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

or \quad N=W\left\lgroup1-\frac{\nu_{B}^{2}}{g \rho} \sin \theta\right\rgroup

Now \quad F_{\max }=\mu_{s} N=\mu_{s} W\left\lgroup1-\frac{\nu_{B}^{2}}{g \rho} \sin \theta\right\rgroup

and for the component not to slide

F<F_{\max }

or\quad\frac{W}{g} \frac{\nu_{B}^{2}}{\rho} \cos \theta<\mu_{s} W\left\lgroup1-\frac{\nu_{B}^{2}}{g \rho} \sin \theta\right\rgroup

or \quad\mu_{s}>\frac{\cos \theta}{\frac{g \rho}{\nu_{B}^{2}}-\sin \theta}

We must determine the values of \theta which maximize the above expression. Thus

\frac{d}{d \theta}\left\lgroup\frac{\cos \theta}{\frac{g \rho}{\nu_{B}^{2}}-\sin \theta}\right\rgroup=\frac{-\sin \theta\left(\frac{g \rho}{\nu_{B}^{2}}-\sin \theta\right)-(\cos \theta)(-\cos \theta)}{\left(\frac{g \rho}{\nu_{B}^{2}}-\sin \theta\right)^{2}}=0

or \quad\sin \theta=\frac{\nu_{B}^{2}}{g \rho} \quad \text { for } \quad \mu_{s}=\left(\mu_{s}\right)_{\min }

Now \quad\sin \theta=\frac{(2.2 \mathrm{\ ft} / \mathrm{s})^{2}}{\left(32.2 \mathrm{\ ft} / \mathrm{s}^{2}\right)\left(\frac{10}{12} \mathrm{\ ft}\right)}=0.180373

or \quad\theta=10.3915^{\circ} \text { and } \theta=169.609^{\circ}

(a) From above,

or \quad\quad\quad\quad\quad\quad\left(\mu_{s}\right)_{\min }=0.1834\blacktriangleleft

(b) We have impending motion

to the left for \quad\quad\quad\quad\quad\quad\theta=10.39^{\circ}\blacktriangleleft

to the right for\quad\quad\quad\quad\quad\quad\theta=169.6^{\circ}\blacktriangleleft