1.73 BIO Dislocated Shoulder. A patient with a dislocated shoulder is put into a traction apparatus as shown in Fig. P1.73. The pulls A

1.73 BIO Dislocated Shoulder. A patient with a dislocated shoulder is put into a traction apparatus as shown in Fig. P1.73. The pulls $\vec{A}$ and $\vec{B}$ have equal magnitudes and must combine to produce an outward traction force of 5.60 N on the patient’s arm. How large should these pulls be?

SOLUTION

The magnitudes of the vectors $\vec{A}$ and $\vec{B}$ are equal: $A=B$.

Then the components of the given vectors where the angles are measured from the +x-axis and rotates toward the +y-axis are,

$R_x=(5.60 \ N) \ cos \ 0^o=5.60 \ N$

$R_y=(5.60 \ N) \ sin \ 0^o=0 \ N$

$A_x=A \ cos \ 32^o$

$A_y=A \ sin \ 32^o$

$B_x=B \ cos \ (360^o-32^o)=A \ cos \ 328^o$

$B_y=B \ sin \ (360^o-32^o)=A \ sin \ 328^o$

Since the resultant vector is $\vec{R}$, then $\vec{R}=\vec{A}+\vec{B}$.

Its x-component is,

$R_x=A_x+B_x $

$5.60 \ N=A \ cos \ 32^o+A \ cos \ 328^o$ where

$A=\frac{5.60 \ N}{cos \ 32^o + cos \ 328^o}=3.30 \ N$.

The y-component is

$R_y=A_y+B_y $

$0=A \ sin \ 32^o+A \ sin \ 328^o$

$A \ sin \ 32^o=-A \ sin \ 328^o$

$0.530 = -0.530$

The pull of vectors $\vec{A}$ and $\vec{B}$ amounts to $3.30 \ N$ and points along the +x-axis as shown in Figure 1. Their y-components has the same magnitude but differs in direction.

Figure 1