Chapter 3.4, Problem 3.104P

Five separate force-couple systems act at the corners of a piece of sheet metal that has been bent into the shape shown. Determine which of these systems is equivalent to a force F = (10 lb)i and a couple of moment M = (15 lb·ft)j + (15 lb·ft)k located at the origin.

To determine

Identify the system having force equivalent $F=\left(10lb\right)i$ and moment equivalent $M=\left(15\text{lb}\cdot \text{ft}\right)j+\left(\left(15\text{lb}\cdot \text{ft}\right)\right)k$ at origin.

Answer to Problem 3.104P

System at corner D is the equivalent force-couple system.

Explanation of Solution

Take counterclockwise torques as positive quantities and clockwise torques as negative quantities.

Refer Figure P3.104.

There are five systems at corners A,D,I,F, and Among these, system at corner F cannot be equivalent. This is because that it does not have force in x-direction. All other systems have force $F=\left(10lb\right)i$. Now, to identify the equivalent system, move all other systems to origin O. This will noT change the forces.

Write the expression to calculate the force part of force couple system.

$R=\sum F$

Here, $R$ is the vector sum of all forces acting on the beam and $F$ denotes any individual force.

Let check the couple part of force-couple system.

Write the expression to calculate the couple part of force couple system.

$M^R=\sum \left(r\times F\right)$

Here, $M^R$ is the resultant moment and $r$ is the length from axis of rotation to the point of application of force.

In the given system, at points A,D,G and I, moments in two directions are directly given. So the expression to find $\sum \left(r\times F\right)$ is as follows.

$\sum \left(r\times F\right)=M_1+M_2+\left(r\times F\right)$

Here, $M_1\text{and}{M}_2$ are the given couple moments

Rewrite the equation for $M^R$ by substituting the above expression for $\sum \left(r\times F\right)$.

$M^R=M_1+M_2+\left(r\times F\right)$ (I)

Write the determinant form to calculate $r\times F$.

$r\times F=\left|\begin{array}{ccc}i& j& k\\ r_i& r_j^{}& r_k\\ F_i& F_j& F_k\end{array}\right|$ (II)

Here, $r_i$ is the component of $r$ in x-direction, $r_j^{}$ is the component of $r$ in y-direction, $r_k$ is the is the component of $r$ in z-direction, $F_i$ is the component of $F$ in x-direction, $F_j$ is the component of $F$ in y-direction, and $F_k$ is the component of $F$ in z-direction.

Conclusion:

Substitute $\left(5\text{lb}\cdot \text{ft}\right)j$ for $M_1$, $\left(15\text{lb}\cdot \text{ft}\right)k$ for $M_2$, $\left(\text{2}\text{ft}\right)k$ for $r$  and $\left(10\text{lb}\right)i$ for $F$ in equation  (I) to find  moment resultant of system at corner A about point O $\left(M_A\right)$ .

$M_A=\left(5\text{lb}\cdot \text{ft}\right)j+\left(15\text{lb}\cdot \text{ft}\right)k+\left(\left(\text{2}\text{ft}\right)k\times \left(10\text{lb}\right)i\right)$ (III)

Solve $\left(\text{2}\text{ft}\right)k\times \left(10\text{lb}\right)i$ using equation (II).

$\begin{array}{c}\left(\text{2}\text{ft}\right)k\times \left(10\text{lb}\right)i=\left|\begin{array}{ccc}i& j& k\\ 0& 0& \text{2}\text{ft}\\ 10\text{lb}& 0& 0\end{array}\right|\\ =\left(0\text{lb}\cdot \text{ft}\right)i-\left(0\text{lb}\cdot \text{ft}-20\text{lb}\cdot \text{ft}\right)j+\left(0\text{lb}\cdot \text{f}\right)k\\ =\left(20\text{lb}\cdot \text{ft}\right)j\end{array}$

Calculate $M_A$ by substituting $\left(\text{2}\text{ft}\right)k\times \left(10\text{lb}\right)i$ for $\left(-20\text{lb}\cdot \text{ft}\right)j$ in equation (III).

$\begin{array}{c}{M}_A=\left(5\text{lb}\cdot \text{ft}\right)j+\left(15\text{lb}\cdot \text{ft}\right)k+\left(20\text{lb}\cdot \text{ft}\right)j\\ =\left(25\text{lb}\cdot \text{ft}\right)j+\left(15\text{lb}\cdot \text{ft}\right)k\end{array}$

Substitute  $\left(-5\text{lb}\cdot \text{ft}\right)j$ for $M_1$, $\left(25\text{lb}\cdot \text{ft}\right)k$ for $M_2$, $\left[\left(4.5\text{ft}\right)i+\left(1\text{ft}\right)j+\left(\text{2}\text{ft}\right)k\right]$$\left(\text{2}\text{ft}\right)k$ for $r$  and $\left(10\text{lb}\right)i$ for $F$ in equation  (I) to find moment resultant of system at corner D about point O $\left(M_D\right)$.

$M_D=\left(-5\text{lb}\cdot \text{ft}\right)j+\left(25\text{lb}\cdot \text{ft}\right)k+\left[\left(4.5\text{ft}\right)i+\left(1\text{ft}\right)j+\left(\text{2}\text{ft}\right)k\right]\times \left(10\text{lb}\right)i$(IV)

Solve $\left[\left(4.5\text{ft}\right)i+\left(1\text{ft}\right)j+\left(\text{2}\text{ft}\right)k\right]\times \left(10\text{lb}\right)i$ using equation (II).

$\begin{array}{c}\left[\left(4.5\text{ft}\right)i+\left(1\text{ft}\right)j+\left(\text{2}\text{ft}\right)k\right]\times \left(10\text{lb}\right)i=\left|\begin{array}{ccc}i& j& k\\ 4.5\text{ft}& 1\text{ft}& \text{2}\text{ft}\\ 10\text{lb}& 0& 0\end{array}\right|\\ =\left(0\text{lb}\cdot \text{f}\right)i-\left(0\text{lb}\cdot \text{ft}-20\text{lb}\cdot \text{ft}\right)j+\left(0\text{lb}\cdot \text{ft}-10\text{lb}\cdot \text{ft}\right)k\\ =\left(20\text{lb}\cdot \text{ft}\right)j-\left(10\text{lb}\cdot \text{ft}\right)k\end{array}$

Calculate $M_D$ by substituting $\left(20\text{lb}\cdot \text{ft}\right)j-\left(10\text{lb}\cdot \text{ft}\right)k$$\left(\text{2}\text{ft}\right)k\times \left(10\text{lb}\right)i$ for $\left[\left(4.5\text{ft}\right)i+\left(1\text{ft}\right)j+\left(\text{2}\text{ft}\right)k\right]\times \left(10\text{lb}\right)i$$\left(-20\text{lb}\cdot \text{ft}\right)j$ in equation (IV).

$\begin{array}{c}{M}_D=\left(-5\text{lb}\cdot \text{ft}\right)j+\left(25\text{lb}\cdot \text{ft}\right)k+\left(20\text{lb}\cdot \text{ft}\right)j-\left(10\text{lb}\cdot \text{ft}\right)k\\ =\left(15\text{lb}\cdot \text{ft}\right)j+\left(15\text{lb}\cdot \text{ft}\right)k\end{array}$

Substitute  $\left(15\text{lb}\cdot \text{ft}\right)i$ for $M_1$, $\left(15\text{lb}\cdot \text{ft}\right)j$ for $M_2$, $\left[\left(0\text{ft}\right)i+\left(0\text{ft}\right)j+\left(0\text{ft}\right)k\right]$ for $r$  and $\left(10\text{lb}\right)i$ for $F$ in equation  (I) to find moment resultant of system at corner G about point O $\left(M_G\right)$.

$\begin{array}{c}{M}_G=\left(15\text{lb}\cdot \text{ft}\right)i+\left(15\text{lb}\cdot \text{ft}\right)j+\left[\left(0\text{ft}\right)i+\left(0\text{ft}\right)j+\left(0\text{ft}\right)k\right]\times \left(10\text{lb}\right)i\\ =\left(15\text{lb}\cdot \text{ft}\right)i+\left(15\text{lb}\cdot \text{ft}\right)j+\left(0\text{ft}\right)i+\left(0\text{ft}\right)j+\left(0\text{ft}\right)k\\ =\left(15\text{lb}\cdot \text{ft}\right)i+\left(15\text{lb}\cdot \text{ft}\right)j\end{array}$

Substitute  $\left(15\text{lb}\cdot \text{ft}\right)j$ for $M_1$, $\left(-5\text{lb}\cdot \text{ft}\right)k$ for $M_2$, $\left[\left(4.5\text{ft}\right)i+\left(1\text{ft}\right)j\right]$ for $r$  and $\left(10\text{lb}\right)j$ for $F$ in equation  (I) to find moment resultant of system at corner I about point O $\left(M_I\right)$.

$M_D=\left(15\text{lb}\cdot \text{ft}\right)j+\left(-5\text{lb}\cdot \text{ft}\right)k+\left(\left[\left(4.5\text{ft}\right)i+\left(1\text{ft}\right)j\right]\times \left(10\text{lb}\right)j\right)$(V)

Solve $\left[\left(4.5\text{ft}\right)i+\left(1\text{ft}\right)j\right]\times \left(10\text{lb}\right)j$ using equation (III).

$\begin{array}{c}\left[\left(4.5\text{ft}\right)i+\left(1\text{ft}\right)j\right]\times \left(10\text{lb}\right)j=\left|\begin{array}{ccc}i& j& k\\ \left(4.5\text{ft}\right)& \left(1\text{ft}\right)& \left(\text{0}\text{ft}\right)\\ \left(10\text{lb}\right)& 0& 0\end{array}\right|\\ =\left(0\text{lb}\cdot \text{f}\right)i-\left(0\text{lb}\cdot \text{ft}\right)j+\left(0\text{lb}\cdot \text{ft}-10\text{lb}\cdot \text{ft}\right)k\\ \left(-10\text{lb}\cdot \text{ft}\right)k\end{array}$

Calculate $M_D$ by substituting $\left(-10\text{lb}\cdot \text{ft}\right)k$ for $\left[\left(4.5\text{ft}\right)i+\left(1\text{ft}\right)j\right]\times \left(10\text{lb}\right)j$$\left(-20\text{lb}\cdot \text{ft}\right)j$ in equation (V).

$\begin{array}{c}{M}_I=\left(15\text{lb}\cdot \text{ft}\right)j-\left(5\text{lb}\cdot \text{ft}\right)k+\left(-10\text{lb}\cdot \text{ft}\right)k\\ =\left(15\text{lb}\cdot \text{ft}\right)j-\left(15\text{lb}\cdot \text{ft}\right)k\end{array}$

From the above calculations, it is found that system at corner D has force equivalent $F=\left(10lb\right)i$ and moment equivalent $M=\left(15\text{lb}\cdot \text{ft}\right)j+\left(\left(15\text{lb}\cdot \text{ft}\right)\right)k$ at origin.

Therefore, System at corner D is the equivalent force-couple system.