A block 'A' of mass 10 kg is placed on wedge 'B' of mass 20 kg. The block is tied with a stretchable string. Friction co-efficient between block and wedge is 0.8 and there is no friction between wedge and the surface (as in figure)
(i) A force F = 2N is applied on block down the plane. Tension in string is equal to –
Text Solution
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(i) : T + f = mg sin θ (m = mass of block)
⇒ T = 60 – 60 = 0
( 
= 64N)
(ii) : T = f + mg sin θ
⇒ f = T – mg sin θ
= 36N
(iii) : Let
a 1 = acc. of wedge w.r.t. ground
a 2 = acc. of block w.r.t. wedge
∴ m a 1 = m (a 2 cos θ – a 1 )
a 2 – 3a 1 = 0 …(i)
T + P s cos θ – mg sin θ – f = ma 2
⇒ a 2 – 
a 1 = 2 …(ii)
From (i) & (ii) a 1 = 
≈ 
m/s 2
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