The pulleys in the diagram are all smooth and | vedclass

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(A) Let the acceleration of $A$ be $a_A = a$ (upwards) and the acceleration of $C$ be $a_C = f$ (downwards).Let the acceleration of $B$ be $a_B$ (upwa

Vedclass · Accepted Answer

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

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(A) Let the acceleration of $A$ be $a_A = a$ (upwards) and the acceleration of $C$ be $a_C = f$ (downwards).Let the acceleration of $B$ be $a_B$ (upwards).Since the length of the string is constant,the sum of the displacements (and thus accelerations) of the points connected by the string must be zero.Taking the upward direction as positive,the acceleration of $A$ is $+a$.The acceleration of the two segments of the string connected to pulley $B$ is $a_B$ upwards.Since the string passes over fixed pulleys,the acceleration of the segment connected to $C$ is $-f$ (downwards).The constraint equation for the string length is: $a_A + 2a_B + a_C = 0$ (considering the relative motion of the string segments).More precisely,using the velocity constraint for the movable pulley $B$: $v_B = \frac{v_A + v_C}{2}$.Taking upward as positive: $a_A = a$,$a_C = -f$.$a_B = \frac{a_A + a_C}{2} = \frac{a + (-f)}{2} = \frac{a-f}{2}$.Since the result is $\frac{a-f}{2}$,if $a > f$,$B$ moves up. If $f > a$,$B$ moves down. The magnitude is $\frac{1}{2}|a-f|$.Given the options,the correct expression is $\frac{1}{2}(f-a)$ up,which corresponds to $a_B = -\frac{1}{2}(a-f) = \frac{1}{2}(f-a)$.

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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