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As the rods are in equilibrium

$ΣF_{X}=0$

$ΣF_{y}=0$

$τ_{net}=0$

For right rod

For right rod

$N_{2}+2mg +Mg=N$ (1)

$N_{1}=f$ (2)

For left rod

For left rod

$N_{2}+N=Mg+2mg $ (3)

From $(1)and(3)N_{1}=f$

$=N_{2}=0=N=Mg+2mg $

Balancing torque about $O$,

$Mg2L cosθ+fLsinθ=NLcosθ$

$2Mgcosθ +fsinθ=(Mg+2mg )cosθ$

$f=(2Mg +2mg )cotθ=f=(2M+m )gcotθ$

Solve any question of Laws of Motion with:-

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