Q63P - University Physics with Modern Physics

Question

A patient with a dislocated shoulder is put into a traction apparatus as shown in Fig. P1.63.The pulls $\overrightarrow{\mathbf{A}}\mathbf{ }\mathbf{and}\mathbf{ }\overrightarrow{\mathbf{B}}$

 have equal magnitudes and must combine to produce an outward traction force of 12.8 N on the patient’s arm.

How large should these pulls be?

Step-by-Step Solution

Answer

The patient’s arm should be pulled is,$A=B=7.55N$.

Step 1: Identification of given data

• The outward traction force is,$R_x=A_{\mathrm{x}}+{\mathrm{B}}_{\mathrm{x}}=128\mathrm{N}$.
• The displacement vector of arm is,$A_x+B_x$.

Step 2: Concept of Traction force

Tactile force, or traction, is the force utilized to induce motion between an object and a tangential surface.

Wet friction is the most typical method of generating motion between two bodies, however shear force applied to the tangential surface may also be utilized.

Step 3: Determine magnitude of pull

The resultant force can be evaluated as,

$\begin{array}{l}{R}_x=A_x+B_x\\ R_x=A\cos 32°+B\cos 32°\end{array}$

Since,$A=B$.

So,

$$\begin{gathered} 2A\cos 32°=R_x \\ A=\frac{R_x}{2\left(\cos 32°\right)} \\ A=\frac{12.8\mathrm{N}}{2\left(\cos 32°\right)} \\ A=7.55\mathrm{N} \end{gathered}$$

The magnitude of force in $A_x$,

$$\begin{gathered} A_x=A\cos 32° \\ {A}_x=7.55\cos 32° \\ {A}_x=6.40\mathrm{N} \end{gathered}$$

Thus, the magnitude of each pull is greater than magnitude pull in x-direction.