Question 2.SP.8: A wall section of precast concrete is temporarily held in pl......
A wall section of precast concrete is temporarily held in place by the cables shown. If the tension is 840 lb in cable AB and 1200 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted by cables AB and AC on stake A.
STRATEGY: This is a problem in adding concurrent forces in space. The simplest approach is to first resolve the forces into components and to then sum the components and find the resultant.
Final Answer
MODELING and ANALYSIS:
Components of the Forces. First resolve the force exerted by each cable on stake A into x, y, and z components. To do this, determine the components and magnitude of the vectors \overrightarrow{A B} \text { and } \overrightarrow{A C}, measuring them from A toward the wall section (Fig. 1). Denoting the unit vectors along the coordinate axes by i, j, k, these vectors are
\begin{array}{ll}\overrightarrow{A B}=-(16 ~\mathrm{ft}) \mathbf{i}+(8 ~\mathrm{ft}) \mathbf{j}+(11 ~\mathrm{ft}) \mathbf{k} & A B=21 ~\mathrm{ft} \\\overrightarrow{A C}=-(16 ~\mathrm{ft}) \mathbf{i}+(8 ~\mathrm{ft}) \mathbf{j}-(16 ~\mathrm{ft}) \mathbf{k} & A C=24 ~\mathrm{ft}\end{array}
Denoting by \boldsymbol{\lambda}_{A B} the unit vector along AB, the tension in AB is
\mathbf{T}_{A B}=T_{A B} \boldsymbol{\lambda}_{A B}=T_{A B} \frac{\overrightarrow{A B}}{A B}=\frac{840 ~\mathrm{lb}}{21 ~\mathrm{ft}} \overrightarrow{A B}
Substituting the expression found for \overrightarrow{A B}, the tension becomes
\mathbf{T}_{A B}=\frac{840 ~\mathrm{lb}}{21 ~\mathrm{ft}}[-(16 ~\mathrm{ft}) \mathbf{i}+(8 ~\mathrm{ft}) \mathbf{j}+(11 ~\mathrm{ft}) \mathbf{k}]
\mathbf{T}_{A B}=-(640 ~\mathrm{lb}) \mathbf{i}+(320 ~\mathrm{lb}) \mathbf{j}+(440 ~\mathrm{lb}) \mathbf{k}
Similarly, denoting by \boldsymbol{\lambda}_{A C} the unit vector along AC, the tension in AC is
\mathbf{T}_{A C}=T_{A C} \boldsymbol{\lambda}_{A C}=T_{A C} \frac{\overrightarrow{A C}}{A C}=\frac{1200 ~\mathrm{lb}}{24 ~\mathrm{ft}} \overrightarrow{A C}
\mathbf{T}_{A C}=-(800~ \mathrm{lb}) \mathbf{i}+(400 ~\mathrm{lb}) \mathbf{j}-(800 ~\mathrm{lb}) \mathbf{k}
Resultant of the Forces. The resultant R of the forces exerted by the two cables is
\mathbf{R}=\mathbf{T}_{A B}+\mathbf{T}_{A C}=-(1440 ~\mathrm{lb}) \mathbf{i}+(720 ~\mathrm{lb}) \mathbf{j}-(360 ~\mathrm{lb}) \mathbf{k}
You can now determine the magnitude and direction of the resultant as
R=\sqrt{R_x^2+R_y^2+R_z^2}=\sqrt{(-1440)^2+(720)^2+(-300)^2}
R = 1650 lb
The direction cosines come from Eqs. (2.33):
\cos \theta_x=\frac{R_x}{R} \quad \cos \theta_y=\frac{R_y}{R} \quad \cos \theta_z=\frac{R_z}{R} (2.33)
\begin{array}{c}\cos \theta_x=\frac{R_x}{R}=\frac{-1440 ~\mathrm{lb}}{1650 ~\mathrm{lb}} \quad \cos \theta_y=\frac{R_y}{R}=\frac{+720~ \mathrm{lb}}{1650~ \mathrm{lb}} \\\cos \theta_z=\frac{R_z}{R}=\frac{-360~ \mathrm{lb}}{1650~ \mathrm{lb}}\end{array}
Calculating each quotient and its arc cosine, the angles are
\theta_x=150.8^{\circ} \quad \theta_y=64.1^{\circ} \quad \theta_z=102.6^{\circ}
REFLECT and THINK: Based on visual examination of the cable forces, you might have anticipated that \theta_x for the resultant should be obtuse and \theta_y should be acute. The outcome of \theta_z was not as apparent.
