Chapter 4.3, Problem 4.105P

PROBLEM 4.105

A 10-ft boom is acted upon by the 840-lb force shown. Determine the tension in each cable and the reaction at the ball-and-socket joint at A.

To determine

The tension in each cable and the reaction at the ball and socket joint at A.

Answer to Problem 4.105P

The tension in cable BD is $\underset{\_}{1100\text{lb}}$ , cable BE is $\underset{\_}{1100\text{lb}}$ , and the reaction in joint A is $\underset{\_}{\left(1200\text{lb}\right)\widehat{i}-\left(560\text{lb}\right)\widehat{j}}$.

Explanation of Solution

The free body diagram corresponding to the system of cable is shown in figure 1 plotted below.

A down ward force of magnitude $840\text{lb}$ acts at point C. The resultant tension experienced in cables are to be found in the problem.

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

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

Here, $\overrightarrow{BD}$ is the position vector connecting point B and point D, $x$ is the x coordinate, y is the y coordinate, and z is the z coordinate.

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

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

Here, $\overrightarrow{BE}$ is the position vector connecting point B and point E, $x$ is the x coordinate , y is the y coordinate, and z is the z coordinate.

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

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

Here, $BD$ is the magnitude of the vector, x is the x coordinate, y is the y coordinate, and z is the z coordinate.

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

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

Here, $BE$ is the magnitude of the vector, $x$ is the x coordinate, $y$ is the y coordinate, $z$ is the z coordinate.

$T_{BD}$ is the tension in the cable BD.

Write the equation to find the tension along cable BD.

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

Here, $T_{BD}$ is the tension vector in the direction of unit vector, $T_{BD}$ is the magnitude of the tension, $\overrightarrow{BD}$ is the position vector, and $BD$ is the magnitude of the position vector.

Write the equation to find the tension along cable BE.

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

Here, $T_{BE}$ is the tension vector along the direction of unit vector, $T_{BE}$ is the magnitude of the tension, $\overrightarrow{BE}$ is the position vector connecting point B and E, $BE$ is the magnitude of the tension.

Write the equation to find the moments at point B due to force at D.

$M_{BD}=r_B\times T_{BD}$ (VII)

Here, $M_{BD}$ is the moment at B due to force at D, $r_{BD}$ is the perpendicular distance between the line of action between forces at B and D, $T_{BD}$ is the tension in the wire.

Write the equation to find the moment at point B due to force at point E.

$M_{BE}=r_B\times T_{BE}$ (VIII)

Here, $M_{BE}$ is the moment of force at point B due to force at point E, $r_B$ is the perpendicular distance between the line of action of force at points B and E, $T_{BE}$ is the tension at cable BE.

Write the equation to find the moment of force at point B due to force at point C.

$M_{BC}=r_C\times T_{BC}$ (IX)

Here, $M_{BC}$ is the moment of force at point B due to force at point C, $r_C$ is the perpendicular distance between the point B and C, $T_{BC}$ is the tension at cable BC.

Write the equation to find the sum of moment of force at point B.

$M_B=M_{BD}+M_{BE}+M_{BC}$ (X)

Here, $M_B$ is the sum of moments at point B, $M_{BD}$ is the moments about cable BD, $M_{BE}$ is the moment about cable BE, and $M_{BC}$ is the moment about cable BE.

Substitute equation (VII), (VIII), and (IX) in equation (X) to find the $M_B$.

$M_B=r_B\times T_{BD}+r_B\times T_{BE}+r_C\times T_{BC}$ (XI)

Substitute $\frac{\overrightarrow{BD}}{BD}$ for $T_{BD}$ , $\frac{\overrightarrow{BE}}{BE}$ for $T_{BE}$ , and $-840$ for $T_{CB}$ in equation (XI) to get $M_B$.

$M_B=r_B\times T_{BD}\left(\frac{\overrightarrow{BD}}{BD}\right)+r_B\times T_{BE}\left(\frac{\overrightarrow{BE}}{BE}\right)+r_C\times T_{BC}\left(-840\right)$ (XII)

Since sum of all external force acting on a body is zero, equate the equation for moments to zero.

$r_B\times T_{BD}\left(\frac{\overrightarrow{BD}}{BD}\right)+r_B\times T_{BE}\left(\frac{\overrightarrow{BE}}{BE}\right)+r_C\times T_{BC}\left(-840\right)=0$ (XIII)

Write the equation to find the sum of x components of force.

$\begin{array}{l}\sum F_x=0\\ A_x-\frac{x}{BD}\left(T_{BD}\right)-\frac{x}{BE}\left(T_{BE}\right)=0\end{array}$ (XIV)

Write the equation to find the sum of y components of force.

$\begin{array}{l}\sum F_y=0\\ A_y+\frac{y}{BD}\left(T_{BD}\right)+\frac{y}{BE}\left(T_{BE}\right)-T_{CB}=0\end{array}$ (XV)

Write the equation to find the sum of z component of forces.

$\begin{array}{l}\sum F_z=0\\ A_z+\frac{z}{BD}\left(T_{BD}\right)-\frac{z}{BE}\left(T_{BE}\right)=0\end{array}$ (XVI)

Conclusion:

Substitute $-6\text{ft}$ for $x$ , $7\text{ft}$ for $y$ , and $6\text{ft}$ for $z$ in equation (I) to get $\overrightarrow{BD}$.

$\overrightarrow{BD}=\left(-6\text{ft}\right)\widehat{i}+\left(7\text{ft}\right)\widehat{j}+\left(6\text{ft}\right)\widehat{k}$

Substitute $-6\text{ft}$ for $x$ , $7\text{ft}$ for $y$ , and $-6\text{ft}$ for $z$ in equation (II) to get $\overrightarrow{BE}$.

$\overrightarrow{BE}=\left(-6\text{ft}\right)\widehat{i}+\left(7\text{ft}\right)\widehat{j}-\left(6\text{ft}\right)\widehat{k}$

Substitute $-6\text{ft}$ for $x$ , $7\text{ft}$ for $y$ , and $6\text{ft}$ for $z$ in equation (III) to get $BD$.

$\begin{array}{c}BD=\sqrt{{\left(-6\text{ft}\right)}^2+{\left(7\text{ft}\right)}^2+{\left(6\text{ft}\right)}^2}\\ =11\text{ft}\end{array}$

Substitute $-6\text{ft}$ for $x$ , $7\text{ft}$ for $y$ , and $-6\text{ft}$ for $z$ in equation (IV) to get $BE$.

$\begin{array}{c}BE=\sqrt{{\left(-6\text{ft}\right)}^2+{\left(7\text{ft}\right)}^2-{\left(6\text{ft}\right)}^2}\\ =11\text{ft}\end{array}$

Substitute $11\text{ft}$ for $BD$ , $\left(-6\text{ft}\right)\widehat{i}+\left(7\text{ft}\right)\widehat{j}+\left(6\text{ft}\right)\widehat{k}$ for $\overrightarrow{BD}$ , $\left(-6\text{ft}\right)\widehat{i}+\left(7\text{ft}\right)\widehat{j}+\left(6\text{ft}\right)\widehat{k}$ for $\overrightarrow{BE}$ , $6\widehat{i}$ for $r_B$ , $6\widehat{i}$ for $r_B$ , and $10\widehat{i}$ for $r_c$ in equation (XIII) to get the value of tensions.

$\begin{array}{l}6\widehat{i}\times \frac{T_{BD}}{11}\left(-6\widehat{i}+7\widehat{j}+6\widehat{k}\right)+6\widehat{i}\times \frac{T_{BE}}{11}\left(-6\widehat{i}+7\widehat{j}-6\widehat{k}\right)+10\widehat{i}\times \left(-840\widehat{j}\right)=0\\ \frac{42}{11}{T}_{BD}\widehat{k}-\frac{36}{11}{T}_{BD}\widehat{j}+\frac{42}{11}{T}_{BE}\widehat{k}+\frac{36}{11}{T}_{BE}\widehat{j}-8400\widehat{k}=0\end{array}$

Equate the coefficients of unit vectors $\widehat{i}$ , $\widehat{j}$ , and $\widehat{k}$ to zero.

Equate the coefficient of unit vector $\widehat{i}$.

$\begin{array}{l}\frac{-36}{11}{T}_{BD}+\frac{36}{11}{T}_{BE}=0\\ T_{BE}=T_{BD}\end{array}$

Equate the coefficients of unit vector $\widehat{k}$.

$\begin{array}{l}\frac{42}{11}{T}_{BD}+\frac{42}{11}{T}_{BE}-8400=0\\ \therefore 2\left(\frac{42}{11}{T}_{BD}\right)=8400\\ \therefore T_{BD}=T_{BE}=1100\text{lb}\\ \end{array}$

Substitute $6$ for $x$ , $11$ for $BD$ , $1100\text{lb}$ for $T_{BD}$ , $11$ for $BE$ , and $1100\text{lb}$ for $T_{BE}$ in equation (XIV) to get $A_x$.

$\begin{array}{l}{A}_x-\frac{6}{11}\left(1100\text{lb}\right)-\frac{6}{11}\left(1100\text{lb}\right)=0\\ \therefore A_x=0\end{array}$

Substitute $7$ for $y$ , $11$ for $BD$ , $1100\text{lb}$ for $T_{BD}$ , $11$ for $BE$ , $840\text{lb}$ for $T_{CB}$ and $1100\text{lb}$ for $T_{BE}$, in equation (XV) to get $A_y$.

$\begin{array}{l}{A}_y+\frac{7}{11}\left(1100\text{lb}\right)+\frac{7}{11}\left(1100\text{lb}\right)-840\text{lb}=0\\ \therefore A_y=-560\text{lb}\end{array}$

Substitute $6$ for $x$ , $11$ for $BD$ , $1100\text{lb}$ for $T_{BD}$ , $11$ for $BE$ , and $1100\text{lb}$ for $T_{BE}$ in equation (XVI) to get $A_z$.

$\begin{array}{l}{A}_z+\frac{6}{11}\left(1100\text{lb}\right)-\frac{6}{11}\left(1100\text{lb}\right)=0\\ \therefore A_z=0\end{array}$

Therefore, the tension in cable BD is $\underset{\_}{1,100\text{lb}}$ , cable BE is $\underset{\_}{1,100\text{lb}}$ , and the reaction in joint A is $\underset{\_}{\left(1200\text{lb}\right)\widehat{i}-\left(560\text{lb}\right)\widehat{j}}$.