91Ó°ÊÓ

Given that when you are lying horizontally, these muscles make an angle of approximately with the femur, and if you raise your legs, the muscles remain approximately horizontal, so the angle increases. Assume for simplicity that these muscles attach to the femur in just one place, 10 cm from the hip joint. For a certain 80-kg person having a leg 90 cm long, the mass of the leg is 15 kg and its center of mass is 44 cm from the his hip joint as measured along the leg. If the person raises his leg above the horizontal, the angle between the abdominal muscles and his femur would also be about $60°$.

Mass of person, m = 80 kg

Length of leg L = 40 m

Torque $\mathrm{Ï„}=FI$

Where Fis force exerted and l is distance.

Applying condition of equilibrium,

Net torque is zero.

$\underset{}{∑}\mathrm{τ}=0$

This gives,

$$\begin{gathered} T=\frac{4.4\left(1.5\mathbf{kg}\right)\left(9.8\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\right)}{\tan \mathrm{θ}} \\ Put\mathrm{θ}=60° \\ T=\frac{4.4\left(1.5\mathbf{kg}\right)\left(9.8\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\right)}{60°}=370\mathbf{N} \end{gathered}$$

For $\mathrm{θ}=60°$

Tension in the muscle is 370 N

When theperson just starts to raise it off the ground .

Then tension is$T=\frac{4.4\left(1.5\mathbf{kg}\right)\left(9.8\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\right)}{\tan 5°}=7400\mathbf{N}$

Hence, the tension is much greater when the person just starts to raise his leg off the ground.

Tension in the leg is inversely proportional to 0, so if$\mathrm{θ}$ approaches zero, then tension will approach to infinity.

If the leg is horizontal then$\mathrm{θ}=0$ , the moment arm for tension is zero, so there is no torque to rotate the leg.

Hence, the person could not raise his leg.