right)}^2}} \\ = \sqrt {5.581} w\\ = 2.36w\end{array}\)

Thus, the magnitude of the force on the vertebra is \(2.36w\).

The angle of inclination of the force on the vertebra can be expressed as:

\(\begin{array}{c}\tan \theta = \frac{{{F_{{\rm{vy}}}}}}{{{F_{{\rm{vx}}}}}}\\\theta = {\tan ^{ - 1}}\left( {\frac{{{F_{{\rm{vy}}}}}}{{{F_{{\rm{vx}}}}}}} \right)\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}\theta = {\tan ^{ - 1}}\left( {\frac{{1.71w}}{{1.63w}}} \right)\\ = 46.37^\circ \end{array}\)

Thus, \(46.37^\circ \) is the angle that the force on the vertebra makes with the horizontal axis.