Chapter 3.4, Problem 3.120P

Two 150-mm-diameter pulleys are mounted on line shaft AD. The belts at B and C lie in vertical planes parallel to the yz plane. Replace the belt forces shown with an equivalent force-couple system at A.

To determine

Replace the given system with its equivalent force-couple system at POINT A.

Answer to Problem 3.120P

The equivalent force is $\left(420\text{N}\right)\widehat{j}-\left(339\text{N}\right)\stackrel{⌢}{k}$ and resultant moment of couple about point A is $\left(1.125\text{N}\cdot \text{m}\right)\widehat{i}+\left(163.9\text{N}\cdot \text{m}\right)\widehat{j}-\left(109.9\text{N}\cdot \text{m}\right)\widehat{k}$.

Explanation of Solution

Refer Fig P3.120. Its vector diagrams showing equivalent force-couple is given below.

Write the formula to calculate the resultant force pulley B.

${\overrightarrow{R}}_B={\overrightarrow{F}}_{B1}+{\overrightarrow{F}}_{B2}$ (I)

Here, ${\overrightarrow{R}}_B$ is the resultant force on pulley B, ${\overrightarrow{F}}_{B1}$ and ${\overrightarrow{F}}_{B2}$ are the forces acting on pulley B.

Write the formula to calculate the net moment of couple on pulley B.

$M_B=r_b\left(-F_{B1}+F_{B2}\right)\widehat{i}$ (II)

Here, $M_B$ is moment of resultant couple about pulley B, $F_{B1}$ and $F_{B2}$ are the magnitudes of forces on pulley B, and $r_b$ is the radius of pulley B.

Write the formula to calculate the resultant force pulley C.

${\overrightarrow{R}}_C={\overrightarrow{F}}_{C1}+{\overrightarrow{F}}_{C2}$ (III)

Here, ${\overrightarrow{R}}_C$ is the resultant force on pulley C, ${\overrightarrow{F}}_{C1}$ and ${\overrightarrow{F}}_{C2}$ are the forces acting on pulley C.

Write the formula to calculate the net moment of couple on pulley B.

$M_C=r_c\left(F_{C2}-F_{C1}\right)\widehat{i}$ (IV)

Here, $M_C$ is moment of resultant couple about pulley C, $F_{C1}$ and $F_{C2}$ are the magnitudes of forces on pulley B, and $r_c$ is the radius of pulley C.

Write the expression to calculate the net force on pulley.

$\overrightarrow{R}={\overrightarrow{R}}_B+{\overrightarrow{R}}_C$ (V)

Here, $\overrightarrow{R}$ is the resultant force vector of whole system.

Write the expression to calculate the resultant moment of couple about point A.

$M_A={\overrightarrow{M}}_B+{\overrightarrow{M}}_C+\left({\overrightarrow{r}}_{B/A}\times {\overrightarrow{R}}_B\right)+\left({\overrightarrow{r}}_{C/A}\times {\overrightarrow{R}}_C\right)$ (VI)

Here, $M_A$ is the net moment of couple of system about point A, $r_{B/A}$ is the distance between end A to pulley B, and $r_{C/A}$ is the distance between pulley C and end A.

Write the determinant form to calculate $r\times F$.

$r\times F=\left|\begin{array}{ccc}i& j& k\\ r_i& r_j& r_k\\ F_i& F_j& F_k\end{array}\right|$ (VII)

Here, $r_i$ is the component of $r$ in x-direction, $r_j^{}$ is the component of $r$ in y-direction, $r_k$ is the is the component of $r$ in z-direction, $F_i$ is the component of $F$ in x-direction, $F_j$ is the component of $F$ in y-direction, and $F_k$ is the component of $F$ in z-direction.

Conclusion:

Substitute $\left(-145\text{N}\right)\left[\cos 20°+\sin 20°\stackrel{⌢}{k}\right]\widehat{j}$ for ${\overrightarrow{F}}_{B1}$, $\left(-215\text{N}\right)\widehat{j}$ for ${\overrightarrow{F}}_{B2}$ in equation (I) to find ${\overrightarrow{R}}_B$.

$\begin{array}{c}{\overrightarrow{R}}_B=\left(-145\text{N}\right)\left[\cos 20°\widehat{j}+\sin 20°\stackrel{⌢}{k}\right]+\left(-215\text{N}\right)\widehat{j}\\ =\left(-351.26\text{N}\right)\widehat{j}+\left(49.59\text{N}\right)\stackrel{⌢}{k}\end{array}$

Substitute $215\text{N}$ for $F_{B2}$, $145\text{N}$ for $F_{B1}$, and $75\text{mm}$ for $r_b$ in equation (II) to find $M_B$.

$\begin{array}{c}{M}_B=\left(75\text{mm}\left(\frac{\text{1}\text{m}}{{\text{10}}^{\text{3}}\text{mm}}\right)\right)\left(-215\text{N}+145\text{N}\right)\widehat{i}\\ =\left(-5.25\text{N}\cdot \text{m}\right)\widehat{i}\end{array}$

Substitute $\left(155\text{N}\right)\left[-\sin 10°\widehat{j}-\cos 10°\stackrel{⌢}{k}\right]$ for ${\overrightarrow{F}}_{C1}$ and $\left(240\text{N}\right)\left[-\sin 10°\widehat{j}-\cos 10°\stackrel{⌢}{k}\right]$ for ${\overrightarrow{F}}_{C2}$ in equation (III) to find ${\overrightarrow{R}}_C$.

$\begin{array}{c}{\overrightarrow{R}}_C=\left(155\text{N}\right)\left[-\sin 10°\widehat{j}-\cos 10°\stackrel{⌢}{k}\right]+\left(240\text{N}\right)\left[-\sin 10°\widehat{j}-\cos 10°\stackrel{⌢}{k}\right]\\ =\left(-68.59\text{N}\right)\widehat{j}-\left(389.00\text{N}\right)\widehat{k}\end{array}$

Substitute $240\text{N}$ for $F_{C2}$, $155\text{N}$ for $F_{C1}$, and $75\text{mm}$ for $r_c$ in equation (IV) to find $M_C$.

$\begin{array}{c}{M}_C=\left(75\text{mm}\left(\frac{\text{1}\text{m}}{{\text{10}}^{\text{3}}\text{mm}}\right)\right)\left(240\text{N}-155\text{N}\right)\widehat{i}\\ =\left(6.375\text{N}\cdot \text{m}\right)\widehat{i}\end{array}$

Substitute $\left(-351.26\text{N}\right)\widehat{j}+\left(49.59\text{N}\right)\stackrel{⌢}{k}$ for ${\overrightarrow{R}}_B$ and $\left(-68.59\text{N}\right)\widehat{j}-\left(389.00\text{N}\right)\widehat{k}$ for ${\overrightarrow{R}}_C$ in equation (V) to find $\overrightarrow{R}$.

$\begin{array}{c}\overrightarrow{R}=\left[\left(-351.26\text{N}\right)\widehat{j}+\left(49.59\text{N}\right)\widehat{k}\right]+\left[\left(-68.59\text{N}\right)\widehat{j}-\left(389.00\text{N}\right)\widehat{k}\right]\\ =\left(420\text{N}\right)\widehat{j}-\left(339\text{N}\right)\widehat{k}\end{array}$

Substitute $\left(-5.25\text{N}\cdot \text{m}\right)\widehat{i}$ for $M_B$, $\left(6.375\text{N}\cdot \text{m}\right)\widehat{i}$ for $M_C$, $\left(225\text{mm}\right)\widehat{i}$ for $r_{B/A}$, $\left(-351.26\text{N}\right)\widehat{j}+\left(49.59\text{N}\right)\stackrel{⌢}{k}$ for ${\overrightarrow{R}}_B$, $\left(450\text{mm}\right)\widehat{i}$ for ${\overrightarrow{r}}_{C/A}$, $\left(-68.59\text{N}\right)\widehat{j}-\left(389.00\text{N}\right)\widehat{k}$ for ${\overrightarrow{R}}_C$ in equation (VI) to find $M_A$.

$M_A=\left\{\begin{array}{c}\left(-5.25\text{N}\cdot \text{m}\right)\widehat{i}+\left(6.375\text{N}\cdot \text{m}\right)\widehat{i}\\ +\left(\left(225\text{mm}\right)\widehat{i}\times \left[\left(-351.26\text{N}\right)\widehat{j}+\left(49.59\text{N}\right)\widehat{k}\right]\right)+\\ \left(\left(450\text{mm}\right)\widehat{i}\times \left[\left(-68.59\text{N}\right)\widehat{j}-\left(389.00\text{N}\right)\widehat{k}\right]\right)\end{array}\right\}$ (IIX)

Calculate $\left(225\text{mm}\right)\widehat{i}\times \left[\left(-351.26\text{N}\right)\widehat{j}+\left(49.59\text{N}\right)\widehat{k}\right]$ using equation (VII).

$\begin{array}{c}\left(225\text{mm}\right)\widehat{i}\times \left[\left(-351.26\text{N}\right)\widehat{j}+\left(49.59\text{N}\right)\widehat{k}\right]=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 225\text{mm}\left(\frac{\text{1}\text{m}}{{\text{10}}^{\text{3}}\text{mm}}\right)& 0\text{m}& 0\text{m}\\ 0\text{N}& -351.26\text{N}& 49.59\text{N}\end{array}\right|\\ =-\left(11.16\text{N}\cdot \text{m}\right)\widehat{j}-\left(79.03\text{N}\cdot \text{m}\right)\widehat{k}\end{array}$ (IX)

Calculate $\left(450\text{mm}\right)\widehat{i}\times \left[\left(-68.59\text{N}\right)\widehat{j}-\left(389.00\text{N}\right)\widehat{k}\right]$ using equation (VII).

$\begin{array}{c}\left(450\text{mm}\right)\widehat{i}\times \left[\left(-68.59\text{N}\right)\widehat{j}-\left(389.00\text{N}\right)\widehat{k}\right]=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 450\text{mm}\left(\frac{\text{1}\text{m}}{{\text{10}}^{\text{3}}\text{mm}}\right)& 0\text{m}& 0\text{m}\\ 0\text{N}& -68.59\text{N}& -389.00\text{N}\end{array}\right|\\ =\left(-175.05\text{N}\cdot \text{m}\right)\widehat{j}\end{array}$ (X)

Rewrite equation (IIX) by substituting equations (IX) and (X).

$\begin{array}{c}{M}_A=\left(-5.25\text{N}\cdot \text{m}\right)\widehat{i}+\left(6.375\text{N}\cdot \text{m}\right)\widehat{i}+\left[-\left(11.16\text{N}\cdot \text{m}\right)\widehat{j}-\left(79.03\text{N}\cdot \text{m}\right)\widehat{k}\right]+\\ \left(-175.05\text{N}\cdot \text{m}\right)\widehat{j}\\ =\left(1.125\text{N}\cdot \text{m}\right)\widehat{i}+\left(163.9\text{N}\cdot \text{m}\right)\widehat{j}-\left(109.9\text{N}\cdot \text{m}\right)\widehat{k}\end{array}$

Therefore, the equivalent force is $\left(420\text{N}\right)\widehat{j}-\left(339\text{N}\right)\stackrel{⌢}{k}$ and resultant moment of couple about point A is $\left(1.125\text{N}\cdot \text{m}\right)\widehat{i}+\left(163.9\text{N}\cdot \text{m}\right)\widehat{j}-\left(109.9\text{N}\cdot \text{m}\right)\widehat{k}$.