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Torque Applications. 1. Sprinting 2. Lower Back 3. Pushups 4. Weight lifting 5. Stability 6. Force Couple. Torque Applications. 1. Sprinting. Hip, knee flexion Swing phase Reduce moment arm. Torque Applications. 1. Sprinting Extreme flexion of the hip and knee

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Torque Applications

- 1. Sprinting
- 2. Lower Back
- 3. Pushups
- 4. Weight lifting
- 5. Stability
- 6. Force Couple

Torque Applications

- 1. Sprinting
- Extreme flexion of the hip and knee
- Reduce moment arm/moment of inertia
- T = F • dMA
- T = I • a
- Where I = m • r2

- Increase pendulum frequency
- f µ √(1/L)
- Where: L = distance from center of mass to axis of rotation

- f µ √(1/L)

- Reduce moment arm/moment of inertia

- Extreme flexion of the hip and knee

Torque Applications

2. Lower back

Torque Applications Find: Farms Diagram

- 3. Pushups - 2nd class lever
- David is performing a pushup. He has a mass of 70 kg and is 175 cm tall. The top (superior) surface of his head is located 19 cm above (superior) to the shoulders. The center of gravity of his body is located at 53% of his height (starting at the inferior end). How much force do the arms have to produce at the shoulders to start a pushup for the ‘down’ position? Ignore the mass of the arms (8.5 kg).
- Given: mDavid = 70 kg marms = 8.5 kg
- Ht = 175 cm dCG = 53% Ht
- D(head + neck) = 19 cm

Farms

dbody-arms

19 cm

Tarms

0.53 • Ht

aTm

Tbody

F(David-arms)

Torque Applications - Pushups

- Formula: T = F • d
- assume that in ‘down” position body is horizontal
- F(body-arms) = (70-8.5 kg) (9.81 m/sec2) = 603.3 N
- dCG = 0.53 • 175 cm = 92.75 cm D(HT-H+N) = 175 - 19 cm = 156 cm
- Solution: Tarms > Tbody
- Farms • D(HT-H+N) > F(body-arms) • dCG
Farms > F(body-arms) • dCG / D(HT-H+N)

- Farms > (603.3 N • 92.75 cm) / (156 cm)
- Farms > 358.7 N or 52.2% of body weight

Farms

dbody-SH

19 cm

Tarms

0.53 • Ht

aTm

Tbody

F(David-arms)

Torque Applications

- Weight lifting
No change in moment arm

Torque Applications

- 5. Stability
• Wider base means greater moment of inertia and resistance to torque (ex. falling over)

F

Torque Applications

- 5. Stability
• More narrow base means smaller moment of inertia and less resistance to torque (ex. falling over)

F

Torque Applications

- 6. Force Couple
- A special case of parallel forces where two forces of equal magnitude are acting in opposite directions but at a distance from each other and on either side of the center of mass or axis of rotation. The result is no linear displacement or deformation, but increased potential for torque.
- Tfc = F • Dfc
- where D is the distance between the forces
- Ex. Spin move in dancing
- both hand pivoting on a golf club handle

Dfc

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