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decentfellow
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Homework Statement
In the system shown in Fig. ##2E.5 (a)##. block ##m_2## is being prevented from descending by pulling ##m_1## to the right with force ##F##. Assuming all the surfaces to be frictionless, Find ##F##
Homework Equations
F.B.D of ##m_1##
From the F.B.D of ##m_1## we get the following equations:-
$$F=m_{1}a_{m_1}\tag{1}$$
$$N_1+m_1g-N_2=0 \implies N_1+m_1g=N_2\tag{2}$$
F.B.D of ##m_2##
From the F.B.D of ##m_2##, we get the following equation:-
$$m_2g-T=m_2a\tag{3}$$
F.B.D of ##m_3##
From the F.B.D of ##m_3## , we get the following set of equations:-
$$m_3g-N_1=0 \implies m_3g=N_1\tag{4}$$
$$T=m_3a\tag{5}$$
The Attempt at a Solution
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From eq. ##(3)## and ##(5)##, we get
$$a=\frac{m_2}{m_2+m_3}g$$
Now, for ##m_2## to not descend, the condition required is $$\vec{a}_{m_3/m_2}=0\implies \vec{a}_{m_3}-\vec{a}_{m_2}=0\implies \vec{a}_{m_3}=\vec{a}_{m_2}$$
From ##(1)##, we have
$$\vec{a}_{m_1}=\frac{\vec{F}}{m_1}$$
$$\therefore F=\frac{m_1m_2}{m_2+m_3}g$$
The book gives the answer as $$F=(m_1+m_2+m_3)\frac{m_2}{m_3}g$$.
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