Chapter 11, Problem 5PS

To determine

To find: the total force exerted on the pin at position A .

Answer to Problem 5PS

  $\Vert F_A\Vert \approx 860.02pounds$

Explanation of Solution

Given information: A precast concrete wall is temporarily kept in its vertical position by

ropes the coordinates of A( 0,0,0), B (8,-10,6) and C (-10,-10,6) respectively. The tensions in AB and AC are 420 pounds and 650 pounds ,respectively .

Calculation:

Locate the origin at point A disregarding the left half of the precast.

Let ${\stackrel{\rightharpoonup }{F}}_A$ be the force at point A so that:

  $\stackrel{\rightharpoonup }{F_{AB}}+\stackrel{\rightharpoonup }{F_{AC}}=\stackrel{\rightharpoonup }{F_A}$ .

Because point A is at(0,0,0), so use this as initial point to find the direction of the forces as follows:

Point B is (8,-10,6) so:

  $\stackrel{\rightharpoonup }{P_{AB}}=(8,-10,6)$ .

The corresponding unit vector is :

  $\begin{array}{l}\stackrel{\rightharpoonup }{F_{AB}}=420\frac{1}{\sqrt{8^2+{(-10)}^2+6^2}}(8i-10j+6k).\\ \stackrel{\rightharpoonup }{F_{AB}}=\frac{3360}{\sqrt{200}}i-\frac{4200}{\sqrt{200}}j+\frac{2520}{\sqrt{200}}k.\end{array}$

Point C is (-10,-10,6)

The corresponding unit is :

  $\begin{array}{l}\stackrel{\rightharpoonup }{F_{AC}}=650\frac{1}{\sqrt{(-{10}^2+{(-10)}^2+6^2}}(-10i-10j+6k).\\ \stackrel{\rightharpoonup }{F_{AC}}=-\frac{6500}{\sqrt{236}}i-\frac{6500}{\sqrt{236}}j+\frac{3900}{\sqrt{236}}k.\end{array}$

The corresponding unit vector for the pin is:

  $\stackrel{\rightharpoonup }{F_A}=F_{Ax}i+F_{Ay}j+F_{Az}k.$

Along x -axis (i-components),

  $\frac{3360}{\sqrt{200}}-\frac{6500}{\sqrt{236}}=\stackrel{\rightharpoonup }{F_{Ax}}.$

Along y -axis (j-components),

  $-\frac{4200}{\sqrt{200}}-\frac{6500}{\sqrt{236}}=\stackrel{\rightharpoonup }{F_{Ay}}.$

Along z -axis (k-components),

  $\frac{2520}{\sqrt{200}}-\frac{3900}{\sqrt{236}}=\stackrel{\rightharpoonup }{F_{Az}}.$

Therefore,

  $\begin{array}{l}\Vert F_A\Vert =\sqrt{F_{Ax}{}^2+F_{Ay}{}^2+F_{Az}{}^2}.\\ \Vert F_A\Vert =\sqrt{{(\frac{3600}{\sqrt{200}}-\frac{6500}{\sqrt{236}})}^2+{(-\frac{4200}{\sqrt{200}}-\frac{6500}{\sqrt{236}})}^2+{(\frac{2520}{\sqrt{200}}+\frac{3900}{\sqrt{236}})}^2}.\\ \Vert F_A\Vert \approx 860.02pounds.\end{array}$