Question

quiz #2 (due 11:59 pm) dislocated shoulder. a patient with a dislocated shoulder is put into a traction apparatus as shown in the figure. the pulls $vec{a}$ and $vec{b}$ have equal magnitudes and must combine to produce and outward traction force of 12.8 n on the patient’s arm. how large should these pulls be? figure p1.61 $vec{a}=a_xhat{i}+a_yhat{j}$ $vec{r}=vec{a}+vec{b}$ $vec{b}=b_xhat{i}+b_yhat{j}$ $vec{r}=12.8nhat{i}$ $\theta = 32^{circ}$ $vec{r}=(a_x + b_x)hat{i}+(a_y + b_y)hat{j}$ $a_y=-b_y$; $a_x = b_x$ $vec{r}=2a_xhat{i}=12.8nhat{i}$ $b_x = a_x=\frac{12.8n}{2}=6.4n$ $a,b =?$

Explanation:

Step1: Analyze the x - components

Since the y - components of $\vec{A}$ and $\vec{B}$ cancel each other out ($A_y=-B_y$) and the x - components add up ($A_x = B_x$), and the resultant force $\vec{R}=\vec{A}+\vec{B}$ has only an x - component with magnitude $R = 12.8N$. So $R=A_x + B_x=2A_x$.

Step2: Solve for $A_x$ and $B_x$

We know that $2A_x=12.8N$, then $A_x = B_x=\frac{12.8N}{2}=6.4N$.

Step3: Find the magnitude of $\vec{A}$ and $\vec{B}$

We know that $\cos\theta=\frac{A_x}{A}$, where $\theta = 32^{\circ}$. So $A=\frac{A_x}{\cos\theta}=\frac{6.4N}{\cos32^{\circ}}$. Since $\cos32^{\circ}\approx0.848$, then $A = B=\frac{6.4N}{0.848}\approx7.55N$.

Answer:

$7.55N$