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

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

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

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

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

\(∴3V_ B+5=0 \)

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

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

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

$$
S=4 \mathrm{ft}
$$

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

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

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

\(∴3 \Delta S_{B}=\Delta S_{C} \)

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

\(∴ \Delta S_ B=-1.33 \mathrm{ft} \)

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

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