Chapter 12, Problem 12.42AP

When a person stands on tiptoe on one foot (a strenuous position), the position of the foot is as shown in Figure P12.32a. The total gravitational force ${\stackrel{\to }{F}}_g$ on the body is supported by the normal force $\stackrel{\to }{n}$ exerted by the floor on the toes of one foot. A mechanical model of the situation is shown in Figure P12.32b, where $\stackrel{\to }{T}$ is the force exerted on the foot by the Achilles tendon and $\stackrel{\to }{R}$ is the force exerted on the foot by the tibia. Find the values of T, R, and θ when F_g = 700 N.

Figure P12.32

To determine

The values of tension $\overrightarrow{T}$, force exerted by the tibia $\overrightarrow{R}$ and the angle $\theta$.

Answer to Problem 12.42AP

The tension in the cable is $1678\text{N}$ and force exerted by the tibia on the foot is $\overrightarrow{R}=2344\text{N}$.

Explanation of Solution

The magnitude of the gravitation al force acting on the toe is $700\text{N}$, the total gravitational force on the body is supported by the normal force exerted by the floor, force exerted on the foot by the Achilles tendon is $\overrightarrow{T}$ and the force exerted by the tibia on the foot is $\overrightarrow{R}$.

The following figure shows the force diagram of the foot.

Figure-(I)

Formula to calculate the net force in horizontal direction is,

    $\overrightarrow{R}\sin 15°-\overrightarrow{T}\sin \theta =0$

Here, $\overrightarrow{R}$ is the force exerted by the tibia on the foot and $\overrightarrow{T}$ is the force exerted by the Achilles tendon on the foot.

Rearrange the above equation to find $\overrightarrow{R}$.

    $\begin{array}{c}\overrightarrow{R}\sin 15°-\overrightarrow{T}\sin \theta =0\\ \overrightarrow{R}=\frac{\overrightarrow{T}\sin \theta }{\sin 15°}\end{array}$        (I)

Formula to calculate the net force in vertical direction is,

    $\left(700\text{N}\right)-\overrightarrow{R}\cos 15°+\overrightarrow{T}\cos \theta =0$

Substitute $\frac{\overrightarrow{T}\sin \theta }{\sin 15°}$ for $\overrightarrow{R}$ in the above equation to find $\overrightarrow{T}$.

    $\begin{array}{c}\left(700\text{N}\right)-\left(\frac{\overrightarrow{T}\sin \theta }{\sin 15°}\right)\cos 15°+\overrightarrow{T}\cos \theta =0\\ \left(700\text{N}\right)-\left(3.732\right)T\sin \theta +\overrightarrow{T}\cos \theta =0\end{array}$        (II)

Formula to calculate the torque about the point $O$ is,

    $-\left(700\text{N}\right)\cos \theta \left(18\text{cm}\right)+\overrightarrow{T}\left(25\text{cm}-18\text{cm}\right)=0$

Rearrange the above equation to find $\overrightarrow{T}$.

    $\begin{array}{c}\overrightarrow{T}\left(7\text{cm}\left(\frac{1\text{cm}}{100\text{cm}}\right)\right)=\left(700\text{N}\right)\cos \theta \left(18\text{cm}\left(\frac{1\text{cm}}{100\text{cm}}\right)\right)\\ \overrightarrow{T}=\left(1800\text{N}\right)\cos \theta \end{array}$        (III)

Substitute $\left(1800\text{N}\right)\cos \theta$ for $\overrightarrow{T}$ in the equation (II) to find $\theta$.

    $\begin{array}{c}\left(700\text{N}\right)-\left(3.732\right)T\sin \theta +\overrightarrow{T}\cos \theta =0\\ \left(700\text{N}\right)-\left(3.732\right)\left(\left(1800\text{N}\right)\cos \theta \right)\sin \theta +\left(\left(1800\text{N}\right)\cos \theta \right)\cos \theta =0\\ \left(700\text{N}\right)-\left(3.732\right)\left(1800\text{N}\right)\cos \theta \sin \theta +\left(1800\text{N}\right){\cos }^2\theta =0\end{array}$

Divide the above equation by $\left(1800\text{N}\right){\cos }^2\theta$.

    $\begin{array}{c}\frac{\left(700\text{N}\right)-\left(3.732\right)\left(1800\text{N}\right)\cos \theta \sin \theta +\left(1800\text{N}\right){\cos }^2\theta }{\left(1800\text{N}\right){\cos }^2\theta }=\frac{0}{\left(1800\text{N}\right){\cos }^2\theta }\\ \frac{\left(700\text{N}\right)}{\left(1800\text{N}\right){\cos }^2\theta }-\frac{\left(3.732\right)\left(1800\text{N}\right)\cos \theta \sin \theta }{\left(1800\text{N}\right){\cos }^2\theta }+\frac{\left(1800\text{N}\right){\cos }^2\theta }{\left(1800\text{N}\right){\cos }^2\theta }=0\\ \left(0.3888\right)\left(1+{\tan }^2\theta \right)-\left(3.732\right)\tan \theta +1=0\\ \left(0.3888\right){\tan }^2\theta -\left(3.732\right)\tan \theta +1.3888=0\end{array}$

Solve the above quadratic equation as follows.

    $\begin{array}{c}\tan \theta =\frac{-\left(3.732\right)\pm \sqrt{{\left(3.732\right)}^2-4\left(0.3888\right)\left(1.3888\right)}}{2\left(0.3888\right)}\\ \tan \theta =9.210\text{or}\text{0}\text{.3878}\\ \theta ={\tan }^{-1}\left(9.210\right)\text{or}{\tan }^{-1}\left(\text{0}\text{.3878}\right)\\ =83.8032°\text{or}21.196°\end{array}$

Since the angle $83.8032°$ for $\theta$ is too large, so this value of $\theta$ is neglected. Thus, the angle is $\theta$ is $21.196°$.

Substitute $21.196°$ for $\theta$ in the equation (III) to find $\overrightarrow{T}$.

    $\begin{array}{c}\overrightarrow{T}=\left(1800\text{N}\right)\cos \theta \\ =\left(1800\text{N}\right)\cos \left(21.196°\right)\\ =1678.224\text{N}\\ \approx 1678\text{N}\end{array}$

Substitute $21.196°$ for $\theta$ and $1678\text{N}$ for $\overrightarrow{T}$ in the equation (I) to find $\overrightarrow{R}$.

    $\begin{array}{c}\overrightarrow{R}=\frac{\overrightarrow{T}\sin \theta }{\sin 15°}\\ =\frac{\left(1678\text{N}\right)\sin \left(21.196°\right)}{\sin 15°}\\ =2344.096\text{N}\\ \approx 2344\text{N}\end{array}$

Conclusion:

Therefore, the tension in the cable is $1678\text{N}$ and force exerted by the tibia on the foot is $\overrightarrow{R}=2344\text{N}$.