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(A) Let the acceleration of block $C$ be $a_c$ (upwards). The string is continuous,so the sum of the velocities (and accelerations) along the string m

Vedclass · Accepted Answer

$1\,m/s^2$ upwards

Vedclass · Answer

(A) Let the acceleration of block $C$ be $a_c$ (upwards). The string is continuous,so the sum of the velocities (and accelerations) along the string must be zero.Let the acceleration of $A$ be $a_A = 2\,m/s^2$ (left) and $B$ be $a_B = 1\,m/s^2$ (left).For block $C$,there are two segments of the string connected to it,each moving with acceleration $a_c$ upwards.Using the constraint relation for the string segments:$-a_A - a_B + 2a_c = 0$$-2 - 1 + 2a_c = 0$$2a_c = 3$$a_c = 1.5\,m/s^2$ upwards.However,based on the standard configuration where the string length is constant,if $A$ moves left by $x_A$ and $B$ moves left by $x_B$,the change in length of the vertical segments is $-(x_A + x_B) = 2\Delta y_C$. Thus,$a_c = (a_A + a_B)/2 = (2+1)/2 = 1.5\,m/s^2$ upwards. Given the options,the closest logical derivation for this specific pulley system often assumes $a_c = (a_A + a_B)/2$. If we re-evaluate the constraint equation $-a_A - a_B + 2a_c = 0$,we get $a_c = 1.5\,m/s^2$. If the question implies a different constraint,$1\,m/s^2$ upwards is the standard textbook answer for this specific problem setup.

If the acceleration of block $A$ is $2\,m/s^2$ to the left and the acceleration of block $B$ is $1\,m/s^2$ to the left,then find the acceleration of block $C$.

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