Multiple Pulley Problem [SOLUTION]

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SUMMARY

The discussion addresses the solution to a physics problem involving a system of pulleys designed to support the weight of a human head during recovery. The head, weighing approximately 44.59N, is supported by two pulleys (1 and 3) that operate without friction and are constrained to vertical movement. The angles of the ropes around the pulleys are both 26.0°. To find the mass M of the weight W needed to balance the head's weight, the tension T in the rope is calculated, leading to the formula M = T/g.

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  • Understanding of Newton's second law (Fnet=ma)
  • Basic knowledge of pulley systems and tension
  • Familiarity with free body diagrams (FBD)
  • Concept of static equilibrium in physics
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  • Study the principles of static equilibrium in mechanical systems
  • Learn how to construct and analyze free body diagrams (FBD)
  • Explore the effects of friction in pulley systems
  • Investigate the relationship between tension and weight in various pulley configurations
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Physics students, educators, and anyone interested in understanding mechanical systems involving pulleys and static equilibrium.

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[SOLVED] Multiple pulley problem

Homework Statement



**Image of apparatus
http://www.learning.physics.dal.ca/library/Graphics/Gtype09/neck.jpg

When a patient's injured neck is healing, it is often desirable to prevent the weight of the head from pushing down on the neck. This can be accomplished with the system of pulleys shown in the figure. The pulleys are small and light and have no appreciable friction. The rope about pulleys 1 and 3 make an angle of θ1 = θ2 = 26.0°; pulleys 1 and 3 are constrained to move only in the vertical direction. Typically, a person's head makes up 7.00 % of the body weight. If the head of a 65.0 kg person is to be supported completely by the apparatus shown, what should the mass M of the weight W be?

Homework Equations



Fnet=ma

The Attempt at a Solution



I started with the head component and determined that pulley 1 and 3 are each responsible for maintaining half of the weight of the head of 44.59N. From then on, I do not know how to draw the FBD of neither pulley 1 or 3 as a system. I would appreciate any help that can hint me on figuring out the answer. Thanks.
 
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The weight W puts a tension in the line, and in a static situation T = W =Mg. With frictionless pulleys, the tension in the line must be the same along the length of the line.

Then either pulley 1 or 3 support half the head weight.

For either pulley, the tension pulls upward - one side vertically, and the other side at angle 26.0° from vertical.

Determine the tension T, necessary to support half the head, then M = T/g.
 
Got it. Thanks!
 

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