Chapter 3.3, Problem 3.79P

Solve part a of Prob. 3.78, assuming that two 15-N vertical forces have been added, one acting upward at A and the other downward at C.

To determine

The net couple acting on the rectangular co-ordinates.

Answer to Problem 3.79P

The net couple acting on the rectangular co-ordinates is $8.78\text{N}\cdot \text{m}$.

Explanation of Solution

Write the equation of the couple acting at the reactangular side $B$ and $C$ (Refer fig P3.78).

$M_1=Fd$

Here, moment of couple at the point $B$ and $C$ is $M_1$, the force acting couple at the point $B$ and $C$ is $F$, and the perpendicular distance from the point $B$ and $C$ to the line of action of force is $d$.

Substitute $25\text{N}$ for $F$ and $0.75\text{m}$ for $d$.

$\begin{array}{c}{M}_1=\left(25\text{N}\right)\left(0.75\text{m}\right)\\ =18.75\text{N}\cdot \text{m}\end{array}$

Write the equation of the couple acting at the reactangular side $D$ and $E$ (Refer fig P3.78).

$M_2=Fd$

Here, moment of couple at the point $D$ and $E$ is $M_2$, the force acting couple at the point $D$ and $E$ is $F$, and the perpendicular distance from the point $D$ and $E$ to the line of action of force is $d$.

Substitute $20\text{N}$ for $F$ and $0.4\text{m}$ for $d$.

$\begin{array}{c}{M}_2=\left(20\text{N}\right)\left(0.4\text{m}\right)\\ =8.0\text{N}\cdot \text{m}\end{array}$

Write the equation of the couple vector acting at the reactangular co-ordinates at point $B$ and $C$ (Refer fig P3.78).

${\overrightarrow{M}}_1=\left(M_1\sin \theta \right)\widehat{i}+\left(M_1\cos \theta \right)\widehat{j}$

Here, the projection angle is $\theta$ represented in x and y axis.

Substitute $18.75\text{N}\cdot \text{m}$ for $M_1$ and $40°$ for $\theta$.

$\begin{array}{c}{\overrightarrow{M}}_1=\left[\left(18.75\text{N}\cdot \text{m}\right)\sin 40°\right]\widehat{i}+\left[\left(18.75\text{N}\cdot \text{m}\right)\cos 40°\right]\widehat{j}\\ =\left(12.0523\text{N}\cdot \text{m}\right)\widehat{i}+\left(14.3633\text{N}\cdot \text{m}\right)\widehat{j}\end{array}$ (I)

Write the equation of the couple vector acting at the reactangular co-ordinates at point $D$ and $E$ (Refer fig P3.78).

${\overrightarrow{M}}_2=-\left(M_2\cos \theta \right)\widehat{j}$

Substitute $8.0\text{N}\cdot \text{m}$ for $M_2$ and $0°$ for $\theta$.

$\begin{array}{c}{\overrightarrow{M}}_2=-\left[\left(8.0\text{N}\cdot \text{m}\right)\cos 0°\right]\widehat{j}\\ =-\left(8.0\text{N}\cdot \text{m}\right)\widehat{j}\end{array}$ (II)

Sketch the moment of additional couple is added to the moment about $A$ of the force applied at $C$ is shown in the Figure 1.

Write the equation of the additional couple vector acting at the point $A$ of the force applied at $C$.

${\overrightarrow{M}}_3=-{\overrightarrow{F}}_y\times \left(d_x-d_z\right)$

Here, the additional couple vector formed by the force at the point $A$ and $C$ is ${\overrightarrow{M}}_3$, force at the point $A$ and $C$ is ${\overrightarrow{F}}_y$, perpendicular distance at the point $A$ to the line of action of force is $d_x$, and perpendicular distance at the point $C$ to the line of action of force is $d_z$.

The negative sign of the force shows that the force acting downward direction.

Substitute $15\text{N}$ for ${\overrightarrow{F}}_y$, $0.4\text{m}$ for $d_x$, and $0.75\text{m}$ for $d_z$.

$\begin{array}{c}{\overrightarrow{M}}_3=-\left(15\text{N}\right)\widehat{j}\times \left[\left(0.4\text{m}\right)\widehat{i}-\left(0.75\text{m}\right)\widehat{k}\right]\\ =-\left(11.25\text{N}\cdot \text{m}\right)\widehat{i}-\left(6\text{N}\cdot \text{m}\right)\widehat{k}\end{array}$ (III)

Conclusion:

Write the equation for the net couple acting on the rectangular co-ordinates.

$\overrightarrow{M}={\overrightarrow{M}}_1+{\overrightarrow{M}}_2+{\overrightarrow{M}}_3$

Adding the equations (I), (II), and (III) to obtained the net couple.

$\begin{array}{c}\overrightarrow{M}=\left(12.0523\text{N}\cdot \text{m}\right)\widehat{i}+\left(14.3633\text{N}\cdot \text{m}\right)\widehat{j}-\left(8.0\text{N}\cdot \text{m}\right)\widehat{j}-\left(11.25\text{N}\cdot \text{m}\right)\widehat{i}-\left(6\text{N}\cdot \text{m}\right)\widehat{k}\\ =\left(0.80\text{N}\cdot \text{m}\right)\widehat{i}+\left(6.36\text{N}\cdot \text{m}\right)\widehat{j}-\left(6\text{N}\cdot \text{m}\right)\widehat{k}\end{array}$

Write the equation for the magnitude of net couple.

$\begin{array}{c}\left|\overrightarrow{M}\right|=\sqrt{{\left(0.80\text{N}\cdot \text{m}\right)}^2+{\left(6.36\text{N}\cdot \text{m}\right)}^2+{\left(-6\text{N}\cdot \text{m}\right)}^2}\\ M=8.780\text{N}\cdot \text{m}\\ \simeq 8.78\text{N}\cdot \text{m}\end{array}$

Therefore, the net couple acting on the rectangular co-ordinates is $8.78\text{N}\cdot \text{m}$.

To determine

The direction of the two forces acting on the point $B$ and $C$ that can be used to form couple.

Answer to Problem 3.79P

The direction of the two forces acting on the point $B$ and $C$ is

${\theta }_x=84.8°,{\theta }_y=43.6°,{\theta }_z=133.1°$.

Explanation of Solution

Write the equation for direction of the moment of couple formed by the forces at point $B$ and $C$ along x axis.

$\cos {\theta }_x=\frac{M_x}{M}$

Here, the angle formed by the moment of couple due to the force located at the point $B$ and $C$ along x axis is ${\theta }_x$ and moment of couple formed along x axis is $M_x$.

Rewrite the equation for ${\theta }_x$.

${\theta }_x={\cos }^{-1}\left(\frac{M_x}{M}\right)$ (IV)

Write the equation for direction of the moment of couple formed by the forces at point $B$ and $C$ along y axis.

$\cos {\theta }_y=\frac{M_y}{M}$

Here, the angle formed by the moment of couple due to the force located at the point $B$ and $C$ along y axis is ${\theta }_y$ and moment of couple formed along y axis is $M_y$.

Rewrite the equation for ${\theta }_y$.

${\theta }_y={\cos }^{-1}\left(\frac{M_y}{M}\right)$ (V)

Write the equation for direction of the moment of couple formed by the forces at point $B$ and $C$ along z axis.

$\cos {\theta }_z=\frac{M_z}{M}$

Here, the angle formed by the moment of couple due to the force located at the point $B$ and $C$ along z axis is ${\theta }_z$ and moment of couple formed along z axis is $M_z$.

Rewrite the equation for ${\theta }_z$.

${\theta }_z={\cos }^{-1}\left(\frac{M_z}{M}\right)$ (VI)

Conclusion:

Substitute $0.80\text{N}\cdot \text{m}$ for $M_x$ and $8.78\text{N}\cdot \text{m}$ for $M$ in equation (IV).

$\begin{array}{c}{\theta }_x={\cos }^{-1}\left(\frac{0.80\text{N}\cdot \text{m}}{8.78\text{N}\cdot \text{m}}\right)\\ =84.8°\end{array}$

Substitute $6.36\text{N}\cdot \text{m}$ for $M_y$ and $8.78\text{N}\cdot \text{m}$ for $M$ in equation (V).

$\begin{array}{c}{\theta }_y={\cos }^{-1}\left(\frac{6.36\text{N}\cdot \text{m}}{8.78\text{N}\cdot \text{m}}\right)\\ =43.6°\end{array}$

Substitute $-6\text{N}\cdot \text{m}$ for $M_z$ and $8.78\text{N}\cdot \text{m}$ for $M$ in equation (VI).

$\begin{array}{c}{\theta }_z={\cos }^{-1}\left(\frac{-6\text{N}\cdot \text{m}}{8.78\text{N}\cdot \text{m}}\right)\\ =133.1°\end{array}$

Therefore, the direction of the two forces acting on the point $B$ and $C$ is

${\theta }_x=84.8°,{\theta }_y=43.6°,{\theta }_z=133.1°$.