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Torque

Torque. Torque: a net “ off-center ” force that causes or tends to cause rotation (i.e., the capability of a force to produce rotation). torque. d. CG. F. Torque. Torque: a rotary force that has a tendency to cause rotation around an axis (twisting force) T or G

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Torque

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  1. Torque Torque: a net “off-center” force that causes or tends to cause rotation (i.e., the capability of a force to produce rotation) torque d CG F

  2. Torque Torque: a rotary force that has a tendency to cause rotation around an axis (twisting force) T or G Units: Englishmetric ft•lb N•m T = F • d d F

  3. Torque • Torque is the angular analog of Force • F = m•a • T = I • a • T = m • r2 • a rod “Choking up on a bat” r1 r1 Axis 1 Axis 2 CG

  4. Need for torque production • (1) movement at every joint • (2) tightening a screw • (3) releasing a discus, shot put

  5. Need for torque production gyroscope effect • (4) throwing a curveball • (5) throwing a spiral - football • angular momentum: L = I • w • (6) double axle - figure skating • (7) wrist snap - golf L

  6. In order to calculate Torque correctly, the force and the distance from the axis of rotation must be perpendicular: T = F • d If F and d are not Then: (1) resolve F into rotary (Fro) and stabilizing (Fs) components and use Fro: T = Fro • d (ex. muscles) (2) calculate a “moment arm” (ex. outside forces) T = F • dMA or F d dMA axis

  7. Moment arm (dMA):shortest (perpendicular) distance between a force’s line of action and an axis of rotation dL - physical distance dMA axis dMA - calculated distance dMA = dL • cos q for free weight where qis deviation from horizontal) q ex. joint dL Lever Rigid, bar-like body Rotates around an axis F or R

  8. A barbell with a weight of 120 N is held by Jody’s hand. The length of her forearm (elbow to hand) is 40 cm. If her forearm is position in an angular position 35° below horizontal, find the torque. Given: FBB = 120 N df = 40 cm q = 35° dMA axis q Find: TBB Formula: TBB = F • d dMA = df • cos q elbow df TBB FBB

  9. Solution: TBB = 120 N • dMA dMA = df • cos q dMA = 40 cm • cos 35° = 40 cm • 0.819 = 32.77 cm TBB = 120 N • dMA TBB = 120 N • 32.77 cm TBB = 3932.4 N•cm TBB = 39.32 N•m dMA axis q elbow df FBB

  10. a Fm

  11. The pectoralis major muscle is contracting with a force of 400N. If the angle of pull = 41°, find the torque produced by the muscle force. The distance (D) between the insertion of the pecs and the center of the shoulder joint is 10 cm • Given: Fm = 400N a = 41° • Find: Fro and Tm • Diagram: • Formuls: sin a = Fro/Fm • Tm = Fro • D shoulder D Fs Fs Fro Fm a a Fm Fro

  12. sin a = Fro/Fm Tm = Fro • DFro = Fm • sin 41° = 400 N • 0.656 = 262.42 N • Tm = Fro • D • Tm = 262.42 • 10 cm = 2624.24 N•cm = 26.24 N•m shoulder D Fs Fs Fro Fm a a Fm Fro

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