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. the tire remains stationary in spite of the three pulls. alex pulls with force of magnitude 206.7 n, and charles pulls with force of magnitude 155.1 n. note that the direction of charles force is not given. what is the magnitude of bettys force?

Explanation:

Step1: Resolve forces in x - direction

Since the tire is stationary, the net force in the x - direction is zero. Let the force exerted by Alex be $F_A = 206.7$ N, the force exerted by Charles be $F_C=155.1$ N and the force exerted by Betty be $F_B$. Assume the x - axis is horizontal and y - axis is vertical. The x - component of Alex's force is $F_{Ax}=-F_A\cos(137^{\circ})$. The x - component of Charles's force is $F_{Cx}=F_C\cos\theta$ and the x - component of Betty's force is $F_{Bx} = 0$. So, $F_{Ax}+F_{Cx}+F_{Bx}=0$.

Step2: Resolve forces in y - direction

The net force in the y - direction is also zero. The y - component of Alex's force is $F_{Ay}=-F_A\sin(137^{\circ})$, the y - component of Charles's force is $F_{Cy}=F_C\sin\theta$ and the y - component of Betty's force is $F_{By}=-F_B$. So, $F_{Ay}+F_{Cy}+F_{By}=0$.
From the x - direction equilibrium $-F_A\cos(137^{\circ})+F_C\cos\theta+0 = 0$, we can express $\cos\theta=\frac{F_A\cos(137^{\circ})}{F_C}$.
From the y - direction equilibrium $-F_A\sin(137^{\circ})+F_C\sin\theta - F_B=0$.
We know that $\sin^{2}\theta+\cos^{2}\theta = 1$. But we can also use the fact that since the tire is in equilibrium, we can use the law of cosines.
Let the forces be vectors $\vec{F}_A$, $\vec{F}_B$ and $\vec{F}_C$. The magnitude of the resultant of $\vec{F}_A$ and $\vec{F}_C$ must be equal to the magnitude of $\vec{F}_B$ because the net force on the tire is zero.
We can use the law of cosines: $F_B^{2}=F_A^{2}+F_C^{2}-2F_AF_C\cos(137^{\circ})$
Substitute $F_A = 206.7$ N and $F_C = 155.1$ N.
$\cos(137^{\circ})\approx - 0.7314$
$F_B^{2}=(206.7)^{2}+(155.1)^{2}-2\times206.7\times155.1\times(- 0.7314)$
$F_B^{2}=42724.89+24056.01+46979.97$
$F_B^{2}=113760.87$
$F_B=\sqrt{113760.87}\approx337.3$ N

Answer:

$337.3$ N