traction force

2012-10-11 11:19 pm
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In treating spinal injuries, it is often necessary to provide some tension along
the spinal column to stretch the backbone. One device for doing this is the
Stryker frame, illustrated in part (a) of the figure. A weight W is attached to the
patient (sometimes around a neck collar, as shown in part (b) of the figure), and
friction between the person's body and the bed prevents sliding.(Figure 1)

1. If the coefficient of static friction between a 83.5 kg patient's body and the
bed is
0.715, what is the maximum traction force along the spinal column that W can
provide without causing the patient to slide?

T = ? N

2. Under the conditions of maximum traction, what is the tension in each cable
attached to the neck collar?

T = ? N

回答 (1)

2012-10-12 12:32 am
✔ 最佳答案
1. Max frictional force = 0.715 x 83.5g N
where g is the acceleration due to gravity (taken to be 10 m/s^2)
hence, max friction = 597 N

Thus, the max traction force = 597 N

2. Let the tension in each cable be T
2T.sin(65) = 597
i.e. T = 597/[2.sin(65)] N = 329 N


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