Question Problem 16.124 - General plane motion: Uniform rod: Equations of motion The 4-kg uniform rod ABD is attached to the crank BC and is fitted with a small wheel that can roll without friction along a vertical slot. Knowing that at the instant shown crank BC rotates with an angular velocity of 6 rad/s clockwise and an angular acceleration of 15 rad/s2 counterclockwise, determine the reaction at A. D 100 mm 200 mm С B 200 mm A The reaction at A is N.
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
Ans\begin{aligned}\omega & =6 \mathrm{rad} / \mathrm{s} \\2 & =15 \mathrm{rad} / \mathrm{s}^{2} \\m & =4 \mathrm{~kg} . \\\left(a_{B}\right)_{n} & =\left(\mathrm{rBC}_{B C}\right) \cdot \omega^{2} \\& =0.1 \times 6^{2}=3.6 \mathrm{~m} / \mathrm{s}^{2} \rightarrow \\\left(A_{B}\right) t & =\left(r_{B}\right) \cdot \alpha=0.1 \times 15=1.5 \mathrm{~m} / \mathrm{s}^{2}+\alpha\end{aligned}a_{A}=a_{G}+a_{A G}\begin{array}{l}\alpha=\sin ^{-}\left(\frac{B C}{A B}\right) \\\alpha=30^{\circ}\end{array}\begin{aligned}\Rightarrow G A & =(1.5+3.6)+(0.2 \alpha<\theta) \\+\rightarrow & \Rightarrow 0=3.6-(0.2 \alpha) \cos \theta \\\alpha & =\frac{3.6}{0.2 \cos 30}=20.78746 \mathrm{rod} / \mathrm{s}^{2}\end{aligned}\text { Sma }=\Sigma(m a) e r r\overline{R_{A}}(0.2 \cos 30)=\bar{I} \cdot \alpha \text {. }(\alpha \bar{I})=\frac{1}{12} m l^{2} \alpha=\underbrace{\frac{1}{I}(4) \times 0.4^{2}}_{\bar{I}} \cdot(20.78746)R_{A}=\frac{1.108664}{0.2 \cos 30^{\circ}}R_{A}=6.400877 \mathrm{~N} \text { Ays } ...