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Torque Applications - PowerPoint PPT Presentation

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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Presentation Transcript

• 1. Sprinting

• 2. Lower Back

• 3. Pushups

• 4. Weight lifting

• 5. Stability

• 6. Force Couple

• 1. Sprinting

Hip, knee flexion

Swing phase

Reduce moment arm

• 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

2. Lower back

• 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

• Find: Farms Diagram

• Farms

dbody-arms

19 cm

Tarms

0.53 • Ht

aTm

Tbody

F(David-arms)

• 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)

• Weight lifting

No change in moment arm

4. Weight lifting

Moment arm changes as cam rotates

Nautilus

4. Weight lifting

Free weight

• 5. Stability

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

F

• 5. Stability

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

F

• 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

• 6. Force Couple - shoulder abduction

supraspinatus

Dfc