Chapter 16.2, Problem 16.127P

The test rig shown was developed to perform fatigue testing on fitness trampolines. A motor drives the 200-mm radius flywheel AB, which is pinned at its center point A, in a counterclockwise direction with a constant angular velocity of 120 rpm. The flywheel is attached to slider CD by the 400-mm connecting rod BC. The mass of the connecting rod BC is 5 kg, and the mass of the link CD and foot is 2 kg. At the instant when θ = 0° and the foot is just above the trampoline, determine the force exerted by pin C on rod BC.

Fig. P16.127

To determine

Find the force exerted by pin C on rod BC for fitness trampoline.

Answer to Problem 16.127P

The force exerted by pin C on rod BC is $\underset{\_}{67.62\text{N}}$.

Explanation of Solution

Given information:

The radius of the flywheel AB is $r=200\text{mm}$.

The mass of the connecting rod BC is $m_{BC}=5\text{kg}$.

The mass of the link CD and foot is $m_{CD}=2\text{kg}$.

The length of the rod BC is $l_{BC}=400\text{mm}$.

The angular velocity is ${\omega }_{AB}=120\text{rpm}$.

The angle is $\theta =0°$.

Calculation:

Convert the unit of angular velocity from $\text{rpm}$ to $\text{rad}/\text{s}$ as shown below.

$\begin{array}{c}{\omega }_{AB}=120\text{rpm}\times \frac{\frac{2\pi }{60}\text{rad}/\text{s}}{1\text{rpm}}\\ =4\pi \text{rad}/\text{s}\end{array}$

Consider the acceleration due to gravity as $g=9.81\text{ft}/{\text{s}}^2$.

Calculate the weight $\left(W\right)$ as shown below.

$W=mg$ (1)

Calculate the weight of rod BC $\left(W_{BC}\right)$ as shown below.

Substitute $5\text{kg}$ for $m$ and $9.81\text{ft}/{\text{s}}^2$ for g in Equation (1).

$\begin{array}{c}{W}_{BC}=5\times 9.81\\ =49.05\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\times \frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\\ =49.05\text{N}\end{array}$

Calculate the weight of link CD $\left(W_{CD}\right)$ as shown below.

Substitute $2\text{kg}$ for $m$ and $9.81\text{ft}/{\text{s}}^2$ for g in Equation (1).

$\begin{array}{c}{W}_{CD}=2\times 9.81\\ =19.62\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\times \frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\\ =19.62\text{N}\end{array}$

Sketch the geometry of the rig as shown in Figure 1.

Refer to Figure 1.

Calculate the angle $\left(\alpha \right)$ as shown below.

$\begin{array}{l}\cos \alpha =\frac{0.2}{0.4}\\ \alpha =60°\end{array}$

Calculate the position vectors $\left(\overrightarrow{r}\right)$ as shown below.

Position of B with respect to A.

${\overrightarrow{r}}_{B/A}=0.2\text{i}$

Position of C with respect to B.

$\begin{array}{c}{\overrightarrow{r}}_{C/B}=-0.4\cos 60°\text{i}+0.4\sin 60°\text{j}\\ =-0.2\text{i}+0.3464\text{j}\end{array}$

Position of G with respect to B.

$\begin{array}{c}{\overrightarrow{r}}_{G/B}=-0.1\text{i}+0.1\sqrt{3}\text{j}\\ =-0.1\text{i}+0.1732\text{j}\end{array}$

Calculate the moment of inertia $\left({\overline{I}}_{BC}\right)$ as shown below.

${\overline{I}}_{BC}=\frac{1}{12}{m}_{BC}{l}_{BC}^2$

Substitute $400\text{mm}$ for $l$ and $5\text{kg}$ for $m$.

$\begin{array}{c}{\overline{I}}_{BC}=\frac{1}{12}\times 5\text{kg}\times {\left(400\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2\\ =0.0667\text{kg}\cdot {\text{m}}^2\end{array}$

Sketch the Free Body Diagram of the rod CD as shown in Figure 2.

Refer to Figure 2.

Apply the Equilibrium of forces along y direction as shown below.

$\begin{array}{l}{\displaystyle \sum F_y}=ma\\ -2g-C_y=m_{CD}{\overline{a}}_{CD,y}\end{array}$

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for g and $2\text{kg}$ for $m_{CD}$.

$\begin{array}{l}-2\times 9.81-C_y=2{\overline{a}}_{CD,y}\\ C_y+2{\overline{a}}_{CD,y}+19.62=0\end{array}$ (2)

Sketch the Free Body Diagram of rod BC as shown in Figure 3.

Refer to Figure 3.

Apply the Equilibrium of forces along x direction as shown below.

$\begin{array}{l}{\displaystyle \sum F_x}=ma\\ C_x+B_x=m_{BC}{\overline{a}}_{BC,x}\end{array}$

Substitute $5\text{kg}$ for $m_{BC}$.

$\begin{array}{l}{C}_x+B_x=5{\overline{a}}_{BC,x}\\ B_x=5{\overline{a}}_{BC,x}-C_x\end{array}$ (3)

Apply the Equilibrium of forces along y direction as shown below.

$\begin{array}{l}{\displaystyle \sum F_y}=ma\\ C_y+B_y-m_{BC}g=m_{BC}{\overline{a}}_{BC,y}\end{array}$

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for g and $2\text{kg}$ for $m_{CD}$.

$\begin{array}{l}{C}_y+B_y-5\times 9.81=5{\overline{a}}_{BC,y}\\ C_y+B_y-5{\overline{a}}_{BC,y}-49.05=0\end{array}$ (4)

Apply the Equilibrium of moment about G as shown below.

$\begin{array}{l}{\displaystyle \sum M_G}=I_G\alpha \\ -0.1\sqrt{3}{C}_x-0.1C_y+0.1\sqrt{3}{B}_x+0.1B_y=I_{BC}{\alpha }_{BC}\\ -0.1732C_x-0.1C_y+0.1732B_x+0.1B_y=I_{BC}{\alpha }_{BC}\end{array}$

Substitute $0.0667\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_{BC}$.

$-0.1732C_x-0.1C_y+0.1732B_x+0.1B_y=0.0667{\alpha }_{BC}$ (5)

Calculate the velocity $\left({\text{v}}_B\right)$ as shown below.

${\text{v}}_B={\text{v}}_A+{\omega }_{AB}\times r_{B/A}$

Substitute $0$ for ${\text{v}}_A$, $4\pi \text{rad}/\text{s}$ for ${\omega }_{AB}$, and $0.2\text{i}$ for $r_{B/A}$.

$\begin{array}{c}{\text{v}}_B=0+4\pi \text{k}\times 0.2\text{i}\\ =0.8\pi \text{j}\end{array}$

Calculate the velocity $\left({\text{v}}_C\right)$ as shown below.

${\text{v}}_C={\text{v}}_B+{\omega }_{BC}\times r_{C/B}$

Substitute $0.8\pi \text{j}$ for ${\text{v}}_B$ and $-0.2\text{i}+0.3464\text{j}$ for $r_{C/B}$.

$\begin{array}{c}{\text{v}}_C\text{j}=0.8\pi \text{j}+{\omega }_{BC}\text{k}\times \left(-0.2\text{i}+0.3464\text{j}\right)\\ =0.8\pi \text{j}-0.2{\omega }_{BC}\text{j}-0.3464{\omega }_{BC}\text{i}\\ =\left(0.8\pi -0.2{\omega }_{BC}\right)\text{j}-0.3464{\omega }_{BC}\text{i}\end{array}$

Resolving the i and j components as shown below.

For i component.

$\begin{array}{l}-0.3464{\omega }_{BC}=0\\ {\omega }_{BC}=0\end{array}$

For j component.

${\text{v}}_C=0.8\pi -0.2{\omega }_{BC}$

Substitute $0$ for ${\omega }_{BC}$.

${\text{v}}_{\text{C}}=0.8\pi \text{m}/\text{s}$

Calculate the relative acceleration $\left({\text{a}}_B\right)$ as shown below.

${\text{a}}_B={\text{a}}_A-{\omega }_{AB}^2{r}_{B/A}+{\alpha }_{AB}\times r_{B/A}$

Substitute $0$ for ${\text{a}}_A$, $4\pi \text{rad}/\text{s}$ for ${\omega }_{AB}$, $0.2\text{i}$ for $r_{B/A}$, and $0$ for ${\alpha }_{AB}$.

$\begin{array}{c}{\text{a}}_B=-{\left(4\pi \right)}^2\times 0.2\text{i}\\ =-3.2{\pi }^2\text{i}\end{array}$

Calculate the relative acceleration $\left({\text{a}}_C\right)$ as shown below.

${\text{a}}_C={\text{a}}_B-{\omega }_{BC}^2{r}_{C/B}+{\alpha }_{BC}\times r_{C/B}$

Substitute $-3.2{\pi }^2\text{i}$ for ${\text{a}}_B$, $0$ for ${\omega }_{BC}$, and $-0.2\text{i}+0.3464\text{j}$ for $r_{C/B}$.

$\begin{array}{c}{\text{a}}_C\text{j}=-3.2{\pi }^2\text{i}-0+{\alpha }_{BC}\text{k}\times \left(-0.2\text{i}+0.3464\text{j}\right)\\ =-3.2{\pi }^2\text{i}-0.2{\alpha }_{BC}\text{j}-0.3464{\alpha }_{BC}\text{i}\\ =-\left(3.2{\pi }^2+0.3464{\alpha }_{BC}\right)\text{i}-0.2{\alpha }_{BC}\text{j}\end{array}$

Resolving i and j components as shown below.

For i component,

$\begin{array}{l}3.2{\pi }^2+0.3464{\alpha }_{BC}=0\\ 0.3464{\alpha }_{BC}=-31.5827\\ {\alpha }_{BC}=-91.17\text{rad}/{\text{s}}^{\text{2}}\end{array}$

For j component,

${\text{a}}_C=-0.2{\alpha }_{BC}$

Substitute $-91.17\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_{BC}$.

$\begin{array}{c}{a}_C={\overline{a}}_{CD,y}=-0.2\times -91.17\\ =18.234\text{m}/{\text{s}}^{\text{2}}\end{array}$

Calculate the relative acceleration $\left({\overline{\text{a}}}_{BC}\right)$ as shown below.

${\overline{\text{a}}}_{BC}={\text{a}}_B-{\omega }_{BC}^2{r}_{G/B}+{\alpha }_{BC}\times r_{G/B}$

Substitute $-3.2{\pi }^2\text{i}$ for ${\text{a}}_B$, $0$ for ${\omega }_{BC}$, $-91.17\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_{BC}$, and $-0.1\text{i}+0.1732\text{j}$ for $r_{G/B}$.

$\begin{array}{c}{a}_{BC,x}\text{i}+a_{BC,y}\text{j}=-3.2{\pi }^2\text{i}-0+\left(-91.17\right)\text{k}\times \left(-0.1\text{i}+0.1732\text{j}\right)\\ =-3.2{\pi }^2\text{i}+9.117\text{j}+15.7906\text{i}\\ =-15.79\text{i}+9.117\text{j}\end{array}$

Resolving i and j components as shown below.

$\begin{array}{c}{a}_{BC,x}=-15.79\text{m}/{\text{s}}^{\text{2}}\\ a_{BC,y}=9.117\text{m}/{\text{s}}^{\text{2}}\end{array}$

Calculate the reaction $\left(C_y\right)$ as shown below.

Substitute $18.234\text{m}/{\text{s}}^2$ for ${\overline{a}}_{CD,y}$ in Equation (2).

$\begin{array}{l}{C}_y+2\times 18.234+19.62=0\\ C_y+56.088=0\\ C_y=-56.088\text{N}\end{array}$

Calculate the reaction $\left(B_y\right)$ as shown below.

Substitute $-56.088\text{N}$ for $C_y$ and $9.117\text{m}/{\text{s}}^2$ for ${\overline{a}}_{BC,y}$ in Equation (4).

$\begin{array}{l}-56.088+B_y-5\times 9.117-49.05=0\\ B_y-150.723=0\\ B_y=150.723\text{N}\end{array}$

Calculate the reaction $\left(C_x\right)$ as shown below.

Substitute $5{\overline{a}}_{BC,x}-C_x$ for $B_x$, $150.723\text{N}$ for $B_y$, $-56.088\text{N}$ for $C_y$, and $-91.17\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_{BC}$ in Equation (5).

$\begin{array}{l}\left[\begin{array}{l}-0.1732C_x-0.1\times \left(-56.088\right)+\\ 0.1732\left(5{\overline{a}}_{BC,x}-C_x\right)+0.1\times 150.723\end{array}\right]=0.0667\times \left(-91.17\right)\\ -0.1732C_x+20.6811+0.866{\overline{a}}_{BC,x}-0.1732C_x=-6.081\\ -0.3464C_x+0.866{\overline{a}}_{BC,x}+26.7621=0\end{array}$

Substitute $-15.79\text{m}/{\text{s}}^{\text{2}}$ for ${\overline{a}}_{BC,x}$.

$\begin{array}{l}-0.3464C_x+0.866\times \left(-15.79\right)+26.7621=0\\ 0.3464C_x=13.08796\\ C_x=37.78\text{N}\end{array}$

Calculate the reaction at C as shown below.

$C=\sqrt{C_x^2+C_y^2}$

Substitute $37.78\text{N}$ for $C_x$ and $-56.088\text{N}$ for $C_y$.

$\begin{array}{c}C=\sqrt{{37.78}^2+{\left(-56.088\right)}^2}\\ =\sqrt{4,573.192144}\\ =67.62\text{N}\end{array}$

Therefore, the force exerted by pin C on rod BC is $\underset{\_}{67.62\text{N}}$.