Chapter 3.2, Problem 3.65P

To determine

The perpendicular distance (d) between rod AB and the line of action of Q.

Answer to Problem 3.65P

The perpendicular distance (d) between rod AB and the line of action of Q is $\underset{\_}{12.69\text{in}\text{.}}$.

Explanation of Solution

Given information:

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

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

The force (Q) is 174 lb.

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

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

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

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

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

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

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

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

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.

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

Calculate the unit vector of AB $\left({\lambda }_{AB}\right)$ 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 Q at the point H.

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

$r_{H/B}=-xi+h_{OH}j$

Substitute 32 in. for x and 17 in. for $h_{OH}$.

$r_{H/B}=-32i+17j$

Calculate the position vector of DH using the relation:

$\stackrel{\rightharpoonup }{DH}=-d_{OC}i-\left[\left(\frac{h_{OA}}{2}+l_{CD}\right)-h_{OH}\right]j-d_{BC}k$

Substitute 16 in. for $d_{OC}$, 30 in. for $h_{OA}$, 23 in. for $l_{CD}$, 17 in. for $h_{OH}$, and 12 in. for $d_{BC}$.

$\begin{array}{l}\stackrel{\rightharpoonup }{DH}=-16i-\left[\left(\frac{30}{2}+23\right)-17\right]j-12k\\ \stackrel{\rightharpoonup }{DH}=-16i-21j-12k\\ DH=\sqrt{{\left(-16\right)}^2+{\left(-21\right)}^2+{\left(-12\right)}^2}\\ DH=29\text{in}\text{.}\end{array}$

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

$Q=Q\frac{\stackrel{\rightharpoonup }{DH}}{DH}$

Substitute 174 lb for P, $-16i-21j-12k$ for $\stackrel{\rightharpoonup }{DH}$, and 29 in. for DH.

$\begin{array}{c}Q=174\times \frac{-16i-21j-12k}{29}\\ =174\times \left(-0.552i-0.724j-0.414k\right)\\ =-96i-125.9j-72k\\ \approx -96i-126j-72k\end{array}$

Calculate the position vector at Q $\left({\lambda }_Q\right)$ using the relation:

${\lambda }_Q=\frac{Q}{Q}$

Substitute $-96i-126j-72k$ for Q and 174 lb for Q.

$\begin{array}{c}{\lambda }_Q=\frac{-96i-126j-72k}{174}\\ =-0.552i-0.724j-0.414k\end{array}$

Calculate the angle between AB and Q $\left(\theta \right)$ using the formula:

$\cos \theta ={\lambda }_{AB}\cdot {\lambda }_Q$

Substitute $0.64i-0.6j-0.48k$ for ${\lambda }_{AB}$ and $-0.552i-0.724j-0.414k$ for ${\lambda }_Q$.

$\begin{array}{l}\cos \theta =\left(0.64i-0.6j-0.48k\right)\cdot \left(-0.552i-0.724j-0.414k\right)\\ \cos \theta =-0.35328+0.4344+0.19872\\ \cos \theta =0.28\\ \theta ={\cos }^{-1}\left(0.28\right)\\ \theta =73.74°\end{array}$

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

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

Substitute $0.64i-0.6j-0.48k$ for ${\lambda }_{AB}$, $-32i+17j$ for $r_{H/B}$, and $-96i-126j-72k$ for Q.

$\begin{array}{c}{M}_{AB}=\left|\begin{array}{ccc}0.64& -0.6& -0.48\\ -32& 17& 0\\ -96& -126& -72\end{array}\right|\\ =\left[0.64\left(-1,224-0\right)\right]-\left[\left(-0.6\right)\left(2,304-0\right)\right]-\left[0.48\left(4,032-\left(-1,632\right)\right)\right]\\ =-783.36+1,382.4-2,718.72\\ =-2,119.7\text{lb}\cdot \text{in}\text{.}\times \frac{1}{12}\frac{\text{ft}}{\text{in}\text{.}}\\ =-176.6\text{lb}\cdot \text{ft}\end{array}$

Calculate the perpendicular distance (d) between rod AB and the line of action of Q:

Take moment about AB:

$M_{AB}=\left(Q\sin \theta \right)d$

Substitute 174 lb for Q and $73.740°$ for $\theta$, and $-176.6\text{lb}\cdot \text{ft}$ for $M_{AB}$.

$\begin{array}{l}-176.6=\left(174\sin 73.740\right)d\\ d=\frac{-176.6}{174\sin 73.740}\\ d=1.0572\text{ft}\times \text{12}\frac{\text{in}\text{.}}{\text{ft}}\\ d=12.69\text{in}\text{.}\end{array}$

Thus, the perpendicular distance (d) between rod AB and the line of action of Q is $\underset{\_}{12.69\text{in}\text{.}}$.