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Question

Figure shows as arrangement of masses hanging from a ceiling. In equilibrium, each rod is horizontal, has negligible mass and extends three times as far to the right of the wire supporting it as to the left. If mass $$m_{4}$$ is $$48\ kg$$ then, ind the magnitude of mass $$m_{1}$$.
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Solution

Given $$m_4=48kg$$
In equilibrium, mass is negligible$$\cfrac{48}{16}=3kg$$
($$\because$$ Each mass has $$4$$ kg
$$\therefore m_1=m_2=m+3=m+4=4\times4=16kg$$)

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Physics

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