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 $4.0kg$, and the doctor has decided to hang a $6.0kg$mass from the rope. The boot is held suspended by the ropes, as shown in $FIGUREP6.41$, 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 u for the upper rope?

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

The extra mass hanged with patient's foot$\left(M\right)=6kg$

The mass of boot$\left(m\right)=4kg$

(a) The value of tension is same throughout the wire, therefore, the tension is due to mass hanging. From the force body diagram

Free body diagram 1

The sum of forces is given by

$\sum _{}{F}_y=T-Mg=0\left(I\right)$

As the body is not accelerating then the net force will be zero

Therefore

$T=Mg\left(II\right)$

Substituting the values of mass (M) and g in the above equation, we can calculate the amount of tension in the rope.

$$\begin{gathered} T=Mg=\left(6.0kg\right)\times \left(9.8m/s^2\right) \\ T=58.8N \end{gathered}$$

(b) To calculate the traction force in horizontal direction, the tension force is needed to cancel with the weight of the boot.

The weight of the boot$F_G=mg\left(III\right)$

Force body diagram $2$

From the body diagram$\left(2\right)$

The forces in the vertical direction are given by

$\sum _{}{F}_y=T\sin \theta -T\sin 15°-mg=0\left(IV\right)$

Solving equation (IV), the angles is obtained as

$\theta ={\sin }^{-1}\left(\frac{T\sin 15°+mg}{T}\right)\left(V\right)$

Also,

$T=Mg\left(VI\right)$

Therefore,

$\theta ={\sin }^{-1}\left(\sin 15°+\frac{m}{M}\right)\left(VII\right)$

Substituting the values of mass of the boot $\left(m\right)$and the extra mass hanged $\left(M\right)$in above equation, the value of angle is given by

$\theta ={\sin }^{-1}\left(\sin 15°+\frac{m}{M}\right)\left(VII\right)$

(c) since all the vertical components of the force are balanced, therefore the traction force is all horizontal,

$\sum _{}{F}_x=T\cos \theta -T\cos 15°-F_{traction}=0\left(IX\right)$

Now taking the value of $\theta =67.7°$from the equation $\left(VIII\right)$and substituting it in equation $\left(IX\right)$

$\sum _{}{F}_x=T\cos 67.7°-T\cos 15°-F_{traction}=0\left(X\right)$

So, traction force is calculated as

$$\begin{gathered} F_{traction}=T\left(\cos 67.7°+\cos 15°\right) \\ {F}_{traction}=\left(58.8\right)\left(1.35\right)=79N \end{gathered}$$