Question

An accident victim with a broken leg is being placed in traction. The patient wears a special boot with a pulley attached to the sole. The foot and boot together have a mass of 4.0 kg, and the doctor has decided to hang a 6.0 kg mass from the rope. The boot is held suspended by the ropes, as shown in FIGURE P6.41, and does not touch the bed.

a. Determine the amount of tension in the rope by using Newton’s laws to analyze the hanging mass.

b. The net traction force needs to pull straight out on the leg. What is the proper angle$\theta$for the upper rope?

c. What is the net traction force pulling on the leg?

Short answer

a. Tension in rope = 58.8N

b. Proper angle $\theta$for the upper rope = 67.5$°$

c. Net traction force pulling on the leg = 79.0N

Step-by-step solution

- Tension

Tension Force : - Force that is transmitted through the a rope, string or wire when pulled by forces acting from opposite sides.

- Given

Mass of foot and boot = 4.0kg

Hanging mass = 6.0kg

Angle($\theta$) = 12$°$

Step 3

a. The amount of tension in rope : -

∑ Fy = T−Mg =0

M = mass of the hanging weight

T = Mg

=6.0×9.81

T =58.8 N

Step 4

b. $\sum _{}^{}$Fy = T sin$\theta$- T sin 12$°$-mg = 0

Solving for the angles

$\theta$= sin−1 $\left(\frac{T\sin 12°+mg}{T}\right)$

T = Mg

$\theta$= sin−1 role="math" localid="1647584331124" $\left(\sin 12°+\begin{array}{c}m\\ M\end{array}\right)$

$\theta$= sin−1 $\left(0.027+\begin{array}{c}4\\ 6.0\end{array}\right)$

$\theta$= sin−1 (0.873)

$\theta$= 60.80$°$

Step 5

$\sum _{}^{}$Fx = T cos 60.80° + T cos 12° − $F_{fraction}$=0

$F_{fraction}$= T (cos 60.80° + cos 12°)

= 58.8 (0.487 + 0.978)

$F_{fraction}$= 86.14N