91Ó°ÊÓ

$$\begin{gathered} {\stackrel{→}{F}}_A=220\mathrm{N} \\ {\stackrel{→}{F}}_C=170\mathrm{N} \end{gathered}$$

The problem is based on Newton’s second law of motion. It states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object.

Formula:

$F_{net}=ma$ …(¾±)

Where, m is the mass of object and is the acceleration of object

Since the tire remains stationary, according to the Newton’s second law, the net force is zero,

Free body diagram,

$$\begin{gathered} {\stackrel{→}{F}}_{net}={\stackrel{→}{F}}_A+{\stackrel{→}{F}}_B+{\stackrel{→}{F}}_C \\ =ma \\ =0\left(\mathrm{net}\mathrm{force}\mathrm{is}\mathrm{zero}\mathrm{because}\mathrm{tire}\mathrm{remains}\mathrm{stationary}\right) \end{gathered}$$

From the above free body diagram,

$F_{net.x}=F_c\cos \mathrm{Ï•}-F_A\cos \mathrm{θ}$ …(¾±¾±)

= 0 …(iii)

Solving for $F_B$for localid="1660895724031" ${\stackrel{→}{F}}_A=220\mathrm{N},{\stackrel{→}{\mathrm{F}}}_{\mathrm{C}}=170\mathrm{N}$and$\mathrm{θ}=47°$, equation (ii) can be written as,

$$\begin{gathered} \cos \mathrm{φ}=\frac{F_A\cos \mathrm{θ}}{F_C} \\ =\frac{\left(220\mathrm{N}\right)\cos 47°}{170\mathrm{N}} \\ =0.88 \end{gathered}$$

Thus, $\mathrm{φ}=28.0°$

Substituting this value in equation (iii) we get,

$$\begin{gathered} F_B=F_A\sin \mathrm{θ}+F_C\sin \mathrm{φ} \\ =\left(220\mathrm{N}\right)\sin 47°+\left(170\mathrm{N}\right)\sin 28° \\ =241\mathrm{N} \end{gathered}$$

Thus,$F_B=241\mathrm{N}$