(II) (a) Calculate the magnitude of the force, FM, required of the "deltoid" muscle to hold up t

(II) (a) Calculate the magnitude of the force, FM, required...

(II) $(a)$ Calculate the magnitude of the force, $F_{\mathrm{M}},$ required of the "deltoid" muscle to hold up the outstretched arm shown in Fig. 56 . The total mass of the arm is 3.3 $\mathrm{kg}$ . (b) Calculate the magnitude of the force $F_{\mathrm{J}}$ exerted by the shoulder joint on the upper arm and the angle (to the horizontal) at which it acts.

(II) $(a)$ Calculate the magnitude of the force, $F_{\mathrm{M}},$ required of
the "deltoid" muscle to hold up the outstretched arm shown in Fig. 56 . The total mass of the arm is 3.3 $\mathrm{kg}$ . (b) Calculate the
magnitude of the force $F_{\mathrm{J}}$ exerted by the shoulder joint on the
upper arm and the angle (to the horizontal) at which it acts.

Key Concepts

Static Equilibrium

This concept states that in a system at rest, the sum of all forces and the sum of all torques (moments) acting on the system must be zero. In the context of biomechanics, particularly in analyzing human limbs, equilibrium conditions are used to determine the forces exerted by muscles and joints to hold a limb stationary.

Torque (Moment of a Force)

Torque is the measure of a force’s tendency to cause rotation about a pivot or axis. It is calculated as the product of the force magnitude and the perpendicular distance from the line of action of the force to the axis of rotation. This concept is fundamental when determining the forces required to balance the weight of a body segment or an object.

Lever Arm

The lever arm, or moment arm, is the perpendicular distance from the pivot point to the line along which a force acts. It plays a critical role in torque calculations, determining how effectively a force can cause rotation. In biomechanical problems involving limbs, varying lever arm lengths significantly affect the forces needed by muscles and joints.

Muscle Force in Biomechanics

This refers to the force exerted by a muscle to perform an action or maintain a posture. In problems involving the human arm, muscle force calculations require balancing the moments produced by the muscle force against those produced by the weight of the limb, taking into account the angles and lever arms involved.

Joint Reaction Force

Joint reaction force is the resultant force that occurs at a joint due to the combination of muscle forces, gravitational forces, and any external loads. It is essential in understanding how joints support loads and maintain stability, as well as in analyzing the forces transmitted through the skeletal structure.

Transcript

00:01 We calculate the torque about the elbow joint, which is the start in the free body diagram.

00:08 So net torque is equal to i alpha.

00:12 And since the r is in equilibrium, alpha is zero and therefore the net torque is zero.

00:18 Now i'm assuming, i'm taking counterclockwise torque to be positive.

00:25 So let's write down the talks about this.

00:31 Elbow joint first.

00:33 So we have a talk due to force fm and this is equal to fm times the distance d, which is the length times the angle between these two, the sign of the angle between these two, so that is sine theta.

00:54 You can also choose to first take the force perpendicular to this distance, to this r and that will be equal to fm sine theta, basically the perpendicular component.

01:14 And this will come out to be fm sine theta and then you can simply multiply this force with the perpendicular distance which is t and you'll arrive at the same expression.

01:29 Now we have another clockwise talk.

01:33 Therefore i'm putting a negative sign here and that comes from mg and the distance is d.

01:41 And here we don't have to include any theta because ng and d are perpendicular to each other.

01:48 So this is equal to 0.

01:51 Now we can solve for fm which will be equal to mg times capital d over small d sine theta.

02:01 Now let me just write down the values of the masses and distances that we have.

02:07 So mass is given to be 3.

02:10 3 kgis, capital d is 0 .24 meters, 0 .24 meter, small d is 0 .12 meter and theta is 15 degree and g is just acceleration due to gravity which is a constant and is equal to 9 .8 meter per second square.

02:43 So this, we plug these values into this equation and we get fm to be 249 .9 newton.

02:58 Now for the second part to find the components of fj, we write newton's second law for both the x and y direction and solve them to get the components of fj in both x and y direction, and then we combine them to find the magnitude of fj.

03:17 So first we start with writing the equation and you turn second law...

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