The pulleys in the diagram are all smooth and | vedclass

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(A) Let $l_1$ be the length of the string segment connected to $A$,$l_2$ be the length of the string segment on either side of the movable pulley $B$,

Vedclass · Accepted Answer

$\frac{1}{2}(f-a)$ up

Vedclass · Answer

(A) Let $l_1$ be the length of the string segment connected to $A$,$l_2$ be the length of the string segment on either side of the movable pulley $B$,and $l_3$ be the length of the string segment connected to $C$. The total length of the string $L$ is constant:$l_1 + 2l_2 + l_3 = L$Taking the second derivative with respect to time $t$ to find accelerations:$\frac{d^2 l_1}{dt^2} + 2\frac{d^2 l_2}{dt^2} + \frac{d^2 l_3}{dt^2} = 0$Let upward direction be positive. The acceleration of $A$ is $a_A = a$. The acceleration of $C$ is $a_C = -f$. The acceleration of the movable pulley $B$ is $a_B$. The rate of change of the length of the segments is related to the accelerations as follows:$\frac{d^2 l_1}{dt^2} = -a_A = -a$$\frac{d^2 l_2}{dt^2} = -a_B$$\frac{d^2 l_3}{dt^2} = -a_C = -(-f) = f$Substituting these into the constraint equation:$-a + 2(-a_B) + f = 0$$2a_B = f - a$$a_B = \frac{1}{2}(f - a)$Since the result is positive,the acceleration of $B$ is $\frac{1}{2}(f - a)$ upwards.

The pulleys in the diagram are all smooth and light. The acceleration of $A$ is $a$ upwards and the acceleration of $C$ is $f$ downwards. The acceleration of $B$ is

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