Chapter 8, Problem 47P

(a)

To determine

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

Answer to Problem 47P

The required force is $330\text{N}$.

Explanation of Solution

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

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

$F_p{l}_p\sin {\theta }_p-m_{w}g{l}_w\sin {\theta }_L-m_{L}g{l}_L\sin {\theta }_L=0$

Here, the force exerted by the patellar tendon is $F_p$, length of patellar tendon is $l_p$, angle of pull by the tendon is ${\theta }_p$, mass of ankle is $m_w$, $g$ is the gravitational acceleration, ${\theta }_L$ is the angle at which leg is holded, $m_L$ mass of lower leg, $l_w$ is the position of ankle weight from knee, $l_L$ is the distance to center of gravity from the knee

Rewrite the above relation in terms of $F_p$.

$F_p=\frac{m_{w}g{l}_w\sin {\theta }_L+m_{L}g{l}_L\sin {\theta }_L}{l_p\sin {\theta }_p}$

Conclusion:

Substitute $3.0\text{kg}$ for $m_w$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $41\text{cm}$ for $l_w$, $30.0°$ for ${\theta }_L$, $5.0\text{kg}$ for $m_L$, $22\text{cm}$ for $l_L$, $10.0\text{cm}$ for $l_p$, and $20.0°$ for ${\theta }_p$ in the above equation to find $F_p$.

$\begin{array}{c}{F}_p=\frac{\left(\left(3.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(41\text{cm}\left(\frac{{\text{10}}^{-\text{2}}\text{m}}{\text{1}\text{cm}}\right)\right)\left(\sin 30.0°\right)\right)+\left(\left(5.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(22\text{cm}\left(\frac{{\text{10}}^{-\text{2}}\text{m}}{\text{1}\text{cm}}\right)\right)\left(\sin 30.0°\right)\right)}{\left(10.0\text{cm}\left(\frac{{\text{10}}^{-\text{2}}\text{m}}{\text{1}\text{cm}}\right)\right)\left(\sin 20.0°\right)}\\ =\frac{11.22}{0.034}\\ =330\text{N}\end{array}$

Therefore, the required force is $330\text{N}$.

(b)

To determine

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

Answer to Problem 47P

The required force is $670\text{N}$.

Explanation of Solution

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

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

$F_p{l}_p\sin {\theta }_p-m_{w}g{l}_w\sin {\theta }_L-m_{L}g{l}_L\sin {\theta }_L=0$

Here, the force exerted by the patellar tendon is $F_p$, length of patellar tendon is $l_p$, angle of pull by the tendon is ${\theta }_p$, mass of ankle is $m_w$, $g$ is the gravitational acceleration, ${\theta }_L$ is the angle at which leg is holded, $m_L$ mass of lower leg, $l_w$ is the position of ankle weight from knee, $l_L$ is the distance to center of gravity from the knee

Rewrite the above relation in terms of $F_p$.

$F_p=\frac{m_{w}g{l}_w\sin {\theta }_L+m_{L}g{l}_L\sin {\theta }_L}{l_p\sin {\theta }_p}$

Conclusion:

Substitute $3.0\text{kg}$ for $m_w$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $41\text{cm}$ for $l_w$, $90.0°$ for ${\theta }_L$, $5.0\text{kg}$ for $m_L$, $22\text{cm}$ for $l_L$, $10.0\text{cm}$ for $l_p$, and $20.0°$ for ${\theta }_p$ in the above equation to find $F_p$.

$\begin{array}{c}{F}_p=\frac{\left(\left(3.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(41\text{cm}\left(\frac{{\text{10}}^{-\text{2}}\text{m}}{\text{1}\text{cm}}\right)\right)\left(\sin 90.0°\right)\right)+\left(\left(5.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(22\text{cm}\left(\frac{{\text{10}}^{-\text{2}}\text{m}}{\text{1}\text{cm}}\right)\right)\left(\sin 90.0°\right)\right)}{\left(10.0\text{cm}\left(\frac{{\text{10}}^{-\text{2}}\text{m}}{\text{1}\text{cm}}\right)\right)\left(\sin 20.0°\right)}\\ =\frac{22.78}{0.034}\\ =670\text{N}\end{array}$

Therefore, the required force is $670\text{N}$.