Chapter 2, Problem 103P

To determine

The magnitude and direction of force exerted on the tibia by the femur.

Answer to Problem 103P

The magnitude of force exerted on the tibia by the femur is $\underset{\_}{281\text{N}}$, and the direction is $\underset{\_}{39.7°\text{below the horizontal to the right}}$.

Explanation of Solution

The free body diagram of the tibia is shown in the Figure 1 below.

Let positive x direction be along the direction of the length and positive y direction is the direction be along the perpendicular to the length of the leg.

From the free body diagram, write the expression for net force in the x direction.

    $F_{\text{net,x}}=m_{\text{L}}g\cos 60.0°+m_{\text{A}}g\cos 60.0°-F_{\text{P}}\cos 20.0°$        (I)

Here, $m_{\text{L}}g$ is the weight of the lower leg, $m_{\text{A}}g$ is the weight of the ankle, $F_{\text{P}}$ is the force exerted by the femur on the tibia.

From the free body diagram, write the expression for net force in the y direction.

    $F_{\text{net,y}}=F_{\text{P}}\sin 20.0°-m_{\text{L}}g\sin 60.0°-m_{\text{A}}g\sin 60.0°$        (II)

Then the magnitude of the sum of the all forces on the tibia can be calculated as follows.

    $F_{\text{net}}=\sqrt{F_{\text{net,x}}^2+F_{\text{net,y}}^2}$        (III)

The angle made by the $F_{\text{net}}$ with respect to the leg can be calculated as follows.

    $\theta ={\tan }^{-1}\left(\frac{F_{\text{net,y}}}{F_{\text{net,x}}}\right)$        (IV)

Conclusion:

Substitute, $5\text{kg}$ for $m_{\text{L}}$, $3\text{kg}$ for $m_{\text{A}}$, $9.80\text{m}/{\text{s}}^2$ for $g$, and $337\text{N}$ for $F_{\text{P}}$ in the equation (I).

    $\begin{array}{c}{F}_{\text{net,x}}=\left(5\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\cos 60.0°+\left(3\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\cos 60.0°-\left(337\text{N}\right)\cos 20.0°\\ =-277.467\text{N}\end{array}$

Substitute, $5\text{kg}$ for $m_{\text{L}}$, $3\text{kg}$ for $m_{\text{A}}$, $9.80\text{m}/{\text{s}}^2$ for $g$, and $337\text{N}$ for $F_{\text{P}}$ in the equation (II).

    $\begin{array}{c}{F}_{\text{net,y}}=\left(337\text{N}\right)\sin 20.0°-\left(5\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\sin 60.0°-\left(3\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\sin 60.0°\\ =47.36\text{N}\end{array}$

Substitute, $47.36\text{N}$ for $F_{\text{net,y}}$, and $-\text{277}\text{.467}\text{N}$ for $F_{\text{net,x}}$ in the equation (III).

    $\begin{array}{c}{F}_{\text{net}}=\sqrt{{\left(-\text{277}\text{.467}\text{N}\right)}^2+{\left(47.36\text{N}\right)}^2}\\ =281\text{N}\end{array}$

Substitute, $47.36\text{N}$ for $F_{\text{net,y}}$, and $-\text{277}\text{.467}\text{N}$ for $F_{\text{net,x}}$ in the equation (IV).

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{47.36\text{N}}{-\text{277}\text{.467}\text{N}}\right)\\ =-9.7°\end{array}$

The magnitude of the force exerted by the femur on the tibia have the magnitude as $F_{\text{net}}$, since the leg is stationary and it is directed opposite to it. So the direction of the force exerted on the tibia by femur can be calculated as follows.

    $30.0°+9.7°=39.7°\text{below the horizontal to the right}$

Therefore, the magnitude of force exerted on the tibia by the femur is $\underset{\_}{281\text{N}}$, and the direction is $\underset{\_}{39.7°\text{below the horizontal to the right}}$.