Chapter 3: (Part A) Biomechanics and Applied Kinesiology Flashcards

1
Q

Rotary motion

A
  • Flexion and extension (sagittal plane)
  • Abduction, adduction (frontal plane)
  • E.g. forearm flexion/extension around the elbow
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2
Q

General plane motion

A

A combination of translatory motion and rotary motion. The object rotates around on an axis, while the axis is translated in space by movement at an adjacent segment

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3
Q

Class I lever

A

Pivot point is in the middle and the force and load are at opposite ends (seesaw)

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4
Q

Class II lever

A

Pivot point is at one end, the force is applied to the opposite end and the load is the middle (wheelbarrow)

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5
Q

Class III lever

A

Pivot point is at one end, the load is located at the opposite end and the force is applied in the middle (person shoveling snow)

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6
Q

Formula for calculating torque

A

R x Ra = F x Fa

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7
Q

Concentric muscle action

A

The muscle is shortening under its own power

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8
Q

Eccentric muscle action

A

The muscle resists as it is forced to lengthen

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9
Q

Isometric muscle action

A

The muscle attempts to contract against a fixed limit or resistance

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10
Q

Isotonic muscle action

A

The muscle shortens against a constant tension (e.g. free-weight exercises)

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11
Q

Isokinetic muscle action

A

Muscle shortens against a resistance that moves at a constant velocity (e.g. KinCom machine)

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12
Q

A client holding a weight in their hand is asked to hold it steady at 90 degrees of elbow flexion. The total weight of the ball and the arm segment is considered to be 15 pounds (R). The elbow flexors (biceps muscle) are considered to act at a perpendicular distance from the elbow joint of 2 inches (Fa). The lever arm length (from the elbow to the center of the ball) is 18 inches (Ra). What is the torque on the elbow flexors in a balanced system (Simply, what is F)?

A

R = 15 pounds Ra = 18 inches F = ?? pounds Fa = 2 inches

R x Ra = F x Fa
15 pounds x 18 inches = ?? x 2 inches
270 = ?? x 2
270/2 = 135

135 pounds = ??
So ?? = 135 pounds

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