Question

Leg traction is applied to a patient's leg as shown in the figure below. If the physician has requested a 57 N force to be applied to the leg, and the angle is θ = 60o , what mass m must be used for the object hanging from the massless cable?

So < represents angle formed shown at 2 tetra split by horizontal.

ceiling
|
|
leg <--
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(mass)

Answer 1

I wish I could understand the figure, but as it is, it looks to me that all the weight added is supported by the ceiling.
Please explain.

Answer 2

You have to make two FBD; 1st for the force being applied to the leg, and 2nd for the mass hanging from cable.

Positive being up and right.
Y direction for FBD #2: T= mg Eq 1
x direction for FBD #1: -F+2Tcos(60)=0 Eq 2
T= F/2cos(60)
T=x

From Eq 1, m=x/g

that's your answer.