Question

When the thigh is fractured, the patient's leg must be kept under traction. One method of doing so is a variation on the Russell traction apparatus. (See the figure.)

Answer 1

i hope theres a picture :)

Answer 2

If the physical therapist specifies that the traction force directed along the leg must be 20.0 N, what must Wbe? yah one sec

Answer 3

im not sure, but the top angle seems redundant to me

Answer 4

Wcos(35) = 20

W = 20sec(35) would be my best guess

Answer 5

is w the same thing as T

Answer 6

yes

Answer 7

i thought the goal was to find (W) which is the mass

Answer 8

find w that produces the tension in the system

Answer 9

of course like i said before, i aint to sure about my physics applications tho

Answer 10

so, if you have anything to offer other than me taking a blind stab at it, feel free to contribute it

Answer 11

so w is what cause the tension

Answer 12

The leg is seeing two tensions.

Therefore, \[2T \cos(35) = W\]

Answer 13

thats where my doubts were :)

Answer 14

so is w the total force ?

Answer 15

These one always make me stop and think twice.

Remember if the system is stationary, the forces must cancel. The first time the wire is diverted towards the leg, we created a horizontal component of the tension force. This horizontal component is created again when we turn away from the leg. Therefore, the force on the leg must be twice the horizontal component such that they cancel.

@nath We have two expressions here.

Answer 16

The first one is\[2T \cos(35) = 20\]and the second\[T = W\]Solve for T, then we know W

Answer 17

why am i able to replace t with w

Answer 18

What forces act in the vertical direction? Tension and weight, right?

Note that because the angles about the pulley attached to the leg are symmetric, their vertical components cancel out.

Answer 19

dont the x force cancel each other out?

Answer 20

The x forces cancel out with the tractive force on the leg.

Let's write out our expressions for summation of forces in the x and y-directions. \[F_x = 0 \rightarrow 2 T \cos(35) = 20\]\[F_y = 0 \rightarrow T \sin(35) - T \sin(35) + T_{ceiling} - W = 0\]The first two terms are at the pulley attached at the leg, the \(T_{ceiling}\) is the tension at the ceiling.

Answer 21

both the net forces of the x and y component are equal to zero

Answer 22

Yes. The leg isn't moving nor is the weight.

Answer 23

so Tceiling =W?