91影视

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 \(\theta = 15^\circ \).

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.

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.

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.

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.

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

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 \(244.24{\rm{ N}}\).

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 \(7.64^\circ \).