The parallel-link mechanism ABCD is used to transport a component I between manufacturing processes at stations E, F, and G by picking it up at a station when θ = 0 and depositing it at the next station when θ = 180°. Knowing that member BC remains horizontal throughout its motion and that links AB

Question

The parallel-link mechanism ABCD is used to transport a component I between manufacturing processes at stations E, F, and G by picking it up at a station when [latex] heta =0[/latex] and depositing it at the next station when [latex] heta ={ 180 }^{ \circ }[/latex] .Knowing that member BC remains horizontal throughout its motion and that links AB and CD rotate at a constant rate in a vertical plane in such a way that [latex] u_{ B } 2.2 \mathrm{\ ft}/\mathrm{s}[/latex], determine (a) the minimum value of the coefficient of static friction between the component and BC if the component is not to slide on BC while being transferred, (b) the values of [latex] heta[/latex] for which sliding is impending.

Accepted Answer

[latex]\begin{gathered}\overset{+}{→}\Sigma F_{x}=m a_{x}: \quad F=\frac{W}{g} \frac{ u_{B}^{2}}{ ho} \cos heta \+\uparrow \Sigma F_{y}=m a_{y}: \quad N-W=-\frac{W}{g} \frac{ u_{B}^{2}}{ ho} \sin heta\end{gathered}[/latex]

or [latex]\quad N=W\left\lgroup1-\frac{ u_{B}^{2}}{g ho} \sin heta ight group[/latex]

Now [latex]\quad F_{\max }=\mu_{s} N=\mu_{s} W\left\lgroup1-\frac{ u_{B}^{2}}{g ho} \sin heta ight group[/latex]

and for the component not to slide

[latex]F

or[latex]\quad\frac{W}{g} \frac{ u_{B}^{2}}{ ho} \cos heta<\mu_{s} W\left\lgroup1-\frac{ u_{B}^{2}}{g ho} \sin heta ight group[/latex]

or [latex]\quad\mu_{s}>\frac{\cos heta}{\frac{g ho}{ u_{B}^{2}}-\sin heta}[/latex]

We must determine the values of [latex] heta[/latex] which maximize the above expression. Thus

[latex]\frac{d}{d heta}\left\lgroup\frac{\cos heta}{\frac{g ho}{ u_{B}^{2}}-\sin heta} ight group=\frac{-\sin heta\left(\frac{g ho}{ u_{B}^{2}}-\sin heta ight)-(\cos heta)(-\cos heta)}{\left(\frac{g ho}{ u_{B}^{2}}-\sin heta ight)^{2}}=0[/latex]

or [latex]\quad\sin heta=\frac{ u_{B}^{2}}{g ho} \quad ext { for } \quad \mu_{s}=\left(\mu_{s} ight)_{\min }[/latex]

Now [latex]\quad\sin heta=\frac{(2.2 \mathrm{\ ft} / \mathrm{s})^{2}}{\left(32.2 \mathrm{\ ft} / \mathrm{s}^{2} ight)\left(\frac{10}{12} \mathrm{\ ft} ight)}=0.180373[/latex]

or [latex]\quad heta=10.3915^{\circ} ext { and } heta=169.609^{\circ}[/latex]

(a) From above,

[latex]\begin{aligned}\left(\mu_{s} ight)_{\min } & =\frac{\cos heta}{\frac{g ho}{ u_{B}^{2}}-\sin heta} \quad ext { where } \quad \sin heta=\frac{ u_{B}^{2}}{g ho} \\left(\mu_{s} ight)_{\min } & =\frac{\cos heta}{\frac{1}{\sin heta}-\sin heta}=\frac{\cos heta \sin heta}{1-\sin ^{2} heta}= an heta \& = an 10.3915^{\circ}\end{aligned}[/latex]

or [latex]\quad\quad\quad\quad\quad\quad\left(\mu_{s} ight)_{\min }=0.1834\blacktriangleleft[/latex]

(b) We have impending motion

to the left for [latex]\quad\quad\quad\quad\quad\quad heta=10.39^{\circ}\blacktriangleleft[/latex]

to the right for[latex]\quad\quad\quad\quad\quad\quad heta=169.6^{\circ}\blacktriangleleft[/latex]

Question 12.62: The parallel-link mechanism ABCD is used to transport a comp...

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