Chapter 3.2, Problem 3.55P

To determine

Find the moment about AB $\left(M_{AB}\right)$.

Answer to Problem 3.55P

The moment about AB $\left(M_{AB}\right)$ is $\underset{\_}{207\text{lb}\cdot \text{ft}}$.

Explanation of Solution

Given information:

The length of the vertical rod CD $\left(l_{CD}\right)$ is 23 in.

The length of the rod AB $\left(l_{AB}\right)$ is 50 in.

The force (P) is 235 lb.

The distance between the points O to B (x) is 32 in.

The distance between the points O to A (y) is 24 in.

The vertical height of OA $\left(h_{OA}\right)$ is 30 in.

The vertical height of OH $\left(h_{OH}\right)$ is 17 in.

The horizontal distance between point O and midpoint C $\left(d_{OC}\right)$ is 16 in.

The horizontal distance between midpoint C and point G over x axis $\left(d_{CG}\right)$ is 21 in.

The horizontal distance between point B and midpoint C $\left(d_{BC}\right)$ is 12 in.

The horizontal distance between midpoint C and point G over z axis (h) is 18 in.

Calculation:

Calculate the position vector of AB using the relation:

$\stackrel{\rightharpoonup }{AB}=xi-h_{OA}j-yk$

Substitute 32 in. for x, 30 in. for $h_{OA}$ and 24 in. for y.

$\begin{array}{c}\stackrel{\rightharpoonup }{AB}=32i-30j-24k\\ AB=\sqrt{{\left(32\right)}^2+{\left(-30\right)}^2+{\left(-24\right)}^2}\\ =50\text{in}\text{.}\end{array}$

Calculate the unit vector of AB using the formula:

${\lambda }_{AB}=\frac{\stackrel{\rightharpoonup }{AB}}{AB}$

Substitute $32i-30j-24k$ for $\stackrel{\rightharpoonup }{AB}$ and 50 in. for AB.

$\begin{array}{c}{\lambda }_{AB}=\frac{32i-30j-24k}{50}\\ =0.64i-0.6j-0.48k\end{array}$

Apply the force P at the point G.

Calculate the position vector of from point G to point B $\left(r_{G/B}\right)$ using the relation:

$r_{G/B}=\left[x-\left(d_{OC}+d_{CG}\right)\right]i+\left(d_{BC}+h\right)k$

Substitute 32 in. for x, 16 in. for $d_{OC}$, 21 in. for $d_{CG}$, 12 in. for $d_{BC}$ and 18 in. for h.

$\begin{array}{c}{r}_{G/B}=\left[32-\left(16+21\right)\right]i+\left(12+18\right)k\\ =5i+30k\end{array}$

Calculate the position vector of DG using the relation:

$\stackrel{\rightharpoonup }{DG}=d_{CG}i-\left(\frac{h_{OA}}{2}+l_{CD}\right)j+hk$

Substitute 21 in. for $d_{CG}$, 30 in. for $h_{OA}$, 23 in. for $l_{CD}$ and 18 in. for h.

$\begin{array}{c}\stackrel{\rightharpoonup }{DG}=21i-\left(\frac{30}{2}+23\right)j+18k\\ =21i-38j+18k\\ DG=\sqrt{{\left(21\right)}^2+{\left(-38\right)}^2+{\left(18\right)}^2}\\ =47\text{in}\text{.}\end{array}$

Calculate the force $\left(P\right)$ using the formula:

$P=P\frac{\stackrel{\rightharpoonup }{DG}}{DG}$

Substitute 235 lb for P, $21i-38j+18k$ for $\stackrel{\rightharpoonup }{DG}$ and 47 in. for DG.

$\begin{array}{c}P=235\times \frac{21i-38j+18k}{47}\\ =235\times \left(0.447i-0.808j+0.383k\right)\\ =105i-189.9j+90k\\ \approx 105i-190j+90k\end{array}$

Calculate the moment about AB $\left(M_{AB}\right)$ using the relation:

$M_{AB}={\lambda }_{AB}\cdot \left(r_{G/B}\times P\right)$

Substitute $0.64i-0.6j-0.48k$ for ${\lambda }_{AB}$, $5i+30k$ for $r_{G/B}$ and $105i-190j+90k$ for P.

$\begin{array}{c}{M}_{AB}=\left|\begin{array}{ccc}0.64& -0.6& -0.48\\ 5& 0& 30\\ 105& -190& 90\end{array}\right|\\ =\left[0.64\left(0-\left(-5700\right)\right)\right]-\left[\left(-0.6\right)\left(450-3150\right)\right]-\left[0.48\left(-950-0\right)\right]\\ =3648-1620+456\end{array}$

$\begin{array}{c}=2484\text{lb}\cdot \text{in}\text{.}\times \frac{1}{12}\frac{\text{ft}}{\text{in}\text{.}}\\ =207\text{lb}\cdot \text{ft}\end{array}$

Therefore, the moment about AB $\left(M_{AB}\right)$ is $\underset{\_}{207\text{lb}\cdot \text{ft}}$.