91Ó°ÊÓ

Given Data:

• The weight of the patient is $m=78.5\mathrm{kg}$
• The coefficient of static friction between the patient’s body and the bed is ${\mathrm{μ}}_3=0.75$
• The angle of each cable from horizontal is $\mathrm{θ}={65}^{∘}$.

The maximum traction force along the spinal column is the same as the static frictional force. The tension in each cable under maximum traction is only due to the vertical components of tension in each cable.

The maximum traction force along the spinal column is calculated as by equating it with static friction force as:

$F={\mathrm{μ}}_{s}mg$

Here $g$is the gravitational acceleration, and its value is $9.8\mathrm{m}/{\mathrm{s}}^2$, $m$is the mass of the patient and ${\mathrm{μ}}_3$is the static friction between the patient’s body and bed.

Substitute all the values in the above equation.

$$\begin{gathered} F=\left(0.75\right)\mathrm{w} \\ F=\left(0.75\right)×\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ F=577\mathrm{N} \end{gathered}$$

Therefore, the maximum traction force along the spinal column without the slide of the patient is 577 N

The equation for vertical equilibrium to find tension in each cable is given as follows:

$$\begin{gathered} F=2T\sin \mathrm{θ} \\ T=\frac{F}{2\sin \mathrm{θ}} \end{gathered}$$

Substitute all the values in the above equation.

$$\begin{gathered} T=\frac{\left(0.75\right)\mathrm{w}}{2\left(\sin {65}^{°}\right)} \\ T=0.41×\mathrm{w} \\ T=0.41×\left(78.5\mathrm{kg}\right)×\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ T=315\mathrm{N} \end{gathered}$$

Therefore, the tension in each cable under maximum traction conditions is 315 N .