Consider the system shown in the figure. Initially the system was in rest. (i) Find the acceleration of block if man climbs the rod with acceleration a (w.r.t. rod) (ii) If the man climb to the top of the rod then find the distance moved by the block.


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Question Number 23224 by Tinkutara last updated on 27/Oct/17

$Consider the system shown in the$ $figure . Initially the system was in rest .$ $(i) Find the acceleration of block if man$ $climbs the rod with acceleration a (w . r . t .$ $rod)$ $(ii) If the man climb to the top of the$ $rod then find the distance moved by the$ $block .$
Commented by mrW1 last updated on 27/Oct/17

$(1)$ $let a_{R} = acceleration of rod (↑) and block (↓)$ $a_{M} = acceleration of man (↑)$ $a_{M} = a + a_{R}$ $F = force between rod and man$ $F - mg = {ma}_{M} = m (a + a_{R})$ $\Rightarrow F = m (g + a + a_{R})$ $(M + m - M) g - F = (M + m + M) a_{R}$ $(M + m - M) g - m (g + a + a_{R}) = (M + m + M) a_{R}$ $- ma = 2 (M + m) a_{R}$ $\Rightarrow a_{R} = - \frac{ma}{2 (M + m)}$ $(2)$ $L = \frac{1}{2} {at}^{2}$ $Δ H = \frac{1}{2} a_{R} t^{2}$ $\Rightarrow Δ H = \frac{a_{R}}{a} \times L$ $\Rightarrow Δ H = - \frac{m}{2 (M + m)} \times L$ $(the block moves up a distance \frac{mL}{2 (M + m)})$
Commented by Tinkutara last updated on 27/Oct/17

Commented by mrW1 last updated on 27/Oct/17

$that depents on which direction is$ $defined as positive . in your book the$ $+ ve direction for the block is upwards,$ $so there is no ‘ ‘ -^{″} . i defined + ve direction$ $downwards, so i have ‘ ‘ -^{″} . the result$ $is the same, i . e . the block moves a$ $distance \frac{mL}{2 (M + m)} upwards .$
Commented by Tinkutara last updated on 28/Oct/17

$OK Thanks .$
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