Question

the angle between the guy wire ac and the mast is 20°. knowing that the tension in ac is 321 lb, determine the x, y, and z components of the force exerted on the boat at c. (include a minus sign if necessary) the component of the force in the x - direction ( f_x ) is (square) lb. the component of the force in the y - direction ( f_y ) is (square) lb. the component of the force in the z - direction ( f_z ) is (square) lb.

Explanation:

Step1: Analyze the force components

First, we know the tension in AC is \( F = 321 \) lb. The angle between the guy wire AC and the mast (y - axis) is \( 20^{\circ} \), and the angle in the x - z plane is \( 40^{\circ} \).

First, find the component of the force perpendicular to the y - axis. Let \( F_{yz} \) be the component of the force in the x - z plane. We use the formula \( F_{yz}=F\sin(20^{\circ}) \), since the angle between the force vector and the y - axis is \( 20^{\circ} \), so the component perpendicular to the y - axis (in x - z plane) is \( F\sin(20^{\circ}) \).

Then, for the x - component: The angle between the projection of the force in the x - z plane and the x - axis is \( 40^{\circ} \). So \( F_x=-F_{yz}\cos(40^{\circ})=-F\sin(20^{\circ})\cos(40^{\circ}) \) (the negative sign is because of the direction, assuming the x - component is in the negative x - direction as per the diagram).

For the y - component: The component along the y - axis is \( F_y = - F\cos(20^{\circ}) \) (negative sign because the force is pulling towards A, opposite to the positive y - direction).

For the z - component: The component along the z - axis is \( F_z = F_{yz}\sin(40^{\circ})=F\sin(20^{\circ})\sin(40^{\circ}) \)

Step2: Calculate \( F_x \)

First, calculate \( \sin(20^{\circ})\approx0.3420 \), \( \cos(40^{\circ})\approx0.7660 \)
\( F_x=- 321\times0.3420\times0.7660\)
\( 321\times0.3420 = 321\times\frac{3420}{10000}=321\times0.342 = 109.782\)
\( 109.782\times0.7660\approx84.0 \)
So \( F_x\approx - 84.0 \) lb (the negative sign indicates direction, but we can also consider the magnitude with sign as per the problem's requirement of including sign)

Step3: Calculate \( F_y \)

\( \cos(20^{\circ})\approx0.9397 \)
\( F_y=-321\times0.9397\approx - 321\times0.94=-301.74\approx - 302 \) lb

Step4: Calculate \( F_z \)

\( \sin(40^{\circ})\approx0.6428 \)
\( F_z = 321\times0.3420\times0.6428\)
\( 321\times0.3420 = 109.782\)
\( 109.782\times0.6428\approx109.782\times0.643\approx70.6 \) lb

Answer:

The component of the force in the x - direction \( F_x\approx\boxed{- 84.0} \) lb.

The component of the force in the y - direction \( F_y\approx\boxed{- 302} \) lb.

The component of the force in the z - direction \( F_z\approx\boxed{70.6} \) lb.