Find the net force acting on pulley 2 assuming all strings and pulleys to be of negligible mass and all surfaces to be frictionless. g=10m/s2.
A
45N
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B
90N
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C
45√2
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D
90√2
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Solution
The correct option is C45√2 Downward force on 5Kg=5g
Downward force on 6Kg=3g as from FBD of 6Kg mass we have ∴ Acceleration of the system is given by, a=Supporting Force−Opposing ForceTotal mass =5g−6gsin30∘20 =5g−3g20=2g20=1m/s2
From the FBD of 5Kg body, we have 5g−T1=5a⟹T1=5(g−a)=45N
Again, from FBD of pulley (2), we have