Answer to Question #241806 in Mechanics | Relativity for Samit

Question

Block B (mass Mb) is sitting on a frictionless table attached to a pulley on both ends. A block C (mass Mc) is sitting on top of block B (frictionless contact) attached to the pulley with a massless string on the left hand side of B. On the right hand pulley a block A (mass Ma) is hanging downwards. What is the acceleration of block A with respect to block C?

Answer 1

Explanations & Calculations

• Refer to the image attached.
• By observation, you can identify that each mass undergoes the same acceleration.
• Then apply F = ma to each one along the direction they accelerate.

$\qquad\qquad \begin{aligned} \small \leftarrow T_1 &=\small m_ca\cdots (1)\\ \small \to T_2-T_1 &=\small m_ba\cdots (2)\\ \small \downarrow m_ag-T_2 &=\small m_aa\cdots(3)\\ \\ \small (1)+(2)+(3),\\ \small m_ag &=\small (m_a+m_b+m_c)a\\ \small a&=\small \frac{m_a}{(m_a+m_b+m_c)}.g \end{aligned}$

• Now to calculate the acceleration of A with respect to C, virtually stop C (apply an imaginary acceleration on C just to cancel out its acceleration & to be fair apply the same component on A & get the resultant) & and get the resultant acceleration as seen by C.
• Then as seen by C, A has two components of accelerations on it, one directing East & the other pointing South. Then the net is,

$\qquad\qquad \begin{aligned} \small a_{net}&=\small \sqrt{a^2+a^2}\\ &=\small \sqrt 2a\,\, ms^{-2} \end{aligned}$