Forces on an inclined plane with a pulley

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SUMMARY

The discussion focuses on calculating the forces acting on a 77.0 kg patient suspended in a raised hospital bed using static and kinetic friction coefficients of 0.800 and 0.500, respectively. The correct approach involves balancing the gravitational force component along the incline with the tension in the wire and the static friction force. The equation should be set up as M(Patient) g sin θ = T + Us * n, where n is the normal force calculated as mg cos θ. This method accurately determines the minimum mass required to keep the patient stationary.

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  • Understanding of Newton's laws of motion
  • Knowledge of static and kinetic friction coefficients
  • Ability to calculate components of gravitational force
  • Familiarity with trigonometric functions in physics
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  • Learn how to derive forces on inclined planes in physics
  • Explore tension in ropes and pulleys in mechanical systems
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This discussion is beneficial for physics students, educators, and anyone involved in mechanics, particularly those studying forces on inclined planes and frictional forces in practical applications.

kmerr98277
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Homework Statement


A 77.0 kg patient is suspended in a raised hospital bed as shown in the figure. The wire is attached to a brace on the patient's neck and pulls parallel to the bed, and the coefficients of kinetic and static friction between the patient and the bed are 0.500 and 0.800, respectively.
P4-80.jpg

My attempt:
mg sinθ = mg sinθ
(77.0)(9.8)(sin50) = m(9.8)(sin90)
578.057 = m9.8
m = 58.985
I'm not sure if I set that up right. Also, I don't know how to factor in the friction to the equation.
 
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kmerr98277 said:

Homework Statement


A 77.0 kg patient is suspended in a raised hospital bed as shown in the figure. The wire is attached to a brace on the patient's neck and pulls parallel to the bed, and the coefficients of kinetic and static friction between the patient and the bed are 0.500 and 0.800, respectively.
P4-80.jpg

My attempt:
mg sinθ = mg sinθ
(77.0)(9.8)(sin50) = m(9.8)(sin90)
578.057 = m9.8
m = 58.985
I'm not sure if I set that up right. Also, I don't know how to factor in the friction to the equation.
I think what the question wants is the lowest mass that keeps the patient stationary right?
Well you are missing something in your solution.. Static friction.
I think you should add the value of static friction to the right side to the incline
So the equation should be M(Patient) g sin θ = T(which equal to mg) + Us * n (equals mg cosθ)
Imagine that patient is not connected to mass. He would fall right? but what is the counter of the force here static friction... Now try to imagine that you are adding small amount of mass each time to balance out the forces. By the equation you will get the right amount of mass and the lowest

 
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