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Two particles a & b,each of mass m are kept stationary by applying a horizontal force F=mg on particle B as shown in fig. then a) tanß = 2tana b) 2T1=5T2 c) rt2T1=rt5T2 d) noe of the above
hi, at point B, T2*cosβ = mg .....(in vertical dirxn).............[1] T2*sinβ = mg........(in horizontal dirxn)........[2] dividing these 2 eqns, we get tanβ = 1 ,.i.e β = ∏/4...... now at pont A, T1*cosα = T2*cosβ + mg........(in vertical dirxn).................[3] => T1*cosα = 2T2*cosβ........(using [1]) in horizontal dirxn T1*sinα = T2*sinβ ........................[4] dividing 4 by 3 ,we get tanα = 0.5*tanβ => 2tanα = tanβ so correct option is (a)........
hi,
at point B,
T2*cosβ = mg .....(in vertical dirxn).............[1]
T2*sinβ = mg........(in horizontal dirxn)........[2]
dividing these 2 eqns, we get
tanβ = 1 ,.i.e β = ∏/4......
now at pont A,
T1*cosα = T2*cosβ + mg........(in vertical dirxn).................[3]
=> T1*cosα = 2T2*cosβ........(using [1])
in horizontal dirxn
T1*sinα = T2*sinβ ........................[4]
dividing 4 by 3 ,we get
tanα = 0.5*tanβ
=> 2tanα = tanβ
so correct option is (a)........
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