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• Notes

$\begin{array}{l}{12-198 . \text { If the end of the cable at } A \text { is pulled down with a }} \\ {\text { speed of } 5 \mathrm{m} / \mathrm{s}, \text { determine the speed at which block } B \text { rises. }}\end{array}$

$+ \downarrow V_{A}=5 \mathrm{m} / \mathrm{s}$

$S_{B}+(2 {S_B}-a)+S_{A}=l$

$3S_ B+S_A=l+2a$

$3 V_ B+V_ A=0+0 \Rightarrow 3 V_B+V_A \downarrow = 0$

$∴3V_ B+5=0$

$V_{B}=\frac{-5}{3} \simeq -1.667 \mathrm{m} / \mathrm{s}$

$∴ V_{B}=1.667 \mathrm{m} / \mathrm{s} \uparrow$

$\begin{array}{l}{12-199 . \text { Determine the displacement of the log if the truck }} \\ {\text { at } C \text { pulls the cable } 4 \mathrm{ft} \text { to the right. }}\end{array}$

$S=4 \mathrm{ft}$

$2S_B+(S_B-S_C)=l$

$3 S_{B}-S_{C}=l$

$3 \Delta S_ B-\Delta S_ C=0$

$∴3 \Delta S_{B}=\Delta S_{C}$

$\Delta S_C=-4 f t \Rightarrow ∴ 3 \Delta S_B=-4$

$∴ \Delta S_ B=-1.33 \mathrm{ft}$

$\Delta S_{B} \simeq 1.33 \mathrm{ft}$