Forces on an inclined plane with a pulley

[kmerr98277]

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.

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.

 

[Biker]

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.

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