posted by Sam .
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.
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.
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
From Eq 1, m=x/g
that's your answer.