91影视

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

$\underset{}{鈭�}{F}_y=T-Mg=0\left(I\right)$

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.

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

(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

$\underset{}{鈭�}{F}_y=T\sin 胃-T\sin 15掳-mg=0\left(IV\right)$

Solving equation (IV), the angles is obtained as

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

Also,

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

Therefore,

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

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

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

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

$\underset{}{鈭�}{F}_x=T\cos 胃-T\cos 15掳-F_{traction}=0\left(IX\right)$

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

$\underset{}{鈭�}{F}_x=T\cos 67.7掳-T\cos 15掳-F_{traction}=0\left(X\right)$

So, traction force is calculated as

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