Question

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Answer 1

Question:-

In the arrangement shown in the figure, the magnitude of acceleration of C(in m/s²) if A and B have acceleration as shown is __________

Answer:-

We can consider either downward or upward motion as positive. In this question, it is more convenient to consider downward motion as positive.

As it is undergoing constraint motion,

${\green{\bigstar}} \boxed{\sf{ \sum(\vec{T}.\vec{a}) = 0}}$

Refer to the attachment for F.B.D.

$\sf \sum(\vec{T}.\vec{a})=0$

$\longrightarrow \sf (\vec{T_A}.\vec{a_A}) + (\vec{T_B}.\vec{a_B}) + (\vec{T_C}.\vec{a_C}) = 0$

$\longrightarrow \sf \vec{T_A}\vec{a_A}cos\theta_1 + \vec{T_B}\vec{a_B}cos\theta_2 + \vec{T_C}\vec{a_C}cos\theta_3 = 0$

$\longrightarrow \small \sf (T a_A cos180^o) + (2T a_B cos180^o) + (T a_C cos0^o) = 0$

$\longrightarrow \sf - Ta_A - 2Ta_B + Ta_C = 0$

$\longrightarrow \sf Ta_C = Ta_A + 2Ta_B$

$\longrightarrow \sf \cancel{T}a_C = \cancel{T}(a_A + 2a_B)$

$\longrightarrow \sf a_C = a_A + 2a_B$

$\longrightarrow \sf a_C = 2 + (2\times 3)$

$\longrightarrow \sf a_C = 2 + 6 \: m/s^2$

$\longrightarrow \sf a_C = 8\:m/s$

As we had considered downward motion as positive, and the value of $\sf a_C$ came positive,

$\longrightarrow \sf \underline{\underline{\sf{\green{ a_C = 8\:m/s^2\: \: downwards }}}}$