Question

Problem 16.89 2 1084 E Review Part A If link CD has an angular velocity of WCD - 10 rad/s, determine the velocity of point E on link BC at the instant 30 (Figure 1) Enter the rand y components of the velocity separated by a comma. Figure 1 of 1 V AXP 11 vec ? (ve), (uk), - 0.3 mm Subm Request Answer 0.0 m Part B Determine the angular velocity of ink AB at the Instant 8 30 Assume the counterclockwise rotation as positive Express your answer with the appropriate units Pearson

please do part a&b and ill do my part by leaving a like. thank you

Answer 1

For CD,

ūc = WCD XTCD

$\Rightarrow \vec{v}_C = (10\hat{k})\times(0.6\hat{j})$

ūc = -61

For AB;

$\vec{v}_B = \vec{\omega}_{AB}\times\vec{r}_{B/A}$

$\Rightarrow \vec{v}_B = (\omega_{AB}\hat{k})\times\left ( \frac{0.6}{tan30^o}\hat{i}+0.6\hat{j} \right )$

$\Rightarrow \vec{v}_B =1.0392\omega_{AB}\hat{j} - 0.6\omega_{AB}\hat{i}$

For BC;

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

$\Rightarrow -6\hat{i} = (1.0392\omega_{AB}\hat{j} - 0.6\omega_{AB}\hat{i})+ (\omega_{BC}\hat{k})\times(0.6\hat{i})$

$\Rightarrow -6\hat{i} = (1.0392\omega_{AB}\hat{j} - 0.6\omega_{AB}\hat{i})+ (0.6\omega_{BC}\hat{j})$

Compare i;

$-6= - 0.6\omega_{AB}$

$\Rightarrow \omega_{AB} = 10\;\;rad/s$

Compare j;

$0= 1.0392\omega_{AB}+ 0.6\omega_{BC}$

$\Rightarrow 0= 1.0392(10)+ 0.6\omega_{BC}$

$\Rightarrow \omega_{BC} = -17.321\;\;rad/s$

For BC;

$\vec{v}_E = \vec{v}_C + \vec{\omega}_{BC}\times\vec{r}_{E/C}$

$\Rightarrow \vec{v}_E = (-6\hat{i}) + (-17.321\hat{k})\times(-0.3\hat{i})$

$\Rightarrow \vec{v}_E = -6\hat{i}+5.196\hat{j}$

$\Rightarrow \vec{v}_E = -6\hat{i}+5.20\hat{j}$

(A)

$(v_E)_x,(v_E)_y \;\;=\;\;\mathbf{-6},\mathbf{5.20}$

(B)

$\omega_{AB} = 10\;\;rad/s$