Question Problem #3 Total = 40 points; You need to DO ALL Problems! This is problem #3 IF slider block C is moving with constant velocity Ve= 3 m/s. Dimensions of the parts are given in the picture. At the instant shown, determine (a) Velocity of point B 10 points (b) The angular velocity of BC 15 points (c) The angular velocity of AB 15 points 0.5 m 1 m 00 c - 3 m/s

First of all\begin{aligned}\overrightarrow{A B} & =0.5[\cos 60 \hat{i}+\sin 60 \hat{j}] \\& =0.25 \hat{i}+0.673 \hat{j} \hat{j} \\\overrightarrow{B C} & =1[\cos 45 \hat{i}-\sin 45 \hat{j}] \\& =0.707 \hat{i}-0.707 \hat{j}\end{aligned}NN, \quad \overrightarrow{V_{C}}=\left(\omega_{A B} \hat{F} \times \overrightarrow{A B}\right)+\left(\omega_{R C} \hat{K} \times \overrightarrow{B C}\right)=\left[\omega_{A B} \hat{k} \times(0.25 \hat{i}+0.433 \hat{j})\right]\begin{array}{l}+ {\left[\omega_{B C} \hat{k} \times(0.707 \hat{i}-0.707 \hat{j})\right] } \\=\left(0.25 \omega_{A S} \hat{j}-0.433 \omega_{A B} \hat{i}\right) \\+\left(0.707 \omega_{B C} \hat{j}+0.707 \omega_{B C} \hat{i}\right)\end{array}\begin{aligned}+3 \hat{j}= & \left(0.707 \omega_{B C \hat{j}}+0.707 \omega_{B C}\right. \\& +\left(0.707\left(\omega_{B C}-0.433 \omega_{A B}\right) \hat{i}\right. \\& +\left(0.25 \omega_{A B}+0.707 \omega_{B C}\right) \hat{j}\end{aligned}Dy equafing both sick,\begin{array}{l}0.707 \omega_{B C}-0.433 \omega_{A S}=0 . \Rightarrow \omega_{B C}=0.612 \omega_{A B} . \\0.25 \omega_{A B}+0.707 \omega_{B C}=-3 \\\Rightarrow \omega_{A B}[0.25+(0.707)(0.612)]=-3 \text {. } \\\Rightarrow \omega_{A S}=\frac{-3}{0.682}=4.4 \mathrm{rad} / \mathrm{rec} \\\text { Thon, } \omega_{B C}=0.612 \times(-4.4)=-2.69 \mathrm{rad} / \mathrm{sec} \\\vec{V}_{B}=(-4.4 \hat{k}) \times(0.25 \hat{i}+0.433 \hat{j}) \\=-1.1 \hat{j}+1.9052 \hat{i} \\\end{array} ...