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 value of the mass held by the hand of the person is\(m = 8.5{\rm{ kg}}\).
• The distance between the shoulder joint and the center of the mass 8.5 kg is \(l = 52{\rm{ cm}}\).

The person holds a mass of 8.5 kg in his palm. The force due to the deltoid muscle acts at an angle of inclination of 15 degrees. The force on the shoulder joint and the total arm mass acts vertically downward.

Apply the equilibrium condition.

The sum of the torques that acts in the clockwise and anti-clockwise directions can be equated to zero. Therefore, the force required by the deltoid muscle can be evaluated with the help of the equilibrium equations.

The arm of the person can be represented as:

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, the value of d is 12 cm, and the value of lis 52 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:

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

Substitute the values in the above equation.

\(\begin{array}{c}{F_{\rm{M}}} = \frac{{3.3{\rm{ kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times 24{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right) + 8.5{\rm{ kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times 52{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)}}{{12{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)\sin 15^\circ }}\\ = \frac{{51.13}}{{0.031}}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}\left( {\frac{{1{\rm{ N}}}}{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)\\ = 1649.36{\rm{ N}}\end{array}\)

Thus, the force exerted by the deltoid muscle is 1649.36 N.