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BIO Injuries to the Spinal Column. In the treatment of
Chapter 8, Problem 4E(choose chapter or problem)
BIO Injuries to the Spinal Column.? 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?
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QUESTION:
BIO Injuries to the Spinal Column.? 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?
ANSWER:Solution 4E Introduction We will first calculate maximum frictional force of the person. The force due to weight must be less than the maximum frictional force of the person to prevent sliding. From this concept we will calculate the maximum possible weight. Knowing the total force acting on the cable, we can now calculate the tension on the cables. Step 1 The maximum possible frictional force is given by F = mg Where is the coefficient of friction, m is the mass of the person and g is the acceleration due to gravity. In the given problem we have m = 78.5 kg And = 0.75 We also know that g = 9.8 m/s2 Using the above values we have 2 F = (0.75)(78.5 kg)(9.8 m/s ) = 757 N So the maximum traction force that can be applied to the person without sliding is 757 N.