Asked 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.
ceiling
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leg <--
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(mass)
So < represents angle formed shown at 2 tetra split by horizontal.
ceiling
|
|
leg <--
|
|
(mass)
Answers
Answered by
MathMate
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.
Please explain.
Answered by
Gharib
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.
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.