Question

In the treatment of spine injuries, it is often necessary to provide tension along the spinal column to stretch the backbone. One device for doing this is the Stryker frame (Fig. E5.4a). A weight W is attached to the patient (sometimes around a neck collar, Fig. E5.4b), and friction between the person’s body and the bed prevents sliding.

(a) If the coefficient of static friction between a 78.5-kg patient’s body and the bed is 0.75, what is the maximum traction force along the spinal column that W can provide without causing the patient to slide?

(b) Under the conditions of maximum traction, what is the tension in each cable attached to the neck collar?

Short answer

(a) The maximum traction force along the spinal column without the slide of the patient is 577 N .

(b) The tension in each cable under maximum traction conditions is 315 N .

Step-by-step solution

Traction Force

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 ${\mu }_3=0.75$
• The angle of each cable from horizontal is $\theta ={65}^{\circ }$.

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.

Determine the maximum traction force along the spinal column without sliding of patient(a)

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

$F={\mu }_{s}mg$

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 ${\mu }_3$is the static friction between the patient’s body and bed.

Substitute all the values in the above equation.

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

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

Determine the tension in each cable under maximum traction condition(b)

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

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

Substitute all the values in the above equation.

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

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