Question

(II) (a) Calculate the magnitude of the force, required of the “deltoid” muscle to hold up the outstretched arm shown in Fig. 9–72. The total mass of the arm is 3.3 kg. (b) Calculate the magnitude of the force exerted by the shoulder joint on the upper arm and the angle (to the horizontal) at which it acts.

Short answer

1. The force required to hold up the outstretched arm is 250.61 N.
2. The magnitude of the force due to the shoulder joint is \(244.24{\rm{ N}}\). The angle that the force due to the shoulder joint makes with the horizontal axis is \(7.64^\circ \).

Step-by-step solution

Identification of given data

The given data can be listed below as:

• The force due to the deltoid muscle is\({F_{\rm{M}}}\).
• The force exerted by the shoulder joint is\({F_{\rm{J}}}\).
• The total mass of the arm is\(m = 3.3{\rm{ kg}}\).
• The acceleration due to gravity is\(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).
• The angle of inclination of the deltoid muscle force is \(\theta = 15^\circ \).

Understanding the mechanism of the arm

The force required to outstretch the arm acts at an angle of 15 degrees. The force acts on the shoulder joint at some inclination. The total mass of the arm acts in the downward direction.

At equilibrium, the net torques about the joint axis can be equated to zero. The torque due to the mass of the arm acts in the clockwise direction. The force required to outstretch the arm acts in the anticlockwise direction.With the help of equilibrium equations, the required forces can be evaluated.

Representation of the forces acting on the arm

The forces acting on the arm can be represented as:

Here, A can be considered as the pivot point as the force due to the shoulder joint is unknown. The value of x is 24 cm and the value of d is 12 cm.

(a) Determination of the force required to hold up the outstretched arm

At equilibrium, the net torques acting on the arm about point A becomes zero.

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

\(\begin{array}{c}\sum {{T_{net}}} = 0\\ - {F_{\rm{M}}}\sin \theta \times d + mg \times x = 0\\{F_{\rm{M}}}\sin \theta \times d = mg \times x\\{F_{\rm{M}}} = \frac{{mgx}}{{d\sin \theta }}\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}{F_{\rm{M}}} = \frac{{3.3{\rm{ kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times 24{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)}}{{12{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)\sin 15^\circ }}\\ = \frac{{7.769}}{{0.031}}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}\left( {\frac{{1{\rm{ N}}}}{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)\\ = 250.61{\rm{ N}}\end{array}\)

Thus, the force required to hold up the outstretched arm is 250.61 N.

(b) Determination of the force required to hold up the outstretched arm

At equilibrium, the forces along the vertical direction can be equated to zero. From the figure, the y-component of force due to the shoulder joint can be expressed as:

\(\begin{array}{c}\sum {{F_{\rm{y}}}} = 0\\ - {F_{{\rm{Jy}}}} + {F_{\rm{M}}}\sin \theta - mg = 0\\{F_{{\rm{Jy}}}} = {F_{\rm{M}}}\sin \theta - mg\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}{F_{{\rm{Jy}}}} = 250.61{\rm{ N}}\left( {\frac{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}{{1{\rm{ N}}}}} \right) \times \sin 15^\circ - 3.3{\rm{ kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2}\;\left( {\frac{{1{\rm{ N}}}}{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)\\ = 32.49{\rm{ N}}\end{array}\)

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

From figure (b), the x-component of force due to the shoulder joint can be expressed as:

\(\begin{array}{c}\sum {{F_x}} = 0\\{F_{{\rm{Jx}}}} - {F_{\rm{M}}}\cos \theta = 0\\{F_{{\rm{Jx}}}} = {F_{\rm{M}}}\cos \theta \end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}{F_{{\rm{Jx}}}} = 250.61{\rm{ N}} \times \cos 15^\circ \\ = 242.07\,{\rm{N}}\end{array}\)

Determination of the resultant magnitude of the force due to the shoulder joint

The resultant force due to the shoulder joint can be expressed as:

\({F_{\rm{J}}} = \sqrt {F_{{\rm{Jx}}}^{\rm{2}} + F_{{\rm{Jy}}}^{\rm{2}}} \)

Substitute the values in the above equation.

\(\begin{array}{c}{F_{\rm{J}}} = \sqrt {{{\left( {242.07\,{\rm{N}}} \right)}^2} + {{\left( {32.49{\rm{ N}}} \right)}^2}} \\ = \sqrt {59653.485} {\rm{ N}}\\ = 244.24{\rm{ N}}\end{array}\)

Thus, the magnitude of the force due to the shoulder joint is \(244.24{\rm{ N}}\).

Determination of the angle of inclination of the force due to the shoulder joint

The angle of inclination due to the shoulder joint can be expressed as:

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

Substitute the values in the above equation.

\(\begin{array}{c}\theta = {\tan ^{ - 1}}\left( {\frac{{32.49{\rm{ N}}}}{{242.07\,{\rm{N}}}}} \right)\\ = 7.64^\circ \end{array}\)

Thus, the angle that the force due to the shoulder joint makes with the horizontal axis is \(7.64^\circ \).