Question 0.4 m Vc = 1.8 m/s Exercise 8.44 Slider C is moving to the right with a velocity of 1.8 m/s and acceleration 0.6 m/s2 as shown. Determine the angular accelera- tion of members AB, CAB and 0.3 m BC, ABC, and components of the acceleration at point B, as for the instant shown. ac = 0.6 m/s2 B

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Transcribed Image Text: 0.4 m Vc = 1.8 m/s Exercise 8.44 Slider C is moving to the right with a velocity of 1.8 m/s and acceleration 0.6 m/s2 as shown. Determine the angular accelera- tion of members AB, CAB and 0.3 m BC, ABC, and components of the acceleration at point B, as for the instant shown. ac = 0.6 m/s2 B

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Instantaneous centers are given below -At this instant, B C will not be rotating It will only haveTranslation. ( \because Its I.center is at infinite with respect to\therefore Velocity of point \left(I_{13}\right) \quad V_{c}=1.8 \mathrm{~m} / \mathrm{s} fixed Link )\begin{array}{l}c=1.8 \mathrm{~m} / \mathrm{s} \text { towards right. } \\\text { and } \alpha_{B C}=0\end{array}Now, we have a_{B}=0.6 \mathrm{~m} / \mathrm{s}^{2} which is also a point on A B. and I t is a tangential acceleration.\begin{array}{l}\therefore a_{B}=(A B) \alpha_{A B} \quad\left\{\because a_{t}=\gamma \alpha\right\} \text {. } \\0.6=0.3 \times \alpha_{A B 3} \text {. } \\\alpha_{A B}=2 \mathrm{rad} / \mathrm{s}^{2} \\\end{array}Now, \quad V_{B}=1.8 \mathrm{~m} / \mathrm{s}=(A B) \omega_{A B} \quad(\because V=\gamma \omega)\begin{array}{l} 1.8=0.3 \omega_{A B} \\\omega_{A B}=6 \text { rad } / \mathrm{s}^{2} \\a_{\text {radial }}=\omega_{A B}^{2}(A B)=36 \times 0.3=10-8 \mathrm{~m} / \mathrm{s}^{2} \\\therefore \text { Components of velocity } \\a_{B}=a_{t}=0.6 \mathrm{~m} / \mathrm{s}^{2} \text { and } a_{r}=10.8 \mathrm{~m} / \mathrm{s}^{2}\end{array}AB is perpendicular to horizontal line or line of motion of the slider. It means BC will be only translating. We got the same from instantaneous center approach.  ...