Question

The brachialis attaches to the forearm .035 m from the elbow at an angle of 32 deg. If the brachialis produces 750 N of force, what is the magnitude of the stabilizing component of the brachialis force

Answer 1

The magnitude of the stabilizing component of the brachialis force is approximately 636.04 N. This is calculated using the cosine of the given angle and the total force generated by the brachialis muscle.

Step-by-step explanation:

The student is asking about the stabilizing component of the brachialis force when the brachialis muscle produces 750 N of force and attaches to the forearm at a certain distance and angle. The stabilizing component is the part of the force that acts perpendicular to the lever arm (in this case, the forearm), which in biomechanics is often referred to as the joint stabilization force. To find the magnitude of this component, we need to take the cos component of the force because it gives us the force acting along the lever arm, which is the stabilizing force. The calculation will be as follows:

F_stabilizing = F_brachialis × cos(angle)
F_stabilizing = 750 N × cos(32°)
F_stabilizing = 750 N × 0.84805 (cosine of 32° approximately)
F_stabilizing = 636.04 N
The magnitude of the stabilizing component of the brachialis force is approximately 636.04 N.

Answer 2

`XY=636N`

Step-by-step explanation:

From the question we are told that:

Distance
`d=0.35m`

Angle
`\theta=32\textdegree`

Force
`F=750N`

Generally the equation for magnitude of the stabilizing component of the brachialis force is mathematically given by

`XY=Fcos\theta`

`XY=750cos 32\textdegree`

`XY=636N`