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)
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.
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.
To find the mass required for leg traction, we'll use the given force and angle.
First, let's resolve the force into its components. The vertical component will be F_vertical = F * sin(theta), and the horizontal component will be F_horizontal = F * cos(theta).
In this case, the vertical component will provide the necessary force for leg traction, so we set it equal to the weight of the mass.
F_vertical = m * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Therefore, m * g = F * sin(theta).
Rearranging the equation, we get:
m = (F * sin(theta)) / g
Now, we can substitute the given values into the equation:
F = 57 N (given)
theta = 60 degrees (given)
g = 9.8 m/s^2 (approximate value)
m = (57 N * sin(60 degrees)) / 9.8 m/s^2
Using a calculator, sin(60 degrees) is approximately 0.866.
m = (57 N * 0.866) / 9.8 m/s^2
Calculating the value,
m ≈ 5 kg (to 2 decimal places)
Therefore, a mass of approximately 5 kg must be used for the object hanging from the massless cable in order to apply a 57 N force to the leg.