Chapter 4.3, Problem 4.107P

PROBLEM 4.107

Solve Problem 4.106 for a = 1.5 m.

PROBLEM 4.106 The 6-m pole ABC is acted upon by a 455-N force as shown. The pole is held by a ball-and-socket joint at A and by two cables BD and BE. For a =3 m, determine the tension in each cable and the reaction at A.

To determine

The tension in each of the cable and the reaction at A.

Answer to Problem 4.107P

The tension in cable BD is $\underset{\_}{525\text{N}}$ , tension in cable BE is $\underset{\_}{105\text{N}}$. Also the reaction at $\underset{\_}{\left(-105\text{N}\right)\widehat{i}+\left(840\text{N}\right)\widehat{j}+\left(140\text{N}\right)\widehat{k}}$.

Explanation of Solution

Refer figure1 shown below. The figure shows that the pole ABC is acted upon by a certain force. The pole is balanced by cables BD and BE about a ball and socket joint A. The forces along the cables can be resolved into its x and y components. $P$ is the force acting on the pole, and $T$ is the tension along the cables.

The sum of moments along line AC is zero at equilibrium.

$\sum M_{AC}=0$ (I)

Here, $\sum M_A$ is the sum of moments along line AC.

Write the equation to find the position vector of point B.

$r_B=y\widehat{j}$ (II)

Here, $r_B$ is the position vector of point B, $y$ is y component of the position vector.

Write the equation to find the position vector of point C.

$r_C=y\widehat{j}$ (III)

Here, $r_C$ is the position vector of point C, $y$ is the y component of the position vector.

Write the equation to find the position vector $\overrightarrow{CF}$.

$\overrightarrow{CF}=x\widehat{i}+y\widehat{j}+z\widehat{k}$ (IV)

Here, $\overrightarrow{CF}$ is the position vector $\overrightarrow{CF}$ , $x$ is the x component of vector, $y$ is the y component of vector, $z$ is the z component of vector

Write the equation to find the magnitude of vector $\overrightarrow{CF}$.

$CF=\sqrt{x^2+y^2+z^2}$ (V)

Here, $CF$ is the magnitude of vector $\overrightarrow{CF}$ , $x$ is the x component of vector, $y$ is the y component of vector, $z$ is the z component of vector

Write the equation to find the position vector $\overrightarrow{BD}$.

$\overrightarrow{BD}=x\widehat{i}+y\widehat{j}+z\widehat{k}$ (VI)

Here, $\overrightarrow{BD}$ is the position vector, $x$ is the x component of vector, $y$ is the y component of vector, $z$ is the z component of vector

Write the equation to find the magnitude of vector $\overrightarrow{BD}$.

$BD=\sqrt{x^2+y^2+z^2}$ (VII)

Here, $BD$ is the magnitude of vector $\overrightarrow{BD}$ , $x$ is the x component of vector, $y$ is the y component of vector, $z$ is the z component of vector

Write the equation to find the position vector $\overrightarrow{BE}$.

$\overrightarrow{BE}=x\widehat{i}+y\widehat{j}+z\widehat{k}$ (VIII)

Here, $\overrightarrow{BE}$ is the position vector , $x$ is the x component of vector, $y$ is the y component of vector, $z$ is the z component of vector

Write the equation to find the magnitude of vector $\overrightarrow{BE}$.

$BE=\sqrt{x^2+y^2+z^2}$ (IX)

Here, $BE$ is the magnitude of vector $\overrightarrow{BE}$ , $x$ is the x component of vector, $y$ is the y component of vector, $z$ is the z component of vector

Write the equation to find the force acting along line CF.

$P=P\frac{\overrightarrow{CF}}{CF}$ (X)

Here, $P$ is the force vector along the line CF, $P$ is the magnitude of force, $\overrightarrow{CF}$ is the position vector along line CF, and $CF$ is the magnitude of the position vector.

Write the equation to find the tension in cable BD.

$T_{BD}=T_{BD}\frac{\overrightarrow{BD}}{BD}$ (XI)

Here, $T_{BD}$ is the tension vector along the cable BD, $T_{BD}$ is the magnitude of tension, $\overrightarrow{BD}$ is the position vector corresponding to cable, $BD$ is the magnitude of position vector.

Write the equation to find the tension in cable BE.

$T_{BE}=T_{BE}\frac{\overrightarrow{BE}}{BE}$ (XII)

Here, $T_{BE}$ is the tension vector along the cable BE, $T_{BE}$ is the magnitude of tension, $\overrightarrow{BE}$ is the position vector corresponding to cable, $BE$ is the magnitude of position vector.

Write the equation to find the sum of moments along the point A.

$\sum M_A=r_B\times T_{BD}+r_B\times T_{BE}+r_C\times P$ (XIII)

Here, $r_B$ is the position vector of point B, $T_{BD}$ is the tension in cable BD, $T_{BE}$ is the tension in cable BE, $r_C$ is the position vector of point C, $P$ is the force at point C.

Since the sum of moments at point A is zero, rewrite equation (XI).

$r_B\times T_{BD}+r_B\times T_{BE}+r_C\times P=0$ (XIV)

Substitute for $r_B$ , $r_C$ , $T_{BD}$ , $T_{BE}$ , $P$ and take the determinant of equation (XII).

$\left|\begin{array}{l}\widehat{i}\widehat{j}\widehat{k}\\ 030\\ 1-2-2\end{array}\right|\frac{T_{BD}}{BD}+\left|\begin{array}{l}\widehat{i}\widehat{j}\widehat{k}\\ 030\\ 1-22\end{array}\right|\frac{T_{BE}}{BE}+\left|\begin{array}{l}\widehat{i}\widehat{j}\widehat{k}\\ 060\\ -3-124\end{array}\right|\frac{P}{CE}=0$ (XV)

Find the determinant value of equation (XIII) and equate the coefficients of unit vectors $\widehat{i}$ , $\widehat{j}$ , and $\widehat{k}$.

Equate the coefficients of unit vector $\widehat{i}$ to zero.

$-2T_{BD}+2T_{BE}+\frac{24}{13}P=0$ (XVI)

Here, $T_{BD}$ is the magnitude of tension in cable BD, $T_{BE}$ is the magnitude of tension in cable BE, $P$ is the magnitude of force.

Equate the coefficient of unit vector $\widehat{k}$ to zero.

$-T_{BD}-T_{BE}+\frac{18}{13}P=0$ (XVII)

Add equations (XIV) to twice of equation (XV) to get the value of tensions in cables.

$\begin{array}{c}-2T_{BD}+2T_{BE}+\frac{24}{13}P+2\left(-T_{BD}-T_{BF}+\frac{18}{13}P\right)=-4T_{BD}+\frac{60}{13}P\\ =0\end{array}$ (XVIII)

The sum of all external forces is zero. Therefore sum of tensions in cables, force at P and S is zero.

$T_{BD}+T_{BE}+P+A=0$ (XIX)

Here, $T_{BD}$ is the tension in cable BD, $T_{BE}$ is the tension in cable BE, $P$ is the force at point P, and A is the magnitude of reaction force at point A.

Conclusion:

Substitute $6\text{m}$ for $y$ in equation (II) to get $r_B$.

$r_B=3\widehat{j}$

Substitute $6\text{m}$ for $y$ in equation (III) to get $r_C$.

$r_C=6\widehat{j}$

Substitute $-1.5\text{m}$ for $x$ , $-6\text{m}$ for $y$ , and $2\text{m}$ for $z$ in equation (IV) to get $\overrightarrow{CF}$.

$\overrightarrow{CF}=-1.5\widehat{i}-6\widehat{j}+2\widehat{k}$

Substitute $-1.5\text{m}$ for $x$ , $-6\text{m}$ for $y$ , and $2\text{m}$ for $z$ in equation (V) to get $CF$.

$\begin{array}{c}CF=\sqrt{{\left(-1.5\text{m}\right)}^2+{\left(-6\text{m}\right)}^2+{\left(2\text{m}\right)}^2}\\ =6.5\text{m}\end{array}$

Substitute $1.5\text{m}$ for $x$ , $-3\text{m}$ for $y$ , and $-3\text{m}$ for $z$ in equation (VI) to get $\overrightarrow{BD}$.

$\overrightarrow{BD}=1.5\widehat{i}-3\widehat{j}-3\widehat{k}$

Substitute $1.5\text{m}$ for $x$ , $-3\text{m}$ for $y$ , and $-3\text{m}$ for $z$ in equation (VII) to get $BD$.

$\begin{array}{c}BD=\sqrt{{\left(1.5\text{m}\right)}^2+{\left(-3\text{m}\right)}^2+{\left(3\text{m}\right)}^2}\\ =4.5\text{m}\end{array}$

Substitute $1.5\text{m}$ for $x$ , $-3\text{m}$ for $y$ , and $3\text{m}$ for $z$ in equation (VIII) to get $\overrightarrow{BE}$

$\overrightarrow{BE}=1.5\widehat{i}-3\widehat{j}+3\widehat{k}$

Substitute $1.5\text{m}$ for $x$ , $-3\text{m}$ for $y$ , and $3\text{m}$ for $z$ in equation (IX) to get $BE$

$\begin{array}{c}BE=\sqrt{{\left(1.5\text{m}\right)}^2+{\left(-3\text{m}\right)}^2+{\left(3\text{m}\right)}^2}\\ =4.5\text{m}\end{array}$

Substitute $\left(-1.5\widehat{i}-6\widehat{j}+2\widehat{k}\right)$ for $\overrightarrow{CF}$ and $6.5\text{m}$ for $CF$ in equation (X) to get $P$.

$\begin{array}{c}P=\frac{P}{7}\left(-1.5\widehat{i}-6\widehat{j}+2\widehat{k}\right)\\ =\frac{P}{13}\left(-3\widehat{i}-12\widehat{j}+4\widehat{k}\right)\end{array}$

Substitute $\left(1.5\widehat{i}-3\widehat{j}-3\widehat{k}\right)$ for $\overrightarrow{BD}$ and $4.5\text{m}$ for $BD$ in equation (XI) to get $T_{BD}$.

$\begin{array}{c}{T}_{BD}=\frac{T_{BD}}{4.5}\left(1.5\widehat{i}-3\widehat{j}-3\widehat{k}\right)\\ =\frac{T_{BD}}{3}\left(\widehat{i}-2\widehat{j}-2\widehat{k}\right)\end{array}$

Substitute $\left(1.5\widehat{i}-2\widehat{j}+2\widehat{k}\right)$ for $\overrightarrow{BE}$ and $4.5\text{m}$ for $BE$ in equation (XII) to get $T_{BE}$.

$T_{BE}=\frac{T_{BE}}{3}\left(\widehat{i}-2\widehat{j}+2\widehat{k}\right)$

Substitute $\frac{P}{13}\left(-3\widehat{i}-12\widehat{j}+4\widehat{k}\right)$ for $P$ , $\frac{T_{BD}}{3}\left(\widehat{i}-2\widehat{j}-2\widehat{k}\right)$ for $T_{BD}$ , and $\frac{T_{BE}}{3}\left(\widehat{i}-2\widehat{j}+2\widehat{k}\right)$ for $T_{BE}$ , $3\widehat{j}$ for $r_B$ , $6\widehat{j}$ for $r_C$ in equation (XIII) to get $\sum M_A$ in determinant form.

$\left|\begin{array}{l}\widehat{i}\widehat{j}\widehat{k}\\ 030\\ 1-2-2\end{array}\right|\frac{T_{BD}}{3}+\left|\begin{array}{l}\widehat{i}\widehat{j}\widehat{k}\\ 030\\ 1-22\end{array}\right|\frac{T_{BE}}{3}+\left|\begin{array}{l}\widehat{i}\widehat{j}\widehat{k}\\ 060\\ -3-124\end{array}\right|\frac{P}{13}=0$

Solve equation (XVIII) to get $T_{BD}$.

$T_{BD}=\frac{15}{13}P$

Substitute $\frac{15}{13}P$ for $T_{BD}$ in equation (XVII) to get $T_{BE}$.

$T_{BE}=\frac{3}{13}P$

Substitute $455\text{N}$ for $P$ in above equation to get $T_{BD}$.

$\begin{array}{c}{T}_{BD}=\frac{15}{13}\left(455\text{N}\right)\\ =525\text{N}\end{array}$

Substitute $455\text{N}$ for $P$ in the above equation to find $T_{BE}$.

$\begin{array}{c}{T}_{BE}=\frac{3}{13}\left(455\text{N}\right)\\ =105\text{N}\end{array}$

Substitute $\frac{525}{3}$ for $T_{BD}$ , $\frac{105}{3}$ for $T_{BE}$ , $\frac{-455}{13}$ for $P$ , and $A_X$ for $A$ in equation (XIX) to get $A_x$.

$\begin{array}{l}\frac{525}{3}+\frac{105}{3}-\frac{455}{13}\left(3\right)+A_x=0\\ A_x=105\text{N}\end{array}$

Similarly substitute $\frac{-525}{3}$ for $T_{{}_{BD}}$ , $\frac{-105}{3}$ for $T_{BE}$ , $-\frac{455}{13}$ for $P$ , and $A_y$ in equation (XIX) to get $A_y$.

$\begin{array}{l}{A}_y=-\frac{525}{3}\left(2\right)-\frac{105}{3}\left(2\right)-\frac{455}{13}\left(12\right)+A_y=0\\ A_y=840\text{N}\end{array}$

Substitute $-\frac{525}{3}$ for $T_{BD}$ , $\frac{105}{3}$ for $T_{BE}$ , $\frac{455}{13}$ for $P$ , and $A_z$ for $A$ in equation (XIX) to get $A_z$.

$\begin{array}{l}\frac{-525}{3}\left(2\right)+\frac{105}{3}\left(2\right)+\frac{455}{13}\left(4\right)+A_z=0\\ A_z=140\text{N}\end{array}$

The vector form of the reaction force is $-\left(105\text{N}\right)\widehat{i}+\left(840\text{N}\right)\widehat{j}+\left(140\text{N}\right)\widehat{k}$

Therefore, the tension in cable BD is $\underset{\_}{525\text{N}}$ , tension in cable BE is $\underset{\_}{105\text{N}}$. Also the reaction at $\underset{\_}{\left(-105\text{N}\right)\widehat{i}+\left(840\text{N}\right)\widehat{j}+\left(140\text{N}\right)\widehat{k}}$.