Chapter 11, Problem 7PS

To determine

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

Answer to Problem 7PS

  $\boxed{204.12N}$, $\boxed{157.11N}$ , $\boxed{225.84N}$

Explanation of Solution

Given information:

A precast concrete wall is temporarily kept in its vertical position by ropes (see figure). Find the total force exerted on the pin at position $A$ . The tensions in $AB$ and $AC$ are $420$ pounds and $650$ pounds, respectively.

Calculation:

Here, we will consider a following vector $a$ that can be expressed as:

  $a=<x,y,z>$

Now, the magnitude of the vector $a$ is:

  $\left|a\right|=\sqrt{x^2+y^2+z^2}$

As, we know that a unit vector is a vector having unit magnitude.

Now, for finding a unit vector $u$ in direction of any vector $a$ then multiply $a$ by the positive scalar $k=\frac{1}{\left|a\right|}$ .

Thus,

  $u=\left(\frac{1}{\left|a\right|}\right)\times a$

  $u=\frac{a}{\left|a\right|}$

Now, we will consider the following figure:

  

Here, two forces $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are acting on the wall, coordinates of $A,B$ and $C$ are:

  $\begin{array}{l}A\left(10,0,0\right)\\ B\left(0,-8,6\right)\\ C\left(0,10,6\right)\end{array}$

The forces are in the vector form in such a way that vector $\overrightarrow{AB}$ is:

  $\overrightarrow{AB}=<0-10,-8-0,6-0>$

  $=<-10,-8,6>$

Vector $\overrightarrow{AC}$ is:

  $\overrightarrow{AC}=<0-10,10-0,6-0>$

  $=<-10,10,6>$

The force vectors are as follows:

We will first find the unit vector in direction of $AB$ :

  $\frac{\overrightarrow{AB}}{AB}$

Thus, force $b$ in direction of $AB$ is:

  $b=\left|b\right|\frac{\overrightarrow{AB}}{AB}$

  $=\left|b\right|\frac{<0,7,-115>}{134.63}$

  $=\left|b\right|<0,0.52,-0.85>$

Now, we will find the unit vector in direction of $AC$ is:

  $\frac{\overrightarrow{AC}}{AC}$

Thus, force $c$ in direction of $AC$ is:

  $c=\left|c\right|\frac{\overrightarrow{AC}}{AC}$

  $=\left|c\right|\frac{<-60,0,-115>}{129.71}$

  $=\left|c\right|<-0.46,0,-0.89>$

Now, find the unit vector in direction of $AD$ is:

  $\frac{\overrightarrow{AD}}{AD}$

Thus, force $d$ in direction of $AD$ is:

  $d=\left|d\right|\frac{\overrightarrow{AD}}{AD}$

  $=\left|d\right|\frac{<45,-65,-115>}{139.55}$

  $=\left|d\right|<0.32,-0.47,-0.82>$

Now, at equilibrium the resultant force acting on crate is zero. So,

  $a+b+c-w=0$

Thus, the equations formed are:

  $\begin{array}{l}-0.46\left|c\right|+0.32\left|d\right|=0\\ 0.52\left|b\right|-0.47\left|d\right|=0\\ -0.85\left|b\right|-0.89\left|c\right|-0.82\left|d\right|=-500\end{array}$

By solving these three equations we will get:

  $\begin{array}{l}\left|b\right|=204.12N\\ \left|c\right|=157.11N\\ \left|d\right|=225.84N\end{array}$

Hence, tensions of supporting cables are: $\boxed{204.12N}$ , $\boxed{157.11N}$ , $\boxed{225.84N}$.