Question

In a two-dimensional tug-of-war, Alex, Betty, and Charles pull horizontally on an automobile tire at the angles shown in the overhead view of Fig. 5-30.The tire remains stationary in spite of the three pulls.

Alex pulls with force ${\overrightarrow{\mathbf{F}}}_{\mathbf{A}}$of magnitude 200N , and Charles pulls with force ${\overrightarrow{\mathbf{F}}}_{\mathbf{c}}$of magnitude 170N . Note that the direction of ${\overrightarrow{\mathbf{F}}}_{\mathbf{c}}$is not given. What is the magnitude of Betty’s ${\overrightarrow{\mathbf{F}}}_{\mathbf{B}}$?

Short answer

The magnitude of Betty’s force is 241N

Step-by-step solution

Given information:

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

Significance of Newton’s second law of motion

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:

$F_{net}=ma$ …(i)

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

 Step 3: Determining the magnitude of Betty’s force. 

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

Free body diagram,

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

From the above free body diagram,

$F_{net.x}=F_c\cos \varphi -F_A\cos \theta$ …(ii)

= 0 …(iii)

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

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

Thus, $\phi =28.0°$

Substituting this value in equation (iii) we get,

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

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