Chapter 3.1, Problem 3.24P

(a)

To determine

Find the moment about O $\left(M_O\right)$ of the force exerted by the wire on corner A.

Answer to Problem 3.24P

The moment about O $\left(M_O\right)$ of the force exerted by the wire on corner A is $\underset{\_}{28.8i+16.2j-28.8k}$.

Explanation of Solution

Given information:

The tension in the wire (F) is 435 N.

The distance between corner D to corner E (a) is 120 mm.

The height of corners CA (h) is 160 mm.

The distance between corner A to origin O (b) is 90 mm.

The distance between corner B to origin O (c) is 120 mm.

Calculation:

Calculate the distance between corner A and corner B (B) using the relation:

$B=b+c$

Substitute 90 mm for b and 120 mm for c.

$\begin{array}{c}B=90+120\\ =210\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}\\ =0.21\text{m}\end{array}$

Calculate the length between corner A to corner E using the relation:

$\stackrel{\rightharpoonup }{AE}=Bi-hj+ak$

Substitute 0.21 m for B, 160 mm for h and 120 mm for a.

$\begin{array}{c}\stackrel{\rightharpoonup }{AE}=0.21i-\left(160\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}\right)j+\left(120\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}\right)k\\ =0.21i-0.16j+0.12k\\ AE=\sqrt{{\left(0.21\right)}^2+{\left(-0.16\right)}^2+{\left(0.12\right)}^2}\\ =0.29\text{m}\end{array}$

Calculate the unit vector $\left({\lambda }_{AE}\right)$ of the corner A to corner E using the formula:

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

Substitute $0.21i-0.16j+0.12k$ for $\stackrel{\rightharpoonup }{AE}$ and 0.29 m for AE.

$\begin{array}{c}{\lambda }_{AE}=\frac{0.21i-0.16j+0.12k}{0.29}\\ =0.7241i-0.5517j+0.4138k\end{array}$

Calculate the force at corner A $\left(F_A\right)$ using the formula:

$F_A=F{\lambda }_{AE}$

Substitute 435 N for F and $0.7241i-0.5517j+0.4138k$ for ${\lambda }_{AE}$.

$\begin{array}{c}{F}_A=435\left(0.7241i-0.5517j+0.4138k\right)\\ =314.9i-239.99j+180k\\ \approx 315i-240j+180k\end{array}$

The expression for the position vector from corner A to corner O $\left(r_{A/O}\right)$ as follows;

$r_{A/O}=-bi+hj$

Substitute 90 mm for b and 160 mm for h.

$\begin{array}{c}{r}_{A/O}=-\left(90\text{mm}\times \frac{\text{1m}}{\text{1000mm}}\right)i+\left(160\text{mm}\times \frac{\text{1m}}{\text{1000mm}}\right)j\\ =-0.09i+0.16j\end{array}$

Calculate the moment about origin O $\left(M_O\right)$ using the relation:

$M_O=r_{A/O}\times F_A$

Substitute $-0.09i+0.16j$ for $r_{A/O}$ and $315i-240j+180k$ for $F_A$.

$\begin{array}{c}{M}_O=\left|\begin{array}{ccc}i& j& k\\ -0.09& 0.16& 0\\ 315& -240& 180\end{array}\right|\\ =\left[28.8-0\right]i-\left[-16.2-0\right]j+\left[21.6-50.4\right]k\\ =28.8i+16.2j-28.8k\end{array}$

Therefore, the moment about O $\left(M_O\right)$ of the force exerted by the wire on corner A is $\underset{\_}{28.8i+16.2j-28.8k}$.

(b)

To determine

Find the moment about O $\left(M_O\right)$ of the force exerted by the wire on corner E.

Answer to Problem 3.24P

The moment about O $\left(M_O\right)$ of the force exerted by the wire on corner E is $\underset{\_}{-28.8i-16.2j+28.8k}$.

Explanation of Solution

Given information:

The tension in the wire (F) is 435 N.

Calculation:

Calculate the force at corner E $\left(F_E\right)$ using the formula:

$F_E=-F_A$

Substitute $315i-240j+180k$ for $F_A$.

$\begin{array}{c}{F}_E=-\left(315i-240j+180k\right)\\ =-315i+240j-180k\end{array}$

The expression for the position vector from corner E to corner O $\left(r_{E/O}\right)$ as follows;

$r_{E/O}=ci+ak$

Substitute 120 mm for c and 120 mm for b.

$\begin{array}{c}{r}_{E/O}=\left(\text{120mm}\times \frac{\text{1m}}{\text{1000mm}}\right)i+\left(120\text{mm}\times \frac{\text{1m}}{\text{1000mm}}\right)k\\ =0.12i+0.12k\end{array}$

Calculate the moment about origin O $\left(M_O\right)$ using the relation:

$M_O=r_{E/O}\times F_E$

Substitute $0.12i+0.12k$ for $r_{E/O}$ and $-315i+240j-180k$ for $F_E$.

$\begin{array}{c}{M}_O=\left|\begin{array}{ccc}i& j& k\\ 0.12& 0& 0.12\\ -315& 240& -180\end{array}\right|\\ =\left[0-28.8\right]i-\left[-21.6-\left(-37.8\right)\right]j+\left[28.8-0\right]k\\ =-28.8i-16.2j+28.8k\end{array}$

Therefore, the moment about O $\left(M_O\right)$ of the force exerted by the wire on corner E is $\underset{\_}{-28.8i-16.2j+28.8k}$.