Chapter 3.3, Problem 3.71P

(a)

To determine

Find the moment of the couple (M) formed by two forces by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples.

Answer to Problem 3.71P

The moment of the couple (M) formed by two forces by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples is $\underset{\_}{6.19\text{N}\cdot \text{m}\left(\text{clockwise}\right)}$.

Explanation of Solution

Given information:

The applied force at point B $\left(F_B\right)$ is 40 N.

The applied force at point C $\left(F_C\right)$ is 40 N.

The length of AB (x) is 390 mm.

The length of BC (y) is 270 mm.

The angle of the inclined lever $\left(\theta \right)$ is $55°$.

The angle of the force acting at point C $\left(\alpha \right)$ is $20°$.

Calculation:

Show the free body diagram of the lever as Figure 1.

Calculate the vertical height of BC $\left(d_1\right)$ by resolving in vertical direction using the relation:

$d_1=y\sin \theta$

Substitute 270 mm for y and $55°$ for $\theta$.

$\begin{array}{c}{d}_1=\left(\text{270mm×}\frac{\text{1}}{\text{1000}}\frac{\text{m}}{\text{mm}}\right)\sin 55°\\ =0.22117\text{m}\end{array}$

Calculate the horizontal height of BC $\left(d_2\right)$ by resolving in horizontal direction using the relation:

$d_2=y\cos \theta$

Substitute 270 mm for y and $55°$ for $\theta$.

$\begin{array}{c}{d}_2=\left(\text{270mm×}\frac{\text{1}}{\text{1000}}\frac{\text{m}}{\text{mm}}\right)\cos 55°\\ =0.154866\text{m}\end{array}$

Calculate the horizontal reaction at C $\left(C_x\right)$ by resolving in horizontal direction using the relation:

$C_x=F_C\times \cos \alpha$

Substitute 40 N for $F_C$ and $20°$ for $\alpha$.

$\begin{array}{c}{C}_x=40\times \cos 20\\ =37.588\text{N}\end{array}$

Calculate the vertical reaction at C $\left(C_y\right)$ by resolving in vertical direction using the relation:

$C_y=F_C\times \sin \alpha$

Substitute 40 N for $F_C$ and $20°$ for $\alpha$.

$\begin{array}{c}{C}_y=40\times \sin 20\\ =13.6808\text{N}\end{array}$

Take the moment about B.

$M=-d_1{C}_x+d_2{C}_y$

Substitute 0.22117 m for $d_1$, 0.154866 m for $d_2$, 37.588 N for $C_x$ and 13.6808 N for $C_y$.

$\begin{array}{c}M=-\left(0.22117\times 37.588\right)+\left(0.154866\times 13.6808\right)\\ =-6.19\text{N}\cdot \text{m}\\ =6.19\text{N}\cdot \text{m}\left(\text{clockwise}\right)\end{array}$

Therefore, the moment of the couple (M) formed by two forces by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples is $\underset{\_}{6.19\text{N}\cdot \text{m}\left(\text{clockwise}\right)}$.

(b)

To determine

Find the moment of the couple (M) formed by two forces by using the perpendicular distance between the two forces.

Answer to Problem 3.71P

The moment of the couple (M) formed by two forces by using the perpendicular distance between the two forces is $\underset{\_}{6.1946\text{N}\cdot \text{m}\left(k\right)\left(\text{clockwise}\right)}$.

Explanation of Solution

Given information:

The applied force at point B $\left(F_B\right)$ is 40 N.

The applied force at point C $\left(F_C\right)$ is 40 N.

The length of AB (x) is 390 mm.

The length of BC (y) is 270 mm.

The angle of the inclined lever $\left(\theta \right)$ is $55°$.

The angle of the force acting at point C $\left(\alpha \right)$ is $20°$.

Calculation:

Calculate the distance (d) between the two forces using the relation:

$d=y\sin \left(\theta -\alpha \right)$

Substitute 270 mm for y, $55°$ for $\theta$ and $20°$ for $\alpha$.

$\begin{array}{c}d=\left(\text{270mm×}\frac{\text{1}}{\text{1000}}\frac{\text{m}}{\text{mm}}\right)\sin \left(55°-20°\right)\\ =0.154866\text{m}\end{array}$

Calculate the moment of the couple (M) formed by two forces by using the perpendicular distance between the two forces using the relation:

$M=Fd\left(-k\right)$

Substitute 40 N for F and 0.154866 m for d.

$\begin{array}{c}M=40\times 0.154866\left(-k\right)\\ =-6.1946\text{N}\cdot \text{m}\left(k\right)\\ =6.1946\text{N}\cdot \text{m}\left(k\right)\left(\text{clockwise}\right)\end{array}$

Therefore, the moment of the couple (M) formed by two forces by using the perpendicular distance between the two forces is $\underset{\_}{6.1946\text{N}\cdot \text{m}\left(k\right)\left(\text{clockwise}\right)}$.

(c)

To determine

Find the moment of the couple (M) formed by summing the moments of two forces about point A.

Answer to Problem 3.71P

The moment of the couple (M) formed by summing the moments of two forces about point A is $\underset{\_}{\left(6.195\text{N}\cdot \text{m}\right)k\left(\text{clockwise}\right)}$.

Explanation of Solution

Given information:

The applied force at point B $\left(F_B\right)$ is 40 N.

The applied force at point C $\left(F_C\right)$ is 40 N.

The length of AB (x) is 390 mm.

The length of BC (y) is 270 mm.

The angle of the inclined lever $\left(\theta \right)$ is $55°$.

The angle of the force acting at point C $\left(\alpha \right)$ is $20°$.

Calculation:

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

$\begin{array}{c}{r}_{B/A}=x\cos \theta i+x\sin \theta j+0k\\ =x\cos \theta i+x\sin \theta j\end{array}$

Substitute 390 mm for x and $55°$ for $\theta$.

$\begin{array}{c}{r}_{B/A}=\left(\text{390mm×}\frac{\text{1}}{\text{1000}}\frac{\text{m}}{\text{mm}}\right)\cos 55°i+\left(\text{390mm×}\frac{\text{1}}{\text{1000}}\frac{\text{m}}{\text{mm}}\right)\sin 55°j\\ =0.39\cos 55°i+0.39\sin 55°j\end{array}$

Calculate the force at B by resolving in horizontal and vertical direction using the relation:

$F_B=-F_B\cos \alpha i-F_B\sin \alpha j+0$

Substitute 40 N for $F_B$ and $20°$ for $\alpha$.

$\begin{array}{c}{F}_B=-40\cos 20°i-40\sin 20°j+0\\ =-40\cos 20°i-40\sin 20°j\end{array}$

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

$\begin{array}{c}{r}_{C/A}=\left(x+y\right)\cos \theta i+\left(x+y\right)\sin \theta j+0k\\ =\left(x+y\right)\cos \theta i+\left(x+y\right)\sin \theta j\end{array}$

Substitute 390 mm for x, 270 mm for y and $55°$ for $\theta$.

$\begin{array}{c}{r}_{C/A}=\left\{\left[\begin{array}{l}\left(\text{390mm×}\frac{\text{1}}{\text{1000}}\frac{\text{m}}{\text{mm}}\right)+\\ \left(\text{270mm×}\frac{\text{1}}{\text{1000}}\frac{\text{m}}{\text{mm}}\right)\end{array}\right]\cos 55°\right\}i+\left\{\left[\begin{array}{l}\left(\text{390mm×}\frac{\text{1}}{\text{1000}}\frac{\text{m}}{\text{mm}}\right)+\\ \left(\text{270mm×}\frac{\text{1}}{\text{1000}}\frac{\text{m}}{\text{mm}}\right)\end{array}\right]\sin 55°\right\}j\\ =0.66\cos 55°i+0.66\sin 55°j\end{array}$

Calculate the force at C by resolving in horizontal and vertical direction using the relation:

$F_C=F_C\cos \alpha i+F_C\sin \alpha j+0$

Substitute 40 N for $F_C$ and $20°$ for $\alpha$.

$\begin{array}{c}{F}_C=40\cos 20°i+40\sin 20°j+0\\ =40\cos 20°i+40\sin 20°j\end{array}$

Calculate the moment of the couple (M) formed by summing the moments of two forces about point A using the relation:

Take the moment about A.

$\begin{array}{c}{\displaystyle \sum M_A}={\displaystyle \sum r_A\times F}\\ M=\left(r_{B/A}\times F_B\right)+\left(r_{C/A}\times F_C\right)\end{array}$

Substitute $0.39\cos 55°i+0.39\sin 55°j$ for $r_{B/A}$, $0.66\cos 55°i+0.66\sin 55°j$ for $r_{C/A}$, $-40\cos 20°i-40\sin 20°j$ for $F_B$ and $40\cos 20°i+40\sin 20°j$ for $F_C$.

$\begin{array}{c}M=\left(0.39\text{m}\right)\left(40\text{N}\right)\left|\begin{array}{ccc}i& j& k\\ \cos 55°& \sin 55°& 0\\ -\cos 20°& -\sin 20°& 0\end{array}\right|+\left(0.66\text{m}\right)\left(40\text{N}\right)\left|\begin{array}{ccc}i& j& k\\ \cos 55°& \sin 55°& 0\\ \cos 20°& \sin 20°& 0\end{array}\right|\\ =\left(0.39\text{m}\right)\left(40\text{N}\right)\left[\left(0\right)-\left(0\right)-\left(-0.1962+0.7698\right)\right]+\left(0.66\text{m}\right)\left(40\text{N}\right)\left[0-0-\left(0.1962-0.7698\right)\right]\\ =\left(8.948-15.143\right)k\end{array}$$\begin{array}{c}=-\left(6.195\text{N}\cdot \text{m}\right)k\\ =\left(6.195\text{N}\cdot \text{m}\right)k\left(\text{clockwise}\right)\end{array}$

Therefore, the moment of the couple (M) formed by summing the moments of two forces about point A is $\underset{\_}{\left(6.195\text{N}\cdot \text{m}\right)k\left(\text{clockwise}\right)}$.