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bulbasaur88
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Problem XIV, page 18 ------>http://www.physics.princeton.edu/~mcdonald/examples/ph101_2006/learning_guide_ph101_2006.pdf
A patient's leg is held in traction as shown in the diagram. The top and bottom pulleys are fixed, while the central pulley is not, but has come to rest at a position determined by the various forces acting on it. Note that one single rope runs between the mass and the pulleys and is fixed to the middle pulley, while a separate strap connects the foot to the center pulley. What is the magnitude and direction of the force on the patient's foot produced by the traction apparatus shown int he diagram?
Also express this force as a vector (using the coordinate system shown). Assume that all three pulleys are frictionless.
What I Understand About the Problem
The middle pulley has NO acceleration in any direction because the foot has come to a rest position.
Forces Acting on the Center Pulley
1. Tension in the direction 40 degrees south of east, T = (2.5)(9.8) = 24.5 N, due to the weight of the box
2. The two tensions, T', in the northeasterly direction
3. The tension caused by the foot in the southwesterly direction, Tf
I realize that in order for the middle pulley to stay in a rest position, all the of the y-components of force must cancel out. Same for the x-components of force.
I also think, that since it is the same single rope that holds the weight of the block, the tension in the whole rope remains the same magnitude; however, since the tension in the two rope segments at the very top share the weight of the block, T' = T/2. Is this a correct assumption?
I honestly do not know how to continue if all of my assumptions are correct up to here, butI would like to be assured if I am going in the right direction before continuing.
Thank you for any help!
Answer: 56.7 N, 19.5 degrees above horizontal
A patient's leg is held in traction as shown in the diagram. The top and bottom pulleys are fixed, while the central pulley is not, but has come to rest at a position determined by the various forces acting on it. Note that one single rope runs between the mass and the pulleys and is fixed to the middle pulley, while a separate strap connects the foot to the center pulley. What is the magnitude and direction of the force on the patient's foot produced by the traction apparatus shown int he diagram?
Also express this force as a vector (using the coordinate system shown). Assume that all three pulleys are frictionless.
What I Understand About the Problem
The middle pulley has NO acceleration in any direction because the foot has come to a rest position.
Forces Acting on the Center Pulley
1. Tension in the direction 40 degrees south of east, T = (2.5)(9.8) = 24.5 N, due to the weight of the box
2. The two tensions, T', in the northeasterly direction
3. The tension caused by the foot in the southwesterly direction, Tf
I realize that in order for the middle pulley to stay in a rest position, all the of the y-components of force must cancel out. Same for the x-components of force.
I also think, that since it is the same single rope that holds the weight of the block, the tension in the whole rope remains the same magnitude; however, since the tension in the two rope segments at the very top share the weight of the block, T' = T/2. Is this a correct assumption?
I honestly do not know how to continue if all of my assumptions are correct up to here, butI would like to be assured if I am going in the right direction before continuing.
Thank you for any help!
Answer: 56.7 N, 19.5 degrees above horizontal