Question 2.SP.7: A tower guy wire is anchored by means of a bolt at A. The te......
A tower guy wire is anchored by means of a bolt at A. The tension in the wire is 2500 N. Determine (a) the components F_x, F_y, \text { and } F_z of the force acting on the bolt and (b) the angles \theta_x, \theta_y, \text { and } \theta_z defining the direction of the force.
STRATEGY: From the given distances, we can determine the length of the wire and the direction of a unit vector along it. From that, we can find the components of the tension and the angles defining its direction.
- Tension in the wire: 2500 N
Final Answer
MODELING and ANALYSIS:
a. Components of the Force. The line of action of the force acting on the bolt passes through points A and B, and the force is directed from A to B. The components of the vector \overrightarrow{A B}, which has the same direction as the force, are
d_x=-40 \mathrm{~m} \quad d_y=+80 \mathrm{~m} \quad d_z=+30 \mathrm{~m}
The total distance from A to B is
A B=d=\sqrt{d_x^2+d_y^2+d_z^2}=94.3 \mathrm{~m}
Denoting the unit vectors along the coordinate axes by i, j, and k, you have
\overrightarrow{A B}=-(40 \mathrm{~m}) \mathbf{i}+(80 \mathrm{~m}) \mathbf{j}+(30 \mathrm{~m}) \mathbf{k}
Introducing the unit vector \lambda=\overrightarrow{A B} / A B (Fig. 1), you can express F in terms of \overrightarrow{A B} as
\mathbf{F}=F \boldsymbol{\lambda}=F \frac{\overrightarrow{A B}}{A B}=\frac{2500 \mathrm{~N}}{94.3 \mathrm{~m}} \overrightarrow{A B}
Substituting the expression for \overrightarrow{A B} gives you
\begin{aligned}\mathbf{F} & =\frac{2500 \mathrm{~N}}{94.3 \mathrm{~m}}[-(40 \mathrm{~m}) \mathbf{i}+(80 \mathrm{~m}) \mathbf{j}+(30 \mathrm{~m}) \mathbf{k}] \\& =-(1060 \mathrm{~N}) \mathbf{i}+(2120 \mathrm{~N}) \mathbf{j}+(795 \mathrm{~N}) \mathbf{k}\end{aligned}
The components of F, therefore, are
F_x=-1060 \mathrm{~N} \quad F_y=+2120 \mathrm{~N} \quad F_z=+795 \mathrm{~N}
b. Direction of the Force. Using Eqs. (2.25), you can write the direction cosines directly (Fig. 2):
\cos \theta_x=\frac{F_x}{F} \quad \cos \theta_y=\frac{F_y}{F} \quad \cos \theta_z=\frac{F_z}{F} (2.25)
\begin{array}{c}\cos \theta_x=\frac{F_x}{F}=\frac{-1060 \mathrm{~N}}{2500 \mathrm{~N}} \quad \cos \theta_y=\frac{F_y}{F}=\frac{+2120 \mathrm{~N}}{2500 \mathrm{~N}} \\\cos \theta_z=\frac{F_z}{F}=\frac{+795 \mathrm{~N}}{2500 \mathrm{~N}}\end{array}
Calculating each quotient and its arc cosine, you obtain
\theta_x=115.1^{\circ} \quad \theta_y=32.0^{\circ} \quad \theta_z=71.5^{\circ}
(Note. You could have obtained this same result by using the components
and magnitude of the vector \overrightarrow{A B} rather than those of the force F.)
REFLECT and THINK: It makes sense that, for a given geometry, only a certain set of components and angles characterize a given resultant force. The methods in this section allow you to translate back and forth between forces and geometry.
