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Solution:-Draw the free body diagram of rod A B\begin{array}{l}U_{A}=V_{B}+V_{A / B} \\V_{B}=1.6 \% 15 \quad 60^{\circ}<0^{\circ} \\{\left[V_{A \downarrow} \downarrow\right]=\left[1.6 \mathrm{~m} / \mathrm{s} \quad 830^{\circ}\right]} \\+\left[V_{A / B} \prod_{40^{\circ}}\right] \\\end{array}Law of sines\frac{V_{A}}{\sin 70^{\circ}}=\frac{V_{\theta 1 B}}{\sin 60^{\circ}}=\frac{1.6 \mathrm{r} / \mathrm{s}}{\sin 50^{\circ}}a)\begin{aligned}\frac{V_{A B B}}{\sin 60^{\circ}} & =\frac{1.6 \mathrm{m-s}}{\sin 50^{\circ}} \text { so } V_{A \mid B}=1.809 \mathrm{~ms}^{2}<, 00 \\A B & =0.5 \mathrm{~m} \\V_{A \mid B} & =(A B) W_{A B}\end{aligned}\begin{array}{l}1 . \text { fo } \mathrm{gm} / \mathrm{s}=(0.5 \mathrm{~m}) \omega_{A B} \\\left.\omega_{A B}=3.618 \mathrm{rdd} / \mathrm{s}\right)^{2}\end{array}b)\begin{array}{l}\frac{M_{A}}{\sin 70^{\circ}}=\frac{1.6 \mathrm{~m} / \mathrm{s}}{\sin 50^{\circ}} \\V_{A}=1.963-15 \mathrm{~V}\end{array}c)\begin{array}{l}\frac{Q_{A}}{\sin 70^{\circ}}=\frac{Q_{A \mid B}}{\sin 60^{\circ}}=\frac{5 \operatorname{sis}^{2}}{\sin 50^{\circ}} \\\frac{d_{A 1 B}}{\sin 60^{\circ}}=\frac{5 m s^{2}}{\sin 50^{\circ}} \\d_{A A_{B}}=5.65 \mathrm{mms}^{2} \\a_{A / B}=(A B) \alpha_{A / B} \\5.65=(0.5) \alpha_{\left.A\right|_{B}} \\\alpha_{A / B}=11.30 \mathrm{~m} / \mathrm{s}^{2} \\\end{array}d)\begin{array}{l}\frac{a_{A}}{\sin 70^{\circ}}=\frac{5 \mathrm{~m} / \mathrm{s}^{2}}{\sin 50^{\circ}} \\a_{A}=6.133 \mathrm{~m} / \mathrm{s}^{2}\end{array}Resuet,a) 3.618 \mathrm{rad} / 2 \mathrm{~m} b) 1.963 \mathrm{~m} / \mathrm{s} \downarrowc) 11.30 \mathrm{~m} / \mathrm{s}^{2} \mathrm{q} d) 6.133 \mathrm{~ms}^{2} d ...