Solved 1 Answer
See More Answers for FREE
Enhance your learning with StudyX
Receive support from our dedicated community users and experts
See up to 20 answers per week for free
Experience reliable customer service
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 ...