Question

Figure 12-85ashows details of a finger in the crimp holdof the climber in Fig. 12-50. A tendon that runs from muscles inthe forearm is attached to the far bone in the finger. Along the way, the tendon runs through several guiding sheaths called pulleys. The A2 pulley is attached to the first finger bone; the A4 pulley is attached to the second finger bone. To pull the finger toward the palm, the forearm muscles pull the tendon through the pulleys, much like strings on a marionette can be pulled to move parts of the marionette. Figure 12-85bis a simplified diagram of the second finger bone, which has length d. The tendon’s pull ${\overrightarrow{\mathbf{F}}}_{\mathbf{t}}$on the bone acts at the point where the tendon enters the A4 pulley, at distance d/3 along the bone. If the force components on each of the four crimped fingers in Fig. 12-50 are ${\mathbf{\text{F}}}_{\mathbf{\text{h}}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{13}\mathbf{.4}\mathbf{\text{\hspace{0.17em}N}}$and ${\mathbf{\text{F}}}_{\mathbf{\text{v}}}\mathbf{=}\mathbf{}\mathbf{162.4}\mathbf{\text{\hspace{0.17em}N}}$, what is the magnitude of${\overrightarrow{\mathbf{F}}}_{\mathbf{t}}$ ? The result is probably tolerable, but if the climber hangs by only one or two fingers, the A2 and A4 pulleys can be ruptured, a common ailment among rock climbers.

Short answer

The magnitude of the force $\overrightarrow{{\text{F}}_{\text{t}}}$ is${\text{F}}_{\text{t}}=175.6\text{\hspace{0.17em}N}$ .

Step-by-step solution

Understanding the given information

$\begin{array}{c}{\text{F}}_{\text{v}}=162.4\text{\hspace{0.17em}N}\\ {\text{F}}_{\text{h}}=13.4\text{\hspace{0.17em}N}\end{array}$

Concept and formula used in the given question

Using the equation for the equilibrium of torque, you can find the magnitude of the tension force. The equation is given below.

${\sum }_{}^{}{\mathrm{\tau }}_{\mathrm{net}}=0$

Calculation of the magnitude of  Ft→ 

Let us consider the given figure of the free body diagram,

Let us assume the pivot point at point A. The equation for the equilibrium of torque can be written as,

${\mathrm{F}}_{\mathrm{t}}\sin 45°\times \frac{\mathrm{d}}{3}-{\mathrm{F}}_{\mathrm{v}}\sin 10°\times \mathrm{d}-{\mathrm{F}}_{\mathrm{h}}\sin 80°\times \mathrm{d}=0$

Now, solving for ${\text{F}}_{\text{t}}$we get,

${\mathrm{F}}_{\mathrm{t}}=\frac{3\times ({\mathrm{F}}_{\mathrm{v}}\sin 10°+{\mathrm{F}}_{\mathrm{h}}\sin 80°)}{\sin 45°}$

Now, by using the values,

${\text{F}}_{\text{v}}=162.4\text{\hspace{0.17em}N}$ and ${\text{F}}_{\mathrm{h}}=13.4\text{\hspace{0.17em}N}$

You get,

${\text{F}}_{\text{t}}=175.6\text{\hspace{0.17em}N}$