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Draw the free body diagram of rod A B as shown below:Draw the force triangle as shown below:The velocity of collar B is\mathbf{v}_{B}=\mathbf{v}_{A}+\mathbf{v}_{A / B} \ldots . . \text { (1) }Here, velocity of collar A is \mathbf{v}_{A}, the velocity of A with respect to B is \mathbf{v}_{A / B}Apply the law of Sine's:\begin{array}{l}\frac{\mathbf{v}_{A}}{\sin 70^{\circ}}=\frac{\mathbf{v}_{A / B}}{\sin 60^{\circ}}=\frac{1.6 \mathrm{~m} / \mathrm{s}}{\sin 50^{\circ}} \ldots \ldots \\\mathbf{v}_{A / B}=\frac{1.6 \mathrm{~m} / \mathrm{s}}{\sin 50^{\circ}}\left(\sin 60^{\circ}\right) \\=1.809 \mathrm{~m} / \mathrm{s}\end{array}a) Calculate the angular velocity of rod A B by using the following equation:v_{A / B}=\left(r_{A B}\right) \omega_{A B}Here, length of A B is r_{A B}Substitute, 1.809 \mathrm{~m} / \mathrm{s} for \mathbf{v}_{A / B}, 0.5 \mathrm{M} for r_{A B}\begin{array}{l}1.809 \mathrm{~m} / \mathrm{s}=(0.5 \mathrm{~m}) \omega_{A B} \\\omega_{A B}=3.618 \mathrm{rad} / \mathrm{s}\end{array}Therefore, the angular velocity, \omega_{A B} is 3.62 \mathrm{rad} / \mathrm{s}b) Calculate the velocity of collar A :From equation (I)\begin{aligned}\mathbf{v}_{A} & =\frac{1.6 \mathrm{~m} / \mathrm{s}}{\sin 50^{\circ}}\left(\sin 70^{\circ}\right) \\& =1.963 \mathrm{~m} / \mathrm{s}\end{aligned}Therefore, the velocity at A is 1.963 \mathrm{~m} / \mathrm{s} \downarrow ...