PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 8, Problem 47P

(a)

To determine

The force should exerted by the patellar tendon to hold the leg at an angle 30.0°.

(a)

Expert Solution
Check Mark

Answer to Problem 47P

The required force is 330N.

Explanation of Solution

Mass of ankle is 3.0kg, length of patellar tendon is 10.0cm, angle of pull by the tendon is 20.0°, mass of lower leg is 5.0kg, distance to center of gravity from the knee is 22cm, and the position of ankle weight from knee is 41cm.

Write the equation for rotational equilibrium at the moment of leg is holded by the patellar tendon.

FplpsinθpmwglwsinθLmLglLsinθL=0

Here, the force exerted by the patellar tendon is Fp, length of patellar tendon is lp, angle of pull by the tendon is θp, mass of ankle is mw, g is the gravitational acceleration, θL is the angle at which leg is holded, mL mass of lower leg, lw is the position of ankle weight from knee, lL is the distance to center of gravity from the knee

Rewrite the above relation in terms of Fp.

Fp=mwglwsinθL+mLglLsinθLlpsinθp

Conclusion:

Substitute 3.0kg for mw, 9.8m/s2 for g, 41cm for lw, 30.0° for θL, 5.0kg for mL, 22cm for lL, 10.0cm for lp, and 20.0° for θp in the above equation to find Fp.

Fp=((3.0kg)(9.8m/s2)(41cm(102m1cm))(sin30.0°))+((5.0kg)(9.8m/s2)(22cm(102m1cm))(sin30.0°))(10.0cm(102m1cm))(sin20.0°)=11.220.034=330N

Therefore, the required force is 330N.

(b)

To determine

The force should exerted by the patellar tendon to hold the leg at an angle 90.0°.

(b)

Expert Solution
Check Mark

Answer to Problem 47P

The required force is 670N.

Explanation of Solution

Mass of ankle is 3.0kg, length of patellar tendon is 10.0cm, angle of pull by the tendon is 20.0°, mass of lower leg is 5.0kg, distance to center of gravity from the knee is 22cm, and the position of ankle weight from knee is 41cm.

Write the equation for rotational equilibrium at the moment of leg is holded by the patellar tendon.

FplpsinθpmwglwsinθLmLglLsinθL=0

Here, the force exerted by the patellar tendon is Fp, length of patellar tendon is lp, angle of pull by the tendon is θp, mass of ankle is mw, g is the gravitational acceleration, θL is the angle at which leg is holded, mL mass of lower leg, lw is the position of ankle weight from knee, lL is the distance to center of gravity from the knee

Rewrite the above relation in terms of Fp.

Fp=mwglwsinθL+mLglLsinθLlpsinθp

Conclusion:

Substitute 3.0kg for mw, 9.8m/s2 for g, 41cm for lw, 90.0° for θL, 5.0kg for mL, 22cm for lL, 10.0cm for lp, and 20.0° for θp in the above equation to find Fp.

Fp=((3.0kg)(9.8m/s2)(41cm(102m1cm))(sin90.0°))+((5.0kg)(9.8m/s2)(22cm(102m1cm))(sin90.0°))(10.0cm(102m1cm))(sin20.0°)=22.780.034=670N

Therefore, the required force is 670N.

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Chapter 8 Solutions

PHYSICS

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