Asked by Jana
I don't need somone to fully answer this questions, but I need help as to where to get started.
Occasionally a severe leg break requires traction to help the bones heal properly. While in traction, the leg is held stationary in the air and is under tension by a series of wires, pulleys and masses. Consider the leg and mass in the traction setup below:
The picture shows a 7.0kg wieght hanging straight down. A pulley redirects the rope horizontally left onto the foot of the leg. Another pully attached to the foot redirects the rope at a 65 degree angle left and up. A third pulley redirects in to the left, and finally a fourth ridirects it down and hold the leg at the knee.
The leg is supported at two locations: A vertical support near the knee and a support attached at the foot which both supports the leg and applies a traction (horizontal pulling force). Determine the total support and traction forces on the leg.
Where do I start?
Occasionally a severe leg break requires traction to help the bones heal properly. While in traction, the leg is held stationary in the air and is under tension by a series of wires, pulleys and masses. Consider the leg and mass in the traction setup below:
The picture shows a 7.0kg wieght hanging straight down. A pulley redirects the rope horizontally left onto the foot of the leg. Another pully attached to the foot redirects the rope at a 65 degree angle left and up. A third pulley redirects in to the left, and finally a fourth ridirects it down and hold the leg at the knee.
The leg is supported at two locations: A vertical support near the knee and a support attached at the foot which both supports the leg and applies a traction (horizontal pulling force). Determine the total support and traction forces on the leg.
Where do I start?
Answers
Answered by
Damon
it is not accelerating
therefore:
Sum of forces in the x direction = 0
Sum of forces in the y direction = 0
Sum of moments around any point = 0
by the way 7 kg down is
7*9.81 Newtons down and tension in the line
therefore:
Sum of forces in the x direction = 0
Sum of forces in the y direction = 0
Sum of moments around any point = 0
by the way 7 kg down is
7*9.81 Newtons down and tension in the line
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